Question
Points A and B have coordinates (1, 1, 2) and (9, m, -6) respectively.
Express AB in terms of m . [2]
The line L , which passes through B, has equation r
Find the value of m . [5]
Consider a unit vector u , such that \(\mathbf{u}=pi-\frac{2}{3}j+\frac{1}{3}k\) , where p > 0 .
Point C is such that BC = 9u .
Find the coordinates of C. [8]
▶️Answer/Explanation
Ans:
- (a)
- valid approach to find \(\overrightarrow{AB}\)
- eg \(\overrightarrow{OB}-\overrightarrow{OA}\) , A – B
- (b)
- valid approach
- one correct equation eg − 3 + 2s = 9, -6 = 24 -5s
- correct value for s eg s = 6
- substituting their s value into their expression/equation to find m
- eg -19 + 6 × 4
- m = 5
- valid approach
- (c)
- valid approach
- correct working to find C
- correct approach to find u (seen anywhere)
- eg \(p^{2}+(-\frac{2}{3})^{2}+(\frac{1}{3})^{2}, \sqrt{p^{2}+\frac{4}{9}+\frac{1}{9}}\)
- recognizing unit vector has magnitude of 1
- eg |u| = 1, \(\sqrt{p^{2}+(-\frac{2}{3})^{2}+(\frac{1}{3}^{2}))}= 1, p^{2} + \frac{5}{9}\)= 1
- correct working
- eg \(p^{2} = \frac{4}{9}, p= \pm \frac{2}{3}\)
- \(p = \frac{2}{3}\)
- substituting their value of p
- eg
Question
Consider the line L1defined by the Cartesian equation \(\frac{x+1}{2}\) = y = 3 – z .
(a) (i) Show that the point (-1, 0, 3) lies on L1 .
(ii) Find a vector equation of L1 . [4]
Consider a second line L2 defined by the vector equation r =
where t ∈ \(\mathbb{R}\) and a ∈ \(\mathbb{R}\).
(b) Find the possible values of a when the acute angle between L1 and L2 is 45°. [8]
It is given that the lines L1 and L2 have a unique point of intersection, A, when a ¹ k .
(c) Find the value of k , and find the coordinates of the point A in terms of a . [7]
▶️Answer/Explanation
Ans:
Question
The vertices of a triangle ABC have coordinates given by A(−1, 2, 3), B(4, 1, 1) and C(3, −2, 2).
a.(i) Find the lengths of the sides of the triangle.
(ii) Find \(\cos {\rm{B\hat AC}}\). [6]
b. (i) Show that \(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \) −7i − 3j − 16k.
(ii) Hence, show that the area of the triangle ABC is \(\frac{1}{2}\sqrt {314} \). [5]
f.The point D on (AB) is such that \(\overrightarrow {{\text{OD}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) where O is the origin.
e.(i) Find the coordinates of D.
(ii) Show that D does not lie between A and B. [5]
▶️Answer/Explanation
Markscheme
a. (i) \(\overrightarrow {{\text{AB}}} = \overrightarrow {{\text{OB}}} – \overrightarrow {{\text{OA}}} = \) 5i – j – 2k (or in column vector form) (A1)
Note: Award A1 if any one of the vectors, or its negative, representing the sides of the triangle is seen.
\(\overrightarrow {{\text{AB}}} = \) |5i – j – 2k|= \(\sqrt {30} \)
\(\overrightarrow {{\text{BC}}} = \) |–i – 3j + k|= \(\sqrt {11} \)
\(\overrightarrow {{\text{CA}}} = \) |–4i + 4j + k|= \(\sqrt {33} \) A2
Note: Award A1 for two correct and A0 for one correct.
(ii) METHOD 1
\(\cos {\text{BAC}} = \frac{{20 + 4 + 2}}{{\sqrt {30} \sqrt {33} }}\) M1A1
Note: Award M1 for an attempt at the use of the scalar product for two vectors representing the sides AB and AC, or their negatives, A1 for the correct computation using their vectors.
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
Note: Candidates who use the modulus need to justify it – the angle is not stated in the question to be acute.
METHOD 2
using the cosine rule
\(\cos {\text{BAC}} = \frac{{30 + 33 – 11}}{{2\sqrt {30} \sqrt {33} }}\) M1A1
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
[6 marks]
\(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
{ – 1}&{ – 3}&1 \\
{ – 4}&4&1
\end{array}} \right|\) A1
\( = \left( {( – 3) \times 1 – 1 \times 4} \right)\)i + \(\left( {1 \times ( – 4) – ( – 1) \times 1} \right)\)j + \(\left( {( – 1) \times 4 – ( – 3) \times ( – 4)} \right)\)k M1A1
= –7i – 3j – 16k AG
(ii) the area of \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} } \right|\) (M1)
\(\frac{1}{2}\sqrt {{{( – 7)}^2} + {{( – 3)}^2} + {{( – 16)}^2}} \) A1
\( = \frac{1}{2}\sqrt {314} \) AG
[5 marks]
attempt at the use of “(r – a)\( \cdot \)n = 0” (M1)
using r = xi + yj + zk, a = \(\overrightarrow {{\text{OA}}} \) and n = –7i – 3j – 16k (A1)
\(7x + 3y + 16z = 47\) A1
Note: Candidates who adopt a 2-parameter approach should be awarded, A1 for correct 2-parameter equations for x, y and z; M1 for a serious attempt at elimination of the parameters; A1 for the final Cartesian equation.
[3 marks]
r = \(\overrightarrow {{\text{OA}}} + t\overrightarrow {{\text{AB}}} \) (or equivalent) M1
r = (–i + 2j + 3k) + t (5i – j – 2k) A1
Note: Award M1A0 if “r =” is missing.
Note: Accept forms of the equation starting with B or with the direction reversed.
[2 marks]
(i) \(\overrightarrow {{\text{OD}}} = \) (–i + 2j + 3k) + t(5i – j – 2k)
statement that \(\overrightarrow {{\text{OD}}} \cdot \overrightarrow {{\text{BC}}} = 0\) (M1)
\(\left( {\begin{array}{*{20}{c}}
{ – 1 + 5t} \\
{2 – t} \\
{3 – 2t}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 3} \\
1
\end{array}} \right) = 0\) A1
\( – 2 – 4t = 0{\text{ or }}t = – \frac{1}{2}\) A1
coordinates of D are \(\left( { – \frac{7}{2},\frac{5}{2},4} \right)\) A1
Note: Different forms of \(\overrightarrow {{\text{OD}}} \) give different values of t, but the same final answer.
(ii) \(t < 0 \Rightarrow \) D is not between A and B R1
[5 marks]
Question
The vertices of a triangle ABC have coordinates given by A(−1, 2, 3), B(4, 1, 1) and C(3, −2, 2).
a.(i) Find the lengths of the sides of the triangle.
(ii) Find \(\cos {\rm{B\hat AC}}\).[6]
(ii) Hence, show that the area of the triangle ABC is \(\frac{1}{2}\sqrt {314} \).[5]
e.(i) Find the coordinates of D.
(ii) Show that D does not lie between A and B. [5]
▶️Answer/Explanation
Markscheme
a.(i) \(\overrightarrow {{\text{AB}}} = \overrightarrow {{\text{OB}}} – \overrightarrow {{\text{OA}}} = \) 5i – j – 2k (or in column vector form) (A1)
Note: Award A1 if any one of the vectors, or its negative, representing the sides of the triangle is seen.
\(\overrightarrow {{\text{AB}}} = \) |5i – j – 2k|= \(\sqrt {30} \)
\(\overrightarrow {{\text{BC}}} = \) |–i – 3j + k|= \(\sqrt {11} \)
\(\overrightarrow {{\text{CA}}} = \) |–4i + 4j + k|= \(\sqrt {33} \) A2
Note: Award A1 for two correct and A0 for one correct.
(ii) METHOD 1
\(\cos {\text{BAC}} = \frac{{20 + 4 + 2}}{{\sqrt {30} \sqrt {33} }}\) M1A1
Note: Award M1 for an attempt at the use of the scalar product for two vectors representing the sides AB and AC, or their negatives, A1 for the correct computation using their vectors.
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
Note: Candidates who use the modulus need to justify it – the angle is not stated in the question to be acute.
METHOD 2
using the cosine rule
\(\cos {\text{BAC}} = \frac{{30 + 33 – 11}}{{2\sqrt {30} \sqrt {33} }}\) M1A1
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
[6 marks]
\(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
{ – 1}&{ – 3}&1 \\
{ – 4}&4&1
\end{array}} \right|\) A1
\( = \left( {( – 3) \times 1 – 1 \times 4} \right)\)i + \(\left( {1 \times ( – 4) – ( – 1) \times 1} \right)\)j + \(\left( {( – 1) \times 4 – ( – 3) \times ( – 4)} \right)\)k M1A1
= –7i – 3j – 16k AG
(ii) the area of \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} } \right|\) (M1)
\(\frac{1}{2}\sqrt {{{( – 7)}^2} + {{( – 3)}^2} + {{( – 16)}^2}} \) A1
\( = \frac{1}{2}\sqrt {314} \) AG
[5 marks]
attempt at the use of “(r – a)\( \cdot \)n = 0” (M1)
using r = xi + yj + zk, a = \(\overrightarrow {{\text{OA}}} \) and n = –7i – 3j – 16k (A1)
\(7x + 3y + 16z = 47\) A1
Note: Candidates who adopt a 2-parameter approach should be awarded, A1 for correct 2-parameter equations for x, y and z; M1 for a serious attempt at elimination of the parameters; A1 for the final Cartesian equation.
[3 marks]
r = \(\overrightarrow {{\text{OA}}} + t\overrightarrow {{\text{AB}}} \) (or equivalent) M1
r = (–i + 2j + 3k) + t (5i – j – 2k) A1
Note: Award M1A0 if “r =” is missing.
Note: Accept forms of the equation starting with B or with the direction reversed.
[2 marks]
(i) \(\overrightarrow {{\text{OD}}} = \) (–i + 2j + 3k) + t(5i – j – 2k)
statement that \(\overrightarrow {{\text{OD}}} \cdot \overrightarrow {{\text{BC}}} = 0\) (M1)
\(\left( {\begin{array}{*{20}{c}}
{ – 1 + 5t} \\
{2 – t} \\
{3 – 2t}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 3} \\
1
\end{array}} \right) = 0\) A1
\( – 2 – 4t = 0{\text{ or }}t = – \frac{1}{2}\) A1
coordinates of D are \(\left( { – \frac{7}{2},\frac{5}{2},4} \right)\) A1
Note: Different forms of \(\overrightarrow {{\text{OD}}} \) give different values of t, but the same final answer.
(ii) \(t < 0 \Rightarrow \) D is not between A and B R1
[5 marks]
Question
Consider the points \({\text{A(1, 0, 0)}}\), \({\text{B(2, 2, 2)}}\) and \({\text{C(0, 2, 1)}}\).
A third plane \({\Pi _3}\) is defined by the Cartesian equation \(16x + \alpha y – 3z = \beta \).
a. Find the vector \(\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} \). [4]
b.Find an exact value for the area of the triangle ABC.[3]
c.Show that the Cartesian equation of the plane \({\Pi _1}\), containing the triangle ABC, is \(2x + 3y – 4z = 2\).[3]
d.A second plane \({\Pi _2}\) is defined by the Cartesian equation \({\Pi _2}:4x – y – z = 4\). \({L_1}\) is the line of intersection of the planes \({\Pi _1}\) and \({\Pi _2}\).
Find a vector equation for \({L_1}\).[5]
e.Find the value of \(\alpha \) if all three planes contain \({L_1}\).[3]
f.Find conditions on \(\alpha \) and \(\beta \) if the plane \({\Pi _3}\) does not intersect with \({L_1}\).[2]
▶️Answer/Explanation
Markscheme
a.\(\overrightarrow {{\rm{CA}}} = \left( \begin{array}{c}1\\ – 2\\ – 1\end{array} \right)\) (A1)
\(\overrightarrow {{\rm{CB}}} = \left( \begin{array}{c}2\\0\\1\end{array} \right)\) (A1)
Note: If \(\overrightarrow {{\text{AC}}} \) and \(\overrightarrow {{\text{BC}}} \) found correctly award (A1) (A0).
\(\overrightarrow {{\rm{CA}}} \times \overrightarrow {{\rm{CB}}} = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ – 2}&{ – 1}\\2&0&1\end{array}} \right|\) (M1)
\(\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)\) A1
[4 marks]
METHOD 1
\(\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} } \right|\) (M1)
\( = \frac{1}{2}\sqrt {{{( – 2)}^2} + {{( – 3)}^2} + {4^2}} \) (A1)
\( = \frac{{\sqrt {29} }}{2}\) A1
METHOD 2
attempt to apply \(\frac{1}{2}\left| {{\text{CA}}} \right|\left| {{\text{CB}}} \right|\sin C\) (M1)
\({\text{CA.CB}} = \sqrt 5 .\sqrt 6 \cos C \Rightarrow \cos C = \frac{1}{{\sqrt {30} }} \Rightarrow \sin C = \frac{{\sqrt {29} }}{{\sqrt {30} }}\) (A1)
\({\text{area}} = \frac{{\sqrt {29} }}{2}\) A1
[3 marks]
METHOD 1
r.\(\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right) = \left( \begin{array}{l}1\\0\\0\end{array} \right) \bullet \left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)\) M1A1
\( \Rightarrow – 2x – 3y + 4z = – 2\) A1
\( \Rightarrow 2x + 3y – 4z = 2\) AG
METHOD 2
\( – 2x – 3y + 4z = d\)
substituting a point in the plane M1A1
\({\text{d}} = – 2\) A1
\( \Rightarrow – 2x – 3y + 4z = – 2\)
\( \Rightarrow 2x + 3y – 4z = 2\) AG
Note: Accept verification that all 3 vertices of the triangle lie on the given plane.
[3 marks]
METHOD 1
\(\left| {\begin{array}{*{20}{c}}{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}}\\2&3&{ – 4}\\4&{ – 1}&{ – 1}\end{array}} \right| = \left( \begin{array}{c} – 7\\ – 14\\ – 14\end{array} \right)\) M1A1
\({\mathbf{n}} = \left( \begin{array}{l}1\\2\\2\end{array} \right)\)
\(z = 0 \Rightarrow y = 0,{\text{ }}x = 1\) (M1)(A1)
\({L_1}:{\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\) A1
Note: Do not award the final A1 if \(\mathbf{r} =\) is not seen.
METHOD 2
eliminate 1 of the variables, eg x M1
\( – 7y + 7z = 0\) (A1)
introduce a parameter M1
\( \Rightarrow z = \lambda \),
\(y = \lambda {\text{, }}x = 1 + \frac{\lambda }{2}\) (A1)
\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent A1
Note: Do not award the final A1 if \(\mathbf{r} =\) is not seen.
METHOD 3
\(z = t\) M1
write x and y in terms of \(t \Rightarrow 4x – y = 4 + t,{\text{ }}2x + 3y = 2 + 4t\) or equivalent A1
attempt to eliminate x or y M1
\(x,{\text{ }}y,{\text{ }}z\) expressed in parameters
\( \Rightarrow z = t\),
\(y = t,{\text{ }}x = 1 + \frac{t}{2}\) A1
\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + t \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent A1
Note: Do not award the final A1 if \(\mathbf{r} =\) is not seen.
[5 marks]
METHOD 1
direction of the line is perpendicular to the normal of the plane
\(\left( \begin{array}{c}16\\\alpha \\ – 3\end{array} \right) \bullet \left( \begin{array}{l}1\\2\\2\end{array} \right) = 0\) M1A1
\(16 + 2\alpha – 6 = 0 \Rightarrow \alpha = – 5\) A1
METHOD 2
solving line/plane simultaneously
\(16(1 + \lambda ) + 2\alpha \lambda – 6\lambda = \beta \) M1A1
\(16 + (10 + 2\alpha )\lambda = \beta \)
\( \Rightarrow \alpha = – 5\) A1
METHOD 3
\(\left| {\begin{array}{*{20}{c}}2&3&{ – 4}\\4&{ – 1}&{ – 1}\\{16}&\alpha &{ – 3}\end{array}} \right| = 0\) M1
\(2(3 + \alpha ) – 3( – 12 + 16) – 4(4\alpha + 16) = 0\) A1
\( \Rightarrow \alpha = – 5\) A1
METHOD 4
attempt to use row reduction on augmented matrix M1
to obtain \(\left( {\begin{array}{*{20}{c}}2&3&{ – 4}\\0&{ – 1}&1\\0&0&{\alpha + 5}\end{array}\left| \begin{array}{c}2\\0\\\beta – 16\end{array} \right.} \right)\) A1
\( \Rightarrow \alpha = – 5\) A1
[3 marks]
\(\alpha = – 5\) A1
\(\beta \ne 16\) A1
[2 marks]
Question
a.Show that the points \({\text{O}}(0,{\text{ }}0,{\text{ }}0)\), \({\text{ A}}(6,{\text{ }}0,{\text{ }}0)\), \({\text{B}}({6,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12} })\), \({\text{C}}({0,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12}})\) form a square.[3]
b.Find the coordinates of M, the mid-point of [OB].[1]
c.Show that an equation of the plane \({\mathit{\Pi }}\), containing the square OABC, is \(y + \sqrt 2 z = 0\).[3]
d.Find a vector equation of the line \(L\), through M, perpendicular to the plane \({\mathit{\Pi }}\).[3]
e.Find the coordinates of D, the point of intersection of the line \(L\) with the plane whose equation is \(y = 0\).[3]
f.Find the coordinates of E, the reflection of the point D in the plane \({\mathit{\Pi }}\).[3]
(i) Find the angle \({\rm{O\hat DA}}\).
(ii) State what this tells you about the solid OABCDE.[6]
▶️Answer/Explanation
Markscheme
a.\(\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6\) (therefore a rhombus) A1A1
Note: Award A1 for two correct lengths, A2 for all four.
Note: Award A1A0 for \(\overrightarrow {{\rm{OA}}} = \overrightarrow {{\rm{CB}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{ or \,\,} } \overrightarrow {{\rm{OC}}} = \overrightarrow {A{\rm{B}}} = \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right)\) if no magnitudes are shown.
\(\overrightarrow {{\rm{OA}}}\,\, {\rm{ g}}\overrightarrow {{\rm{OC}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{g}}\left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = 0 \) (therefore a square) A1
Note: Other arguments are possible with a minimum of three conditions.
[3 marks]
\({\text{M}}\left( {3,{\text{ }} – \frac{{\sqrt {24} }}{2},{\text{ }}\frac{{\sqrt {12} }}{2}} \right)\left( { = \left( {3,{\text{ }} – \sqrt 6 ,{\text{ }}\sqrt 3 } \right)} \right)\) A1
[1 mark]
METHOD 1
\(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OC}}} = \)\(\left( \begin{array}{l}6\\0\\0\end{array} \right) \times \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = \left( \begin{array}{c}0\\ – 6\sqrt {12} \\ – 6\sqrt {24} \end{array} \right)\left( { = \left( \begin{array}{c}0\\ – 12\sqrt 3 \\ – 12\sqrt 6 \end{array} \right)} \right)\) M1A1
Note: Candidates may use other pairs of vectors.
equation of plane is \( – 6\sqrt {12} y – 6\sqrt {24} z = d\)
any valid method showing that \(d = 0\) M1
\(\mathit{\Pi} :y+\sqrt{2z}=0\) AG
METHOD 2
equation of plane is \(ax + by + cz = d\)
substituting O to find \(d = 0\) (M1)
substituting two points (A, B, C or M) M1
eg
\(6a = 0,{\text{ }} – \sqrt {24} b + \sqrt {12} c = 0\) A1
\(\mathit{\Pi} :y+\sqrt{2z}=0\) AG
[3 marks]
\(\boldsymbol{r} = \left( \begin{array}{c}3\\ – \sqrt 6 \\\sqrt 3 \end{array} \right) + \lambda \left( \begin{array}{l}0\\1\\\sqrt 2 \end{array} \right)\) A1A1A1
Note: Award A1 for r = , A1A1 for two correct vectors.
[3 marks]
Using \(y = 0\) to find \(\lambda \) M1
Substitute their \(\lambda \) into their equation from part (d) M1
D has coordinates \(\left( {{\text{3, 0, 3}}\sqrt 3 } \right)\) A1
[3 marks]
\(\lambda \) for point E is the negative of the \(\lambda \) for point D (M1)
Note: Other possible methods may be seen.
E has coordinates \(\left( {{\text{3, }} – 2\sqrt 6 ,{\text{ }} – \sqrt 3 } \right)\) A1A1
Note: Award A1 for each of the y and z coordinates.
[3 marks]
(i) \(\overrightarrow {{\text{DA}}} {\text{ g}}\overrightarrow {{\text{DO}}} = \)\(\left( \begin{array}{c}3\\0\\ – 3\sqrt 3 \end{array} \right){\rm{g}}\left( \begin{array}{c} – 3\\0\\ – 3\sqrt 3 \end{array} \right) = 18\) M1A1
\(\cos {\rm{O\hat DA}} = \frac{{18}}{{\sqrt {36} \sqrt {36} }} = \frac{1}{2}\) M1
hence \({\rm{O\hat DA}} = 60^\circ \) A1
Note: Accept method showing OAD is equilateral.
(ii) OABCDE is a regular octahedron (accept equivalent description) A2
Note: A2 for saying it is made up of 8 equilateral triangles
Award A1 for two pyramids, A1 for equilateral triangles.
(can be either stated or shown in a sketch – but there must be clear indication the triangles are equilateral)
[6 marks]
Question
a. Consider the plane \({\mathit{\Pi} _1}\), parallel to both lines \({L_1}\) and \({L_2}\). Point C lies in the plane \({\mathit{\Pi} _1}\).
The line \({L_3}\) has vector equation \(\boldsymbol{r} = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}k\\1\\ – 1\end{array} \right)\).
The plane \({\mathit{\Pi} _2}\) has Cartesian equation \(x + y = 12\).
The angle between the line \({L_3}\) and the plane \({\mathit{\Pi} _2}\) is 60°.
Given the points A(1, 0, 4), B(2, 3, −1) and C(0, 1, − 2) , find the vector equation of the line \({L_1}\) passing through the points A and B.[2]
The line \({L_2}\) has Cartesian equation \(\frac{{x – 1}}{3} = \frac{{y + 2}}{1} = \frac{{z – 1}}{{ – 2}}\).
Show that \({L_1}\) and \({L_2}\) are skew lines.[5]
Find the Cartesian equation of the plane \({\Pi _1}\).
(i) Find the value of \(k\).
(ii) Find the point of intersection P of the line \({L_3}\) and the plane \({\mathit{\Pi} _2}\).[7]
▶️Answer/Explanation
Markscheme
a.direction vector \(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{c}1\\3\\ – 5\end{array} \right)\) or \(\overrightarrow {{\rm{BA}}} = \left( \begin{array}{c} – 1\\ – 3\\5\end{array} \right)\) A1
\(\boldsymbol{r} = \left( \begin{array}{l}1\\0\\4\end{array} \right) + t\left( \begin{array}{c}1\\3\\ – 5\end{array} \right)\) or \(\boldsymbol{r} = \left( \begin{array}{c}2\\3\\ – 1\end{array} \right) + t\left( \begin{array}{c}1\\3\\ – 5\end{array} \right)\) or equivalent A1
Note: Do not award final A1 unless ‘\(\boldsymbol{r} = {\text{K}}\)’ (or equivalent) seen.
Allow FT on direction vector for final A1.
[2 marks]
both lines expressed in parametric form:
\({L_1}\):
\(x = 1 + t\)
\(y = 3t\)
\(z = 4 – 5t\)
\({L_2}\):
\(x = 1 + 3s\)
\(y = – 2 + s\) M1A1
\(z = – 2s + 1\)
Notes: Award M1 for an attempt to convert \({L_2}\) from Cartesian to parametric form.
Award A1 for correct parametric equations for \({L_1}\) and \({L_2}\).
Allow M1A1 at this stage if same parameter is used in both lines.
attempt to solve simultaneously for x and y: M1
\(1 + t = 1 + 3s\)
\(3t = – 2 + s\)
\(t = – \frac{3}{4},{\text{ }}s = – \frac{1}{4}\) A1
substituting both values back into z values respectively gives \(z = \frac{{31}}{4}\)
and \(z = \frac{3}{2}\) so a contradiction R1
therefore \({L_1}\) and \({L_1}\) are skew lines AG
[5 marks]
finding the cross product:
\(\left( \begin{array}{c}1\\3\\ – 5\end{array} \right) \times \left( \begin{array}{c}3\\1\\ – 2\end{array} \right)\) (M1)
= – i – 13j – 8k A1
Note: Accept i + 13j + 8k
\( – 1(0) – 13(1) – 8( – 2) = 3\) (M1)
\( \Rightarrow – x – 13y – 8z = 3\) or equivalent A1
[4 marks]
(i) \((\cos \theta = )\frac{{\left( \begin{array}{c}k\\1\\ – 1\end{array} \right) \bullet \left( \begin{array}{l}1\\1\\0\end{array} \right)}}{{\sqrt {{k^2} + 1 + 1} \times \sqrt {1 + 1} }}\) M1
Note: Award M1 for an attempt to use angle between two vectors formula.
\(\frac{{\sqrt 3 }}{2} = \frac{{k + 1}}{{\sqrt {2({k^2} + 2)} }}\) A1
obtaining the quadratic equation
\(4{(k + 1)^2} = 6({k^2} + 2)\) M1
\({k^2} – 4k + 4 = 0\)
\({(k – 2)^2} = 0\)
\(k = 2\) A1
Note: Award M1A0M1A0 if \(\cos 60^\circ \) is used \((k = 0{\text{ or }}k = – 4)\).
(ii) \(r = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}2\\1\\ – 1\end{array} \right)\)
substituting into the equation of the plane \({\Pi _2}\):
\(3 + 2\lambda + \lambda = 12\) M1
\(\lambda = 3\) A1
point P has the coordinates:
(9, 3, –2) A1
Notes: Accept 9i + 3j – 2k and \(\left( \begin{array}{l}9\\3\\- 2\end{array} \right)\).
Do not allow FT if two values found for k.
[7 marks]
Question
Consider the triangle \(ABC\). The points \(P\), \(Q\) and \(R\) are the midpoints of the line segments [\(AB\)], [\(BC\)] and [\(AC\)] respectively.
Let \(\overrightarrow {{\text{OA}}} = {{a}}\), \(\overrightarrow {{\text{OB}}} = {{b}}\) and \(\overrightarrow {{\text{OC}}} = {{c}}\).
a. Find \(\overrightarrow {{\text{BR}}} \) in terms of \({{a}}\), \({{b}}\) and \({{c}}\).[2]
(ii) Find a vector equation of the line that passes through \(A\) and \(Q\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\mu \).
(iii) Hence show that \(\overrightarrow {{\text{OG}}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) given that \(G\) is the point where [\(BR\)] and [\(AQ\)] intersect.[9]
A point \(X\) is such that [\(GX\)] is perpendicular to the plane \(ABC\).
Given that the tetrahedron \(ABCX\) has volume \({\text{12 unit}}{{\text{s}}^{\text{3}}}\), find possible coordinates of \(X\).[9]
▶️Answer/Explanation
Markscheme
a.\(\overrightarrow {{\text{BR}}} = \overrightarrow {{\text{BA}}} + \overrightarrow {{\text{AR}}} \;\;\;\left( { = \overrightarrow {{\text{BA}}} + \frac{1}{2}\overrightarrow {{\text{AC}}} } \right)\) (M1)
\( = ({{a}} – {{b}}) + \frac{1}{2}({{c}} – {{a}})\)
\( = \frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}\) A1
[2 marks]
(i) \({{\text{r}}_{{\text{BR}}}} = {{b}} + \lambda \left( {\frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = \frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}}} \right)\) A1A1
Note: Award A1A0 if the \({\text{r}} = \) is omitted in an otherwise correct expression/equation.
Do not penalise such an omission more than once.
(ii) \(\overrightarrow {{\text{AQ}}} = – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}\) (A1)
\({{\text{r}}_{{\text{AQ}}}} = {{a}} + \mu \left( { – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}} \right)\) A1
Note: Accept the use of the same parameter in (i) and (ii).
(iii) when \(\overrightarrow {{\text{AQ}}} \) and \(\overrightarrow {{\text{BP}}} \) intersect we will have \({{\text{r}}_{{\text{BR}}}} = {{\text{r}}_{{\text{AQ}}}}\) (M1)
Note: If the same parameters are used for both equations, award at most M1M1A0A0M1.
\(\frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}} = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}\)
attempt to equate the coefficients of the vectors \({{a}}\), \({{b}}\) and \({{c}}\) M1
\(\left. {\begin{array}{*{20}{c}} {\frac{\lambda }{2} = 1 – \mu } \\ {1 – \lambda = \frac{\mu }{2}} \\ {\frac{\lambda }{2} = \frac{\mu }{2}} \end{array}} \right\}\) (A1)
\(\lambda = \frac{2}{3}\) or \(\mu = \frac{2}{3}\) A1
substituting parameters back into one of the equations M1
\(\overrightarrow {{\text{OG}}} = \frac{1}{2} \bullet \frac{2}{3}{{a}} + \left( {1 – \frac{2}{3}} \right){{b}} + \frac{1}{2} \bullet \frac{2}{3}{{c}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) AG
Note: Accept solution by verification.
[9 marks]
\(\overrightarrow {{\text{CP}}} = \frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}\) (M1)A1
so we have that \({{\text{r}}_{{\text{CP}}}} = {{c}} + \beta \left( {\frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}} \right)\) and when \(\beta = \frac{2}{3}\) the line passes through
the point \(G\) (ie, with position vector \(\frac{1}{3}({{a}} + {{b}} + {{c}})\)) R1
hence [\(AQ\)], [\(BR\)] and [\(CP\)] all intersect in \(G\) AG
[3 marks]
\(\overrightarrow {{\text{OG}}} = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 7 \\ { – 5} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right)\) A1
Note: This independent mark for the vector may be awarded wherever the vector is calculated.
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 6} \\ { – 6} \\ { – 6} \end{array}} \right)\) M1A1
\(\overrightarrow {{\text{GX}}} = \alpha \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right)\) (M1)
volume of Tetrahedron given by \(\frac{1}{3} \times {\text{Area ABC}} \times {\text{GX}}\)
\( = \frac{1}{3}\left( {\frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|} \right) \times {\text{GX}} = 12\) (M1)(A1)
Note: Accept alternative methods, for example the use of a scalar triple product.
\( = \frac{1}{6}\sqrt {{{( – 6)}^2} + {{( – 6)}^2} + {{( – 6)}^2}} \times \sqrt {{\alpha ^2} + {\alpha ^2} + {\alpha ^2}} = 12\) (A1)
\( = \frac{1}{6}6\sqrt 3 |\alpha |\sqrt 3 = 12\)
\( \Rightarrow |\alpha | = 4\) A1
Note: Condone absence of absolute value.
this gives us the position of \(X\) as \(\left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right) \pm \left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ 4 \end{array}} \right)\)
\({\text{X}}(6,{\text{ }}8,{\text{ }}3)\) or \(( – 2,{\text{ }}0,{\text{ }} – 5)\) A1
Note: Award A1 for either result. [9 marks]Total [23 marks]
Question
Consider the triangle \(ABC\). The points \(P\), \(Q\) and \(R\) are the midpoints of the line segments [\(AB\)], [\(BC\)] and [\(AC\)] respectively.
Let \(\overrightarrow {{\text{OA}}} = {{a}}\), \(\overrightarrow {{\text{OB}}} = {{b}}\) and \(\overrightarrow {{\text{OC}}} = {{c}}\).
a.Find \(\overrightarrow {{\text{BR}}} \) in terms of \({{a}}\), \({{b}}\) and \({{c}}\).[2]
(i) Find a vector equation of the line that passes through \(B\) and \(R\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\lambda \).
(ii) Find a vector equation of the line that passes through \(A\) and \(Q\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\mu \).
(iii) Hence show that \(\overrightarrow {{\text{OG}}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) given that \(G\) is the point where [\(BR\)] and [\(AQ\)] intersect.[9]
A point \(X\) is such that [\(GX\)] is perpendicular to the plane \(ABC\).
Given that the tetrahedron \(ABCX\) has volume \({\text{12 unit}}{{\text{s}}^{\text{3}}}\), find possible coordinates
of \(X\).[9]
▶️Answer/Explanation
Markscheme
a.\(\overrightarrow {{\text{BR}}} = \overrightarrow {{\text{BA}}} + \overrightarrow {{\text{AR}}} \;\;\;\left( { = \overrightarrow {{\text{BA}}} + \frac{1}{2}\overrightarrow {{\text{AC}}} } \right)\) (M1)
\( = ({{a}} – {{b}}) + \frac{1}{2}({{c}} – {{a}})\)
\( = \frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}\) A1 [2 marks]
(i) \({{\text{r}}_{{\text{BR}}}} = {{b}} + \lambda \left( {\frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = \frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}}} \right)\) A1A1
Note: Award A1A0 if the \({\text{r}} = \) is omitted in an otherwise correct expression/equation.
Do not penalise such an omission more than once.
(ii) \(\overrightarrow {{\text{AQ}}} = – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}\) (A1)
\({{\text{r}}_{{\text{AQ}}}} = {{a}} + \mu \left( { – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}} \right)\) A1
Note: Accept the use of the same parameter in (i) and (ii).
(iii) when \(\overrightarrow {{\text{AQ}}} \) and \(\overrightarrow {{\text{BP}}} \) intersect we will have \({{\text{r}}_{{\text{BR}}}} = {{\text{r}}_{{\text{AQ}}}}\) (M1)
Note: If the same parameters are used for both equations, award at most M1M1A0A0M1.
\(\frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}} = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}\)
attempt to equate the coefficients of the vectors \({{a}}\), \({{b}}\) and \({{c}}\) M1
\(\left. {\begin{array}{*{20}{c}} {\frac{\lambda }{2} = 1 – \mu } \\ {1 – \lambda = \frac{\mu }{2}} \\ {\frac{\lambda }{2} = \frac{\mu }{2}} \end{array}} \right\}\) (A1)
\(\lambda = \frac{2}{3}\) or \(\mu = \frac{2}{3}\) A1
substituting parameters back into one of the equations M1
\(\overrightarrow {{\text{OG}}} = \frac{1}{2} \bullet \frac{2}{3}{{a}} + \left( {1 – \frac{2}{3}} \right){{b}} + \frac{1}{2} \bullet \frac{2}{3}{{c}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) AG
Note: Accept solution by verification.
[9 marks]
\(\overrightarrow {{\text{CP}}} = \frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}\) (M1)A1
so we have that \({{\text{r}}_{{\text{CP}}}} = {{c}} + \beta \left( {\frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}} \right)\) and when \(\beta = \frac{2}{3}\) the line passes through
the point \(G\) (ie, with position vector \(\frac{1}{3}({{a}} + {{b}} + {{c}})\)) R1
hence [\(AQ\)], [\(BR\)] and [\(CP\)] all intersect in \(G\) AG
[3 marks]
\(\overrightarrow {{\text{OG}}} = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 7 \\ { – 5} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right)\) A1
Note: This independent mark for the vector may be awarded wherever the vector is calculated.
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 6} \\ { – 6} \\ { – 6} \end{array}} \right)\) M1A1
\(\overrightarrow {{\text{GX}}} = \alpha \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right)\) (M1)
volume of Tetrahedron given by \(\frac{1}{3} \times {\text{Area ABC}} \times {\text{GX}}\)
\( = \frac{1}{3}\left( {\frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|} \right) \times {\text{GX}} = 12\) (M1)(A1)
Note: Accept alternative methods, for example the use of a scalar triple product.
\( = \frac{1}{6}\sqrt {{{( – 6)}^2} + {{( – 6)}^2} + {{( – 6)}^2}} \times \sqrt {{\alpha ^2} + {\alpha ^2} + {\alpha ^2}} = 12\) (A1)
\( = \frac{1}{6}6\sqrt 3 |\alpha |\sqrt 3 = 12\)
\( \Rightarrow |\alpha | = 4\) A1
Note: Condone absence of absolute value.
this gives us the position of \(X\) as \(\left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right) \pm \left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ 4 \end{array}} \right)\)
\({\text{X}}(6,{\text{ }}8,{\text{ }}3)\) or \(( – 2,{\text{ }}0,{\text{ }} – 5)\) A1
Note: Award A1 for either result.
[9 marks]
Total [23 marks]
Question
a.Two planes have equations
\[{\Pi _1}:{\text{ }}4x + y + z = 8{\text{ and }}{\Pi _2}:{\text{ }}4x + 3y – z = 0\]
Let \(L\) be the line of intersection of the two planes.
B is the point on \({\Pi _1}\) with coordinates \((a,{\text{ }}b,{\text{ }}1)\).
The point P lies on \(L\) and \({\rm{A\hat BP}} = 45^\circ \).
Find the cosine of the angle between the two planes in the form \(\sqrt {\frac{p}{q}} \) where \(p,{\text{ }}q \in \mathbb{Z}\). [4]
(ii) Show that the point \({\text{A }}(1,{\text{ }}0,{\text{ }}4)\) lies on both planes.
(iii) Write down a vector equation of \(L\).[6]
c.Given the vector \(\overrightarrow {{\text{AB}}} \) is perpendicular to \(L\) find the value of \(a\) and the value of \(b\).[5]
d.Show that \({\text{AB}} = 3\sqrt 2 \).[1]
e.Find the coordinates of the two possible positions of \(P\). [5]
▶️Answer/Explanation
Markscheme
Note: Throughout the question condone vectors written horizontally.
a.angle between planes is equal to the angles between the normal to the planes (M1)
\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) = 18\) (A1)
let \(\theta \) be the angle between the normal to the planes
\(\cos \theta = \frac{{18}}{{\sqrt {18} \sqrt {26} }} = \sqrt {\frac{{18}}{{26}}} {\text{ }}\left( {{\text{or equivalent, for example }}\sqrt {\frac{{324}}{{468}}} {\text{ or }}\sqrt {\frac{9}{{13}}} } \right)\) M1A1
[4 marks]
Note: Throughout the question condone vectors written horizontally.
(i) METHOD 1
\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right)\) M1A1
which is a multiple of \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) R1AG
Note: Allow any equivalent wording or \(\left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right) = 4\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\), do not allow \(\left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\).
METHOD 2
let \(z = t\) (or equivalent)
solve simultaneously to get M1
\(y = t – 4,{\text{ }}x = 3 – 0.5t\) A1
hence direction vector is \(\left( {\begin{array}{*{20}{c}} { – 0.5} \\ 1 \\ 1 \end{array}} \right)\)
which is a multiple of \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) R1AG
METHOD 3
\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = – 4 + 2 + 2 = 0\) M1A1
\(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = – 4 + 6 – 2 = 0\) A1
Note: If only one scalar product is found award M0A0A0.
(ii) \({\Pi _1}:{\text{ }}4 + 0 + 4 = 8\) and \({\Pi _2}:{\text{ }}4 + 0 – 4 = 0\) R1
(iii) r \( = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) A1A1
Note: A1 for “r \( = \)” and a correct point on the line, A1 for a parameter and a correct direction vector.
[6 marks]
Note: Throughout the question condone vectors written horizontally.
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {a – 1} \\ b \\ { – 3} \end{array}} \right)\) (A1)
\(\left( {\begin{array}{*{20}{c}} {a – 1} \\ b \\ { – 3} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = 0\) M1
Note: Award M0 for \(\left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = 0\).
\( – a + 1 + 2b – 6 = 0 \Rightarrow a – 2b = – 5\) A1
lies on \({\Pi _1}\) so \(4a + b + 1 = 8 \Rightarrow 4a + b = 7\) M1
\(a = 1,{\text{ }}b = 3\) A1
[5 marks]
Note: Throughout the question condone vectors written horizontally.
\({\text{AB}} = \sqrt {{0^2} + {3^2} + {{( – 3)}^2}} = 3\sqrt 2 \) M1AG
[1 mark]
Note: Throughout the question condone vectors written horizontally.
METHOD 1
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{AP}}} } \right| = 3\sqrt 2 \) (M1)
\(\overrightarrow {{\text{AP}}} = t\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) (A1)
\(\left| {3t} \right| = 3\sqrt 2 \Rightarrow t = \pm \sqrt 2 \) (M1)A1
\({\text{P}}\left( {1 – \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) and \(\left( {1 + \sqrt 2 ,{\text{ }} – 2\sqrt 2 ,{\text{ }}4 – 2\sqrt 2 } \right)\) A1
[5 marks]
METHOD 2
let P have coordinates \((1 – \lambda ,{\text{ }}2\lambda ,{\text{ }}4 + 2\lambda )\) M1
\(\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 3 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BP}}} = \left( {\begin{array}{*{20}{c}} { – \lambda } \\ {2\lambda – 3} \\ {3 + 2\lambda } \end{array}} \right)\) A1
\(\cos 45^\circ = \frac{{\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} }}{{\left| {{\text{BA}}} \right|\left| {{\text{BP}}} \right|}}\) M1
Note: Award M1 even if AB rather than BA is used in the scalar product.
\(\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} = 18\)
\(\frac{1}{{\sqrt 2 }} = \frac{{18}}{{\sqrt {18} \sqrt {9{\lambda ^2} + 18} }}\)
\(\lambda = \pm \sqrt 2 \) A1
\({\text{P}}\left( {1 – \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) and \(\left( {1 + \sqrt 2 ,{\text{ }} – 2\sqrt 2 ,{\text{ }}4 – 2\sqrt 2 } \right)\) A1
Note: Accept answers given as position vectors.
[5 marks]
Question
A line \(L\) has equation \(\frac{{x – 2}}{p} = \frac{{y – q}}{2} = z – 1\) where \(p,{\text{ }}q \in \mathbb{R}\).
A plane \(\Pi \) has equation \(x + y + 3z = 9\).
Consider the different case where the acute angle between \(L\) and \(\Pi \) is \(\theta \)
where \(\theta = \arcsin \left( {\frac{1}{{\sqrt {11} }}} \right)\).
a.Show that \(L\) is not perpendicular to \(\Pi \).[3]
Given that \(L\) lies in the plane \(\Pi \), find the value of \(p\) and the value of \(q\).[4]
(i) Show that \(p = – 2\).
(ii) If \(L\) intersects \(\Pi \) at \(z = – 1\), find the value of \(q\).[11]
▶️Answer/Explanation
Markscheme
EITHER
n \( = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right)\) and d \( = \left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\) A1A1
and n \( \ne \) kd R1
OR
n \( \times \) d \( = \left( {\begin{array}{*{20}{c}} { – 5} \\ {3p – 1} \\ {2 – p} \end{array}} \right)\) M1A1
the vector product is non-zero for \(p \in \mathbb{R}\) R1
THEN
\(L\) is not perpendicular to \(\Pi \) AG
[3 marks]
METHOD 1
\((2 + p\lambda ) + (q + 2\lambda ) + 3(1 + \lambda ) = 9\) M1
\((q + 5) + (p + 5)\lambda = 9\) (A1)
\(p = – 5\) and \(q = 4\) A1A1
METHOD 2
direction vector of line is perpendicular to plane, so
\(\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 0\) M1
\(p = – 5\) A1
\((2,{\text{ }}q,{\text{ }}1)\) is common to both \(L\) and \(\Pi \)
either \(\left( {\begin{array}{*{20}{c}} 2 \\ q \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 9\) or by substituting into \(x + y + 3z = 9\) M1
\(q = 4\) A1
[4 marks]
(i) METHOD 1
\(\alpha \) is the acute angle between n and L
if \(\sin \theta = \frac{1}{{\sqrt {11} }}\) then \(\cos \alpha = \frac{1}{{\sqrt {11} }}\) (M1)(A1)
attempting to use \(\cos \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\) or \(\sin \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\) M1
\(\frac{{p + 5}}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{1}{{\sqrt {11} }}\) A1A1
\({(p + 5)^2} = {p^2} + 5\) M1
\(10p = – 20\) (or equivalent) A1
\(p = – 2\) AG
METHOD 2
\(\alpha \) is the angle between n and L
if \(\sin \theta = \frac{1}{{\sqrt {11} }}\) then \(\sin \alpha = \frac{{\sqrt {10} }}{{\sqrt {11} }}\) (M1)A1
attempting to use \(\sin \alpha = \frac{{\left| {n \times d} \right|}}{{\left| n \right|\left| d \right|}}\) M1
\(\frac{{\sqrt {{{( – 5)}^2} + {{(3p – 1)}^2} + {{(2 – p)}^2}} }}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{{\sqrt {10} }}{{\sqrt {11} }}\) A1A1
\({p^2} – p + 3 = {p^2} + 5\) M1
\( – p + 3 = 5\) (or equivalent) A1
\(p = – 2\) AG
(ii) \(p = – 2\) and \(z = – 1 \Rightarrow \frac{{x – 2}}{{ – 2}} = \frac{{y – q}}{2} = – 2\) (A1)
\(x = 6\) and \(y = q – 4\) (A1)
this satisfies \(\Pi \) so \(6 + q – 4 – 3 = 9\) M1
\(q = 10\) A1
[11 marks]
Examiners report
Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate \(\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\), with some scripts seen involving needlessly convoluted arguments.
Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate \(\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\), with some scripts seen involving needlessly convoluted arguments.
Part (c) often proved problematic, as some candidates unsurprisingly used the sine (or cosine) of an incorrect angle, and few consequent marks were then available. Some good clear solutions were seen, occasionally complete with diagrams in the cases of the thoughtful candidates who were able to ‘work through’ the question rather than just apply a standard vector result.
Question
A line \(L\) has equation \(\frac{{x – 2}}{p} = \frac{{y – q}}{2} = z – 1\) where \(p,{\text{ }}q \in \mathbb{R}\).
A plane \(\Pi \) has equation \(x + y + 3z = 9\).
Consider the different case where the acute angle between \(L\) and \(\Pi \) is \(\theta \)
where \(\theta = \arcsin \left( {\frac{1}{{\sqrt {11} }}} \right)\).
a. Show that \(L\) is not perpendicular to \(\Pi \).[3]
b.Given that \(L\) lies in the plane \(\Pi \), find the value of \(p\) and the value of \(q\).[4]
(i) Show that \(p = – 2\).
(ii) If \(L\) intersects \(\Pi \) at \(z = – 1\), find the value of \(q\).[11]
▶️Answer/Explanation
Markscheme
EITHER
a.n \( = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right)\) and d \( = \left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\) A1A1
and n \( \ne \) kd R1
OR
n \( \times \) d \( = \left( {\begin{array}{*{20}{c}} { – 5} \\ {3p – 1} \\ {2 – p} \end{array}} \right)\) M1A1
the vector product is non-zero for \(p \in \mathbb{R}\) R1
THEN
\(L\) is not perpendicular to \(\Pi \) AG
[3 marks]
METHOD 1
\((2 + p\lambda ) + (q + 2\lambda ) + 3(1 + \lambda ) = 9\) M1
\((q + 5) + (p + 5)\lambda = 9\) (A1)
\(p = – 5\) and \(q = 4\) A1A1
METHOD 2
direction vector of line is perpendicular to plane, so
\(\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 0\) M1
\(p = – 5\) A1
\((2,{\text{ }}q,{\text{ }}1)\) is common to both \(L\) and \(\Pi \)
either \(\left( {\begin{array}{*{20}{c}} 2 \\ q \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 9\) or by substituting into \(x + y + 3z = 9\) M1
\(q = 4\) A1
[4 marks]
(i) METHOD 1
\(\alpha \) is the acute angle between n and L
if \(\sin \theta = \frac{1}{{\sqrt {11} }}\) then \(\cos \alpha = \frac{1}{{\sqrt {11} }}\) (M1)(A1)
attempting to use \(\cos \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\) or \(\sin \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\) M1
\(\frac{{p + 5}}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{1}{{\sqrt {11} }}\) A1A1
\({(p + 5)^2} = {p^2} + 5\) M1
\(10p = – 20\) (or equivalent) A1
\(p = – 2\) AG
METHOD 2
\(\alpha \) is the angle between n and L
if \(\sin \theta = \frac{1}{{\sqrt {11} }}\) then \(\sin \alpha = \frac{{\sqrt {10} }}{{\sqrt {11} }}\) (M1)A1
attempting to use \(\sin \alpha = \frac{{\left| {n \times d} \right|}}{{\left| n \right|\left| d \right|}}\) M1
\(\frac{{\sqrt {{{( – 5)}^2} + {{(3p – 1)}^2} + {{(2 – p)}^2}} }}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{{\sqrt {10} }}{{\sqrt {11} }}\) A1A1
\({p^2} – p + 3 = {p^2} + 5\) M1
\( – p + 3 = 5\) (or equivalent) A1
\(p = – 2\) AG
(ii) \(p = – 2\) and \(z = – 1 \Rightarrow \frac{{x – 2}}{{ – 2}} = \frac{{y – q}}{2} = – 2\) (A1)
\(x = 6\) and \(y = q – 4\) (A1)
this satisfies \(\Pi \) so \(6 + q – 4 – 3 = 9\) M1
\(q = 10\) A1
[11 marks]
Question
Consider the lines \({l_1}\) and \({l_2}\) defined by
\({l_1}:\) r \( = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ a \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)\) and \({l_2}:\frac{{6 – x}}{3} = \frac{{y – 2}}{4} = 1 – z\) where \(a\) is a constant.
Given that the lines \({l_1}\) and \({l_2}\) intersect at a point P,
a. find the value of \(a\);[4]
b.determine the coordinates of the point of intersection P.[2]
▶️Answer/Explanation
Markscheme
METHOD 1
a.\({l_1}:\)r \( = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ a \end{array}} \right) = \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = – 3 + \beta } \\ {y = – 2 + 4\beta } \\ {z = a + 2\beta } \end{array}} \right.\) M1
\(\frac{{6 – ( – 3 + \beta )}}{3} = \frac{{( – 2 + 4\beta ) – 2}}{4} \Rightarrow 4 = \frac{{4\beta }}{3} \Rightarrow \beta = 3\) M1A1
\(\frac{{6 – ( – 3 + \beta )}}{3} = 1 – (a + 2\beta ) \Rightarrow 2 = – 5 – a \Rightarrow a = – 7\) A1
METHOD 2
\(\left\{ {\begin{array}{*{20}{l}} { – 3 + \beta = 6 – 3\lambda } \\ { – 2 + 4\beta = 4\lambda + 2} \\ {a + 2\beta = 1 – \lambda } \end{array}} \right.\) M1
attempt to solve M1
\(\lambda = 2,{\text{ }}\beta = 3\) A1
\(a = 1 – \lambda – 2\beta = – 7\) A1
[4 marks]
\(\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ { – 7} \end{array}} \right) + 3 \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)\) (M1)
\( = \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { – 1} \end{array}} \right)\) A1
\(\therefore {\text{P}}(0,{\text{ 10, }} – 1)\)
[2 marks]
Question
The points A and B are given by \({\text{A}}(0,{\text{ }}3,{\text{ }} – 6)\) and \({\text{B}}(6,{\text{ }} – 5,{\text{ }}11)\).
The plane Π is defined by the equation \(4x – 3y + 2z = 20\).
a.Find a vector equation of the line L passing through the points A and B.[3]
b.Find the coordinates of the point of intersection of the line L with the plane Π.[3]
▶️Answer/Explanation
Markscheme
a.\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right)\) (A1)
r = \(\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { – 6} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right)\) or r = \(\left( {\begin{array}{*{20}{c}} 6 \\ { – 5} \\ {11} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right)\) M1A1
Note: Award M1A0 if r = is not seen (or equivalent).
[3 marks]
substitute line L in \(\Pi :4(6\lambda ) – 3(3 – 8\lambda ) + 2( – 6 + 17\lambda ) = 20\) M1
\(82\lambda = 41\)
\(\lambda = \frac{1}{2}\) (A1)
r = \(\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { – 6} \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ { – 1} \\ {\frac{5}{2}} \end{array}} \right)\)
so coordinate is \(\left( {3,{\text{ }} – 1,{\text{ }}\frac{5}{2}} \right)\) A1
Note: Accept coordinate expressed as position vector \(\left( {\begin{array}{*{20}{c}} 3 \\ { – 1} \\ {\frac{5}{2}} \end{array}} \right)\). [3 marks]
Question
The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.
It is given that \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \).
The position vectors \(\mathop {{\text{OA}}}\limits^ \to \), \(\mathop {{\text{OB}}}\limits^ \to \), \(\mathop {{\text{OC}}}\limits^ \to \) and \(\mathop {{\text{OD}}}\limits^ \to \) are given by
a = i + 2j − 3k
b = 3i − j + pk
c = qi + j + 2k
d = −i + rj − 2k
where p , q and r are constants.
The point where the diagonals of ABCD intersect is denoted by M.
The plane \(\Pi \) cuts the x, y and z axes at X , Y and Z respectively.
a.i.Explain why ABCD is a parallelogram.[1]
a.ii.Using vector algebra, show that \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \).[3]
b.Show that p = 1, q = 1 and r = 4.[5]
c.Find the area of the parallelogram ABCD.[4]
d.Find the vector equation of the straight line passing through M and normal to the plane \(\Pi \) containing ABCD.[4]
e.Find the Cartesian equation of \(\Pi \).[3]
f.i.Find the coordinates of X, Y and Z.[2]
f.ii.Find YZ.[2]
▶️Answer/Explanation
Markscheme
a.i.a pair of opposite sides have equal length and are parallel R1
hence ABCD is a parallelogram AG
[1 mark]
attempt to rewrite the given information in vector form M1
b − a = c − d A1
rearranging d − a = c − b M1
hence \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) AG
Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.
[3 marks]
EITHER
b.use of \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \) (M1)
\(\left( \begin{gathered}
2 \hfill \\
– 3 \hfill \\
p + 3 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q + 1 \hfill \\
1 – r \hfill \\
4 \hfill \\
\end{gathered} \right)\) A1A1
OR
use of \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) (M1)
\(\left( \begin{gathered}
– 2 \hfill \\
r – 2 \hfill \\
1 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q – 3 \hfill \\
2 \hfill \\
2 – p \hfill \\
\end{gathered} \right)\) A1A1
THEN
attempt to compare coefficients of i, j, and k in their equation or statement to that effect M1
clear demonstration that the given values satisfy their equation A1
p = 1, q = 1, r = 4 AG
[5 marks]
attempt at computing \(\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to \) (or equivalent) M1
\(\left( \begin{gathered}
– 11 \hfill \\
– 10 \hfill \\
– 2 \hfill \\
\end{gathered} \right)\) A1
area \( = \left| {\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to } \right|\left( { = \sqrt {225} } \right)\) (M1)
= 15 A1
[4 marks]
valid attempt to find \(\mathop {{\text{OM}}}\limits^ \to = \left( {\frac{1}{2}\left( {a + c} \right)} \right)\) (M1)
\(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
– \frac{1}{2} \hfill \\
\end{gathered} \right)\) A1
the equation is
r = \(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
– \frac{1}{2} \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
11 \hfill \\
10 \hfill \\
2 \hfill \\
\end{gathered} \right)\) or equivalent M1A1
Note: Award maximum M1A0 if ‘r = …’ (or equivalent) is not seen.
[4 marks]
attempt to obtain the equation of the plane in the form ax + by + cz = d M1
11x + 10y + 2z = 25 A1A1
Note: A1 for right hand side, A1 for left hand side.
[3 marks]
putting two coordinates equal to zero (M1)
\({\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)\) A1
[2 marks]
\({\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}} \) M1
\( = \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)\) A1
[4 marks]
Question
Given the points A(1,1,1) B(2,2,2) C(5,5,5) D(4,5,6)
(a) Show that A,B,C are collinear
(b) Find the vector equation of the line AD
(c) Find the cosine of the angle BAD
▶️Answer/Explanation
Ans
(a) AB =\(\begin{pmatrix}
1\\1
\\ 1\end{pmatrix}\) BC = \(\begin{pmatrix}
3\\3
\\ 3\end{pmatrix}\), Hence BC = 3AB. Since BC//AB and they have a common point A,B,C are collinear. (Notice: You can also consider AC = 4AB)
(b) r = \(\begin{pmatrix}
1\\1
\\ 1\end{pmatrix}\) + \(\lambda \begin{pmatrix}
3\\4
\\ 5\end{pmatrix}\)
(c) cos BAD = \(\frac{AB.AD}{|AB|.|AD|}=\frac{12}{\sqrt{150}}\)
Question
Consider the line r = \(\begin{pmatrix}
2\\0
\\ 3\end{pmatrix}\) + \(\lambda \begin{pmatrix}
2\\1
\\ 2\end{pmatrix}\)
(a) Check if the points A(8,3,8) and B(8,3,9) lie in the line
(b) Find two points on the line, 6 units far from B.
(c) Find two points on the line, \(\sqrt{89}\) units far from the origin.
(d) Find two points on the line, \(\sqrt{54}\) units far from C(1,0,2).
(e) Find the point on the line which is closest to the point D(0,1,0).
(f) Find the distance between the line and the point D(0,1,0).
(f) Find the reflection D of the point D in the line.
▶️Answer/Explanation
Ans
(a) Only B(8,3,9) lies in the line
For the following questions we notice that a point in the line has the form P(2+2λ,λ,3+2λ).
(b) d(A,P) = 6 \(\Rightarrow λ=1\) or 5. The points are (4,1,5) and (12,5,13)
(c) d(O,P) = \(sqrt{89} \Rightarrow λ = 2, λ = -\frac{38}{9}\) The points are (6,2,7) and \((-\frac{58}{9},\frac{38}{9},-\frac{49}{9})\)
(c) d(C,P) = \(\sqrt{54} \Rightarrow λ=2,λ=-\frac{26}{9}\) The points are (6,2,7) and \((-\frac{58}{9}, -\frac{26}{9},-\frac{25}{9})\)
(e) For the foot P, DP ⊥ line \(\Rightarrow \begin{pmatrix}
2+2\lambda \\ \lambda -1
\\ 3+2\lambda\end{pmatrix}\) \(\begin{pmatrix}
2\\1
\\ 2\end{pmatrix}=0 \Rightarrow \lambda =-1\). The point is (0,-1,1)
(f) For P(0,-1,1) (from (e), d(D,P)= \(\sqrt{5}\)
(g) For D(0,1,0), P(0,-1,1) it is D'(0,-3,2)(since P is the midpoint of DD)
Question
Express the following lines of the 3D space in the form r = a + λb
(a) \(\frac{x-2}{3}=\frac{y+4}{7}=\frac{z-1}{5}\)
(b) \(\frac{x+1}{3}=5-y=\frac{2z-1}{5}\)
(c) \(\frac{x-2}{3}=\frac{y+4}{7}, z=10\)
(d) \(\frac{x-2}{3}=\frac{y+4}{0}=\frac{z+1}{1}\)
(e) x – axis (f) y – axis (g) z – axis
▶️Answer/Explanation
Ans
Question
Given two non-zero vectors a and b show that
(a) |a|=|b| if and only if a+b ⊥ a-b
(b) a ⊥ b if and only if |a+b|=|a-b|
(c) Give the geometric interpretation
(i) of the result (a) (ii) of the result (b) (iii) of the results (a) and (b) together.
▶️Answer/Explanation
Ans
Given two non-zero vectors a and b show that
(a) |a|=|b| \(\Leftrightarrow |a|^2 =|b|^2 \Leftrightarrow a^2 = b^2 \Leftrightarrow a^2-b^2 = 0 (a+b). (a-b)=0 \Leftrightarrow a+b ⊥ a-b\)
(b) |a+b| = |a-b| \(\Leftrightarrow |a+b|^2 = |a-b|^2 \Leftrightarrow (a+b)^2 = (a-b)^2\)
\(\Leftrightarrow a^2 + 2a.b + b^2 = a^2 – 2a.b +b^2 \Leftrightarrow 4a.b = 0 \Leftrightarrow a.b = 0 \Leftrightarrow a ⊥ b\)
(c)
(i) a parallelogram is a rhombus if and only if the diagonals are equal
(ii) a parallelogram is a rectangle if and only if the diagonals bisect each other
(iii) a parallelogram is a square if and only if the diagonals are equal and bisect each other
Question
The line \(l_1\) and \(l_2\) have equations
\(r_1=\begin{pmatrix}
1\\4
\\ 3\end{pmatrix}+\lambda \begin{pmatrix}
2\\1
\\ 2\end{pmatrix}\) and \(r_2=\begin{pmatrix}
4\\4
\\ 5\end{pmatrix}\)+ \(\mu \begin{pmatrix}
1\\2
\\ 2\end{pmatrix}\)
respectively, where \(\lambda\) and \(\mu\) are parameters.
(a) Find the acute angle between \(l_1\) and \(\l_2\)
(b) Find the acute angle between \(l_1\) and z-axis
(c) Find the point of intersection of \(l_1\) and \(l_2\)
(d) Find the equations of the two bisector lines between \(l_1\) and \(l_2\)
(e) Find the reflection of point A(1,4,3) in line \(l_2\)
(f) Find the reflection of line \(l_1\) in line \(l_2\)
▶️Answer/Explanation
Ans
(a) 27.3 (b) 48.2 (c) P(5,6,7)
(d) They pass through P(5,6,7). Since the direction vectors have the same magnitude, for the bisector lines we can consider the direction vectors \(d_1 \pm d_2\)
(otherwise we would add and subtract the unit vectors of \(d_1, d_2\)
(e) The foot is P(29/9, 22/9, 31/9) so the reflection of A is A'(49/9,8/9,35/9)
(f) it is the line passing through P(5,6,7) and A'(49/9,8/9,35/9)
Question
Two bodies are moving according to the equations
(a) Find the velocities and the speeds of the bodies.
(b) Find the point of intersection of their paths
(c) Do they collide or not?
▶️Answer/Explanation
Ans
Two bodies are moving according to the equations
(b) Point of intersection (5,6,7)
(c) They do not collide sine at the point of intersection \(t_1=2,t_2=1\)
Question
Find the angle between the vectors v = i + j + 2k and w = 2i + 3j + k. Give your answer in radians.
▶️Answer/Explanation
Ans
v . w = 2 + 3 +2 = 7
|v| = \(\sqrt{6}\) |w| = \(\sqrt{14}\)
θ = arccos \((\frac{v.w}{|v||w|)\) = arccos \((\frac{7}{\sqrt{6}\sqrt{14}})\)=0.702 radians.
Question
A triangle has its vertices at A(–1, 3, 2), B(3, 6, 1) and C(–4, 4, 3).
(a) Show that \(\overrightarrow{AB}.\overrightarrow{AC}\) = –10.
(b) Show that, to three significant figures, cosBAC = –0.591.
▶️Answer/Explanation
Ans
(a) Finding correct vectors \(\overrightarrow{AB}\) = \(\begin{pmatrix}
4\\3
\\ -1\end{pmatrix}\) \(\overrightarrow{AC}\) = \(\begin{pmatrix}
-3\\1
\\ 1\end{pmatrix}\)
Substituting correctly in scalar product \(\overrightarrow{AB}.\overrightarrow{AC}\) = 4(-3)+3(1)-1(1)
=-10
(b) |\(\overrightarrow{AB}\)| = \(\sqrt{26}\) |\(overrightarrow{AC\)|=\(\sqrt{11}\)
Attempting to use scalar product formula, cos BAC = \(\frac{-10}{\sqrt{26}\sqrt{11}}\)
=-0.591 (to 3 s.f)
Question
The vectors a, b and c are defined by \(a= \begin{pmatrix}
3\\2
\\ -1\end{pmatrix}\), \(b = \begin{pmatrix}
1\\5
\\ 2\end{pmatrix}\), and \(c=\begin{pmatrix}
2\\y
\\ 3\end{pmatrix}\)
Given that c is perpendicular to 2a – b, find the value of y.
▶️Answer/Explanation
Ans
Question
Consider the vectors 
Let s = (a ⋅b)c + d , where s is perpendicular to a .
Find an expression for λ in terms of µ.
▶️Answer/Explanation
Ans
a•b=3
\(s=3\begin{pmatrix}
3\\1
\\ \lambda\end{pmatrix}\)+\(\begin{pmatrix}
\mu\\-2
\\ 1\end{pmatrix}\)
\(s=\begin{pmatrix}
9+\mu\\1
\\ 3\lambda+1\end{pmatrix}\)
s•a \(=0\Rightarrow \begin{pmatrix}
9+\mu\\1
\\ 3\lambda + 1\end{pmatrix}\)\(\bullet \begin{pmatrix}
2\\3
\\ -1\end{pmatrix}=0\)
\(18+2\mu+3-3\lambda-1=0 (20+2\mu=3\lambda)\)
\(\lambda = \frac{20+2\mu}{3}\)
Question
Let α be the angle between the vectors a and b, where a=(cos θ)i + (sin θ)j, b=(sin θ)i + (cos θ)j and 0 < θ < . Express α in terms of θ.
▶️Answer/Explanation
Ans
Method 1: Let the angle be α, then cosα \(=\frac{a.b}{|a||b|}\)
\(=\frac{2sin\theta cos\theta}{(1)(1)}\)
\(=sin 2 \theta =cos(\frac{\pi}{2}-20)\)
α = \(\frac{\pi}{2}\)– 2θ or α = arcos(sin 2θ)
Question
The position vectors of points P and Q are \(\begin{pmatrix}
2\\-3
\\ 1\end{pmatrix}\) and \(\begin{pmatrix}
2\\2
\\ -4\end{pmatrix}\) respectively. The origin is at O.
Find (a) the angle POQ; (b) the area of the triangle OPQ.
▶️Answer/Explanation
Ans
(a) \(cos POQ=\frac{\overrightarrow{OP}.\overrightarrow{OQ}}{|\overrightarrow{OP}||\overrightarrow{OQ}|}=\frac{-6}{\sqrt{14}\sqrt{24}})=-\frac{6}{\sqrt{336}}\)
POQ=109o (1.90 radians)
(b) METHOD 1
area \(\Delta POQ=\frac{1}{2}|\overrightarrow{PO}||\overrightarrow{OQ}| sin POQ=8.66(Accept 8.67)\)
METHOD 2 (this requires the cross product; it will be learned later on)
area \(\Delta POQ=\frac{1}{2}|\overrightarrow{PO}\times \overrightarrow{OQ}|=\frac{1}{2}|10i+10j+10k|=\sqrt{75}(=5\sqrt{3})\)
Question
The angle between the vector a = i – 2j + 3k and the vector b = 3i – 2j + mk is 30°.
Find the values of m.
▶️Answer/Explanation
Ans
Question
The rectangle box shown in the diagram has dimensions 6 cm × 4 cm × 3 cm.
Find, correct to the nearest one-tenth of a degree, the size of the angle AHC.
▶️Answer/Explanation
Ans
AH = 5 cm, HC = 3\(\sqrt{5}\) cm, AC = 2\(\sqrt{13}\) cm
\(cos AHC=\frac{AH^2+CH^2-AC^2}{2(AH)(CH)}=\frac{25+45-52}{31\sqrt{5}}\)
i.e AHC = 74.4o (to the nearest one-tenth of a degree)
Question
Three distinct non-zero vectors are given by \(\overrightarrow{OA}=a,\overrightarrow{OB}=b\) and \(\overrightarrow{OC}=c\).
If \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow {BC} and \(\overrightarrow{OB}\) is perpendicular to CA, show that OC is perpendicular to \(\overrightarrow{AB}\).
▶️Answer/Explanation
Ans
\(\overrightarrow{BC}=c-b\)
\(\overrightarrow{CA}=a-c\)
⇒ a • (c – b) = 0 and b • (a – c) = 0
⇒ a • c = a • b and a • b = b • c
⇒ a • c = b • c
⇒ b • c – a • c = 0 ⇒c • (b – a) = 0
⇒ \(\overrightarrow{OC}\) is perpendicular to \(\overrightarrow{AB}\), as b≠a
Question
Given two non-zero vectors u and v such that |u + v| = |u – v|, find the value of u ⋅ v.
▶️Answer/Explanation
Ans
|u+v|=|u-v| \(\Rightarrow |u+v|^2=|u-v|^2\)
\(\Rightarrow (u+v).(u+v)=(u-v).(u-v)\)
\(\Rightarrow |u|^2 + 2u.v +|v|^2 = |u|^2 – 2u.v +|v|^2\)
\(\Rightarrow 4u.v = 0 \Rightarrow u.v = 0\)
Question
The lines l1 and l2 have equations
respectively, where \(\lambda\) and \(\mu\) are parameters.
(a) Show that \(l_1\) passes through the point (2,-7,4).
(b) Determine whether the lines \(l_1\) and \(l_2\) intersect.
▶️Answer/Explanation
Ans
(a) \(l_1 r=\begin{pmatrix}
4+\lambda\\3+5\lambda
\\ -2\lambda\end{pmatrix}\)
for \(l_1\) for x=2, \(\lambda=-2\)
\(\Rightarrow y=-7\)
\(\Rightarrow z=4\)
Therefore point fits on line.
(b) \(4+\lambda=2\) Eq(1)
\(3+5\lambda=-1+2\mu\) Eq(2)
\(-2k=3-3\mu\) Eq(3)
From Eq(1), \(\lambda=-2\)
From Eq(2), \(3-10=-1+2\mu\)
\(-7=-1+2\mu\)
\(\mu=-3\)
Substituting in Eq(3)
\(\Rightarrow 4 = 3 + 9\)
\(\Rightarrow \) lines do not intersect
Question
The vector equations of the lines L1 and L2 are given by
L1: r = i + j + k + λ(i + 2j + 3k); L2: r = i + 4j + 5k + µ(2i + j + 2k).
The two lines intersect at the point P. Find the position vector of P.
Extra Question
Find the equation of a line that bisects the angle between L1 and L2
▶️Answer/Explanation
Ans
Solve 1 + λ = 1 + 2µ, 1 + 2λ = 4 + µ, 1 + 3λ = 5 + 2µ
Solving, λ = 2, (or µ = 1).
P has position vector 3i + 5j + 7k; or \(\begin{pmatrix}
3\\5
\\ 7\end{pmatrix}\)
Extra question(for the bisector lines)
The direction unit vectors are: \(d_1=\begin{pmatrix}
1/\sqrt{14}\\2/\sqrt{14}
\\ 3/\sqrt{14}\end{pmatrix}\),\( d_2=\begin{pmatrix}
2/3\\1/3
\\ 2/3\end{pmatrix}\)
The lines are: \(r=\begin{pmatrix}
3\\5
\\ 7\end{pmatrix}\)+\(\lambda \begin{pmatrix}
1/\sqrt{14}+2/3\\2/\sqrt{14}+1/3
\\ 3/\sqrt{14}+2/3\end{pmatrix}\) and \(r=\begin{pmatrix}
3\\5
\\ 7\end{pmatrix}\)+\(\lambda\begin{pmatrix}
1/\sqrt{14}-2/3\\ 2/\sqrt{14}-1/3
\\ 3/\sqrt{14}-2/3\end{pmatrix}\)
Question
Consider the four points A(1, 4, –1), B(2, 5, –2), C(5, 6, 3) and D(8, 8, 4). Find the point of intersection of the lines (AB) and (CD).
▶️Answer/Explanation
Ans
Line (AB) is \(\begin{pmatrix}
x\\y
\\ z\end{pmatrix}\)=\(\begin{pmatrix}
1\\4
\\ -1\end{pmatrix}\)+\(\lambda\begin{pmatrix}
1\\1
\\ -1\end{pmatrix}\) Line (CD) is \(\begin{pmatrix}
x\\y
\\ z\end{pmatrix}\)=\(\begin{pmatrix}
5\\6
\\ 3\end{pmatrix}\)+\( \mu \begin{pmatrix}
3\\2
\\ 1\end{pmatrix}\)
at point of intersection of two lines
1 + λ = 5 + 3µ
4 + λ = 6 + 2µ
–1 – λ = 3 + µ
solving simultaneously any two of these three equations gives
λ = –2 and µ = –2 (only one value required).
⇒ point of intersection (–1, 2, 1)
Question
Point A(3, 0, –2) lies on the line r = 3i – 2k + λ(2i – 2j + k), where λ is a real parameter. Find the coordinates of one point which is 6 units from A, and on the line.
▶️Answer/Explanation
Ans
The direction of the line is v = 2i – 2j + k and |v|= 3.
Therefore, the position vector of any point on the line 6 units from A is
3i – 2k ± 2v = 7i – 4j or –i + 4j – 4k,
giving the point (7, –4, 0) or (–1, 4, –4).
OR
Any point on the line has the form P(3+2λ,-2λ,-2+λ). The distance between A and B must be 6.
This will give two values for λ: λ=±2. Thus the point is (7, –4, 0) or (–1, 4, –4).
Question
Consider the points A (1, 3, –17) and B (6, – 7, 8) which lie on the line l.
(a) Find an equation of line l, giving the answer in parametric form.
(b) The point P is on l such that OP is perpendicular to l. Find the coordinates of P.
▶️Answer/Explanation
Ans
(a) \(\overrightarrow{AB}=\begin{pmatrix}
5\\-10
\\ 25\end{pmatrix}\). Direction vector of line is 1: –2: 5.
(Accept any multiple of 1: –2: 5)
Therefore the equation of l in parametric form is
x = λ + l, y = –2λ + 3, z = 5λ – 17
(or x = λ + 6, y = –2λ – 7, z = 5λ + 8, or any equivalent parametric form)
(b) P on l => P can be written as (p + 1, –2p + 3, 5p – 17).
\(\overrightarrow{OP}\) ⊥ \(l=>\begin{pmatrix}
p+1\\-2p+3
\\ 5p-17\end{pmatrix}\).\(\begin{pmatrix}
1\\-2
\\ 5\end{pmatrix}\)\(=0\)
p + 1 + 4p – 6 + 25p – 85 = 0
30p = 90 => p = 3
Therefore P is (4, –3, –2)
Question
The line L is given by the parametric equations x = 1 – λ, y = 2 – 3λ, z = 2. Find the coordinates of the point on L that is nearest to the origin.
METHOD A (consider that the vector OP is perpendicular to the line)
METHOD B (use the formula of the distance d between the origin and any point on L; then use the derivative of d – or even better of d2 – to find the minimum)
▶️Answer/Explanation
Ans
EITHER
The position vector for the point nearest to the origin is perpendicular to the direction of the line. At that point:
\(\begin{pmatrix}1-\lambda\\ 2-3\lambda
\\ 2\end{pmatrix}\)\(\bullet \begin{pmatrix}
-1\\-3
\\ 0\end{pmatrix}=0\)
Therefore, \(10\lambda – 7=0\) Therefore, \(\lambda =\frac{7}{10}\)
OR
Let s be the distance from the origin to a point on the line, then
s2 = (1 – λ)2 + (2 – 3λ)2 + 4= 10λ2 – 14λ + 9
\(\frac{d(s^2)}{d\lambda}=20\lambda -14\) For minimum \(\frac{d(s^2)}{d\lambda}=0.\Rightarrow \lambda = \frac{7}{10}\)
THEN
\(x=\frac{3}{10}, y=-\frac{1}{10}\) The point is \((\frac{3}{10},\frac{-1}{10},2)\).
Question
Calculate the shortest distance from the point A(0, 2, 2) to the line
r→ = 5i→ + 9 →j + 6k→ + t(i→ + 2 →j + 2k→ ) where t is a scalar.
Extra questions
(b) Write down the coordinates of the point on the line that is nearest to A
(c) The point A is reflected in the line. Find the coordinates of the corresponding point A΄
▶️Answer/Explanation
Ans
\(\overrightarrow{AB}.\overrightarrow{v}=0\) Therefore, \(\begin{pmatrix}
t+5\\2t+7
\\ 2t+4 \end{pmatrix}\)\(\begin{pmatrix}
1\\2
\\ 2\end{pmatrix}=0\)
t + 5 + 4t + 14 + 4t + 8 = 0, giving t = –3
Then \(\overrightarrow{AB}=\begin{pmatrix}
2\\1
\\ -2\end{pmatrix}\), and the required distance AB = 3.
Extra questions
(b) for t = –3 we obtain B(2,3,0)
(c) B (2,3,0) is the midpoint of AA΄. Since we know A(0,2,2), we obtain A΄(4,4,-2)
Question
Consider the points A(1, −1, 4), B(2, − 2, 5) and O(0, 0, 0).
(a) Calculate the cosine of the angle between \(\overrightarrow{OA}\) and \(\overrightarrow{AB}\).
(b) Find a vector equation of the line L1 which passes through A and B.
The line L2 has equation r = 2i + 4j + 7k + t(2i + j + 3k), where t∈\(\mathbb{R}\).
(c) Show that the lines L1 and L2 intersect and find the coordinates of their point of intersection.
▶️Answer/Explanation
Ans
Use of \(cos\theta=\frac{\overrightarrow{OA}\bullet{\overrightarrow{AB}}}{|\overrightarrow{OA}||\overrightarrow{OA}|}\)
\(\overrightarrow{AB}=i-j+k\) \(|\overrightarrow{AB}|=\sqrt{3}\) and \(|\overrightarrow{OA}|=3\sqrt{2}\)
\(\overrightarrow{OA}\bullet \overrightarrow{AB}=6\)
substituting gives \(cos\theta=\frac{2}{\sqrt{6}}(=\frac{\sqrt{6}}{3})\) or equivalent
(b) L1: r = OA + sAB or equivalent
L1: r = i − j + 4k + s(i − j + k) or equivalent
(c) Equating components and forming equations involving s and t
1 + s = 2 + 2t, −1 − s = 4 + t, 4 + s = 7 + 3t
Having two of the above three equations
Attempting to solve for s or t
Finding either s = −3 or t = −2
Explicitly showing that these values satisfy the third equation
Point of intersection is (−2, 2, 1)
Question
The triangle ABC has vertices at the points A(–l, 2, 3), B(–l, 3, 5) and C(0, –1, 1).
(a) Find the size of the angle θ between the vectors \(\overrightarrow{AB}\) and \(overrightarrow{AC}\).
(b) Hence, or otherwise, find the area of triangle ABC.
Let l1 be the line parallel to \(\overrightarrow{AB}\) which passes through D(2, –1, 0) and l2 be the line parallel to \(\overrightarrow{AB}\) which passes through E(–l, 1, 1).
(c) (i) Find the equations of the lines l1 and l2 .
(ii) Hence show that l1 and l2 do not intersect.
(d) Find the shortest distance between l1 and l2.
▶️Answer/Explanation
Ans
(a) Given the points A(–1, 2, 3), B(–1, 3, 5) and C(0, –1, 1),
then \(\overrightarrow{AB}=\begin{pmatrix}
0\\1
\\ 2\end{pmatrix}\),\( \overrightarrow{AC}=\begin{pmatrix}
1\\-3
\\ -2\end{pmatrix}\) and \(|\overrightarrow{AB}|=\sqrt{5},|\overrightarrow{AC}|=\sqrt{14}\) is given by
\(\theta = arccos (\frac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|})=arccos(\frac{-7}{\sqrt{5}\sqrt{14}})=147^o(3s.f)or2.56rad\)
(b) Area \(=\frac{1}{2}|\overrightarrow{AB}||\overrightarrow{AC}| sin\theta or \frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\)
Area=2.29 units2 (accept 2.28, 2.30 and \(\frac{\sqrt{21}}{2}\))
(c) (i) The parametric equations of \(l_1\) and \(l_2\) are
l1: x = 2, y = –1 + λ, z = 2λ
l2: x = –1 + µ, y = 1 – 3µ, z = 1 – 2µ
(ii) To test for a point of intersection we use the system of equations:
2 = –1 + µ …..(1)
–1 + λ = 1 – 3µ ……(2)
2λ = 1 – 2µ ………(3)
Then \(\mu=3, \lambda = -7\) from (1) and (2)
Substituting into 3 gives RHS = –14, LHS = –5
Therefore the system of equations has no solution and the lines do not intersect.
(d) Consider random points P(2,-1+λ,2λ) on l1 and Q(-1+µ,1-3µ, 1-2µ) on l2.
Then PQ = \(\begin{pmatrix}
-3+\mu\\2-3\mu-\lambda
\\ 1-2\mu-2\lambda\end{pmatrix}\)
PQ is perpendicular to both lines: so PQ. \(\begin{pmatrix}
0\\1
\\ 2\end{pmatrix}=0\) and PQ. \(\begin{pmatrix}
1\\-3
\\ -2\end{pmatrix}=0\)
That is
(2−3µ−λ)+2(1−2µ−2λ) = 0⇒ 5λ+7µ= 4
(−3+µ)−3(2−3µ−λ)−2(1−2µ−2λ) = 0⇒ 7λ+14µ=11
This gives \(\lambda = -1, \mu = 9/7. PQ=\(\begin{pmatrix}
-12/7\\-6/7
\\ 3/7\end{pmatrix}\). |PQ| = \(\frac{\sqrt{1890}}{7}=1.96\)
Question
(a) A curve is defined by the implicit equation 2xy + 6x2 – 3y2 = 6.
Show that the tangent at the point A with coordinates \((1,\frac{2}{3})\), and at point B.
Find, in the form \(r=\begin{pmatrix}
a\\b\end{pmatrix}\)+ \(s\begin{pmatrix}
c\\d\end{pmatrix}\)
(i) the equation of the tangent at A;
(ii) the equation of the normal at B.
(c) Find the acute angle between the tangent at A and the normal at B.
▶️Answer/Explanation
Ans
(a) 2xy + 6x2 – 3y2 = 6
\(\Rightarrow 2x\frac{dy}{dx}+2y+12x-6y\frac{dy}{dx}=0\)
\(\Rightarrow 2\frac{dy}{dx}+\frac{4}{3}+12-4\frac{dy}{dx}=0\)
\(\Rightarrow 2\frac{dy}{dx}=\frac{40}{3} \Rightarrow \frac{dy}{dx} = \frac{20}{3}\)
(b) (i) The equation of the tangent at A is \(r=\begin{pmatrix}
1\\\frac{2}{3}\end{pmatrix}\) + \(s\begin{pmatrix}
3\\ 20\end{pmatrix}\)
(ii) Substituting x = 1 2y + 6 – 3y2 = 6
\(\Rightarrow 3y^2 – 2y = 0\)
\(\Rightarrow y(3y – 2) = 0\)
\(\Rightarrow y = 0, \frac{2}{3}\)
Gradient at B satisfies \(2\frac{dy}{dx}+0+12=0\)
\(\Rightarrow \frac{dy}{dx}=-6\)
Gradient of the normal is \(\frac{1}{6}\)
Hence equation of the normal at B is \(r=\begin{pmatrix}
1\\0\end{pmatrix}\)+\(t\begin{pmatrix}
6\\1\end{pmatrix}\)
(c) \(cos\theta =\frac{18+20}{\sqrt{409}\sqrt{37}} (=0.309)\)
\(\Rightarrow \theta = 1.26\) rads or 72.0o.