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 \(\overrightarrow {{\text{OA}}} \) and \(\overrightarrow {{\text{AB}}} \).
(b) Find a vector equation of the line \({L_1}\) which passes through A and B.
The line \({L_2}\) has equation r = 2i + 4j + 7k + t(2i + j + 3k), where \(t \in \mathbb{R}\) .
(c) Show that the lines \({L_1}\) and \({L_2}\) intersect and find the coordinates of their point of intersection.
(d) Find the Cartesian equation of the plane which contains both the line \({L_2}\) and the point A.
▶️Answer/Explanation
Markscheme
(a) Use of \(\cos \theta = \frac{{\overrightarrow {{\text{OA}}} \cdot \overrightarrow {{\text{AB}}} }}{{\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{AB}}} } \right|}}\) (M1)
\({\overrightarrow {{\text{AB}}} }\) = i − j + k A1
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt 3 \) and \(\left| {\overrightarrow {{\text{OA}}} } \right| = 3\sqrt 2 \) A1
\(\overrightarrow {{\text{OA}}} \cdot \overrightarrow {{\text{AB}}} = 6\) A1
substituting gives \(\cos \theta = \frac{2}{{\sqrt 6 }}\,\,\,\,\,\left( { = \frac{{\sqrt 6 }}{3}} \right)\,\,\,\,\,\)or equivalent M1 N1
[5 marks]
(b) \({L_1}\) : r = \(\overrightarrow {{\text{OA}}} + s\overrightarrow {{\text{AB}}} \,\,\,\,\,\) or equivalent (M1)
\({L_1}\) : r = i − j + 4k + s(i − j + k)\(\,\,\,\,\,\)or equivalent A1
Note: Award (M1)A0 for omitting “ r = ” in the final answer.
[2 marks]
(c) Equating components and forming equations involving s and t (M1)
1 + s = 2 + 2t , −1 − s = 4 + t , 4 + s = 7 + 3t
Having two of the above three equations A1A1
Attempting to solve for s or t (M1)
Finding either s = −3 or t = −2 A1
Explicitly showing that these values satisfy the third equation R1
Point of intersection is (−2, 2, 1) A1 N1
Note: Position vector is not acceptable for final A1.
[7 marks]
(d) METHOD 1
\(r = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 1} \\
4
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
2 \\
1 \\
3
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
{ – 3} \\
3 \\
{ – 3}
\end{array}} \right)\) (A1)
\(x = 1 + 2\lambda – 3\mu {\text{ , }}y = – 1 + \lambda + 3\mu {\text{ and }}z = 4 + 3\lambda – 3\mu \) M1A1
Elimination of the parameters M1
\(x + y = 3\lambda {\text{ so }}4(x + y) = 12\lambda {\text{ and }}y + z = 4\lambda + 3{\text{ so }}3(y + z) = 12\lambda + 9\)
\(3(y + z) = 4(x + y) + 9\) A1
Cartesian equation of plane is 4x + y − 3z = −9 (or equivalent) A1 N1
[6 marks]
METHOD 2
EITHER
The point (2, 4, 7) lies on the plane.
The vector joining (2, 4, 7) and (1, −1, 4) and 2i + j + 3k are parallel to the plane. So they are perpendicular to the normal to the plane.
(i − j + 4k) − (2i + 4j + 7k) = −i − 5j − 3k (A1)
\(n = \left| {\begin{array}{*{20}{c}}
i&j&k \\
{ – 1}&{ – 5}&{ – 3} \\
2&1&3
\end{array}} \right|\) M1
= −12i − 3j + 9k\(\,\,\,\,\,\)or equivalent parallel vector A1
OR
\({L_1}\) and \({L_2}\) intersect at D(−2, 2, 1)
\(n = \left| {\begin{array}{*{20}{c}}
i&j&k \\
2&1&{ – 3} \\
{ – 3}&3&{ – 3}
\end{array}} \right|\) M1
= −12i − 3j + 9k\(\,\,\,\,\,\)or equivalent parallel vector A1
THEN
r\( \cdot \)n = (i − j + 4k)\( \cdot \)(−12i − 3j + 9k) M1
= 27 A1
Cartesian equation of plane is 4x + y −3z = −9 (or equivalent) A1 N1
[6 marks]
Total [20 marks]
Examiners report
Most candidates scored reasonably well on this question. The most common errors were: Using OB rather than AB in (a); omitting the r = in (b); failure to check that the values of the two parameters satisfied the third equation in (c); the use of an incorrect vector in (d). Even when (d) was correctly answered, there was usually little evidence of why a specific vector had been used.
Question
(a) Show that the two planes
\[{\pi _1}:x + 2y – z = 1\]
\[{\pi _2}:x + z = – 2\]
are perpendicular.
(b) Find the equation of the plane \({\pi _3}\) that passes through the origin and is perpendicular to both \({\pi _1}\) and \({\pi _2}\).
▶️Answer/Explanation
Markscheme
(a) for using normal vectors (M1)
\(\left( {\begin{array}{*{20}{c}}
1 \\
2 \\
{ – 1}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
1
\end{array}} \right) = 1 – 1 = 0\) M1A1
hence the two planes are perpendicular AG
(b) METHOD 1
EITHER
\(\left| {\begin{array}{*{20}{c}}
i&j&k \\
1&2&{ – 1} \\
1&0&1
\end{array}} \right| = \) 2i – 2j– 2k M1A1
OR
if \(\left( {\begin{array}{*{20}{c}}
a \\
b \\
c
\end{array}} \right)\) is normal to \({\pi _3}\), then
\(a + 2b – c = 0\) and \(a + c = 0\) M1
a solution is a = 1, b = –1, c = –1 A1
THEN
\({\pi _3}\) has equation \(x – y – z = d\) (M1)
as it goes through the origin, d = 0 so \({\pi _3}\) has equation \(x – y – z = 0\) A1
Note: The final (M1)A1 are independent of previous working.
METHOD 2
\(r = \left( {\begin{array}{*{20}{c}}
0 \\
0 \\
0
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
1 \\
2 \\
{ – 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1 \\
0 \\
1
\end{array}} \right)\) A1(A1)A1A1
[7 marks]
Examiners report
Although many candidates were successful in answering this question, a surprising number showed difficulties in working with normal vectors. In part (b) there were several candidates who found the cross product of the vectors but were unable to use it to write the equation of the plane.
Question
a.Consider the vectors a = 6i + 3j + 2k, b = −3j + 4k.
(i) Find the cosine of the angle between vectors a and b.
(ii) Find a \( \times \) b.
(iii) Hence find the Cartesian equation of the plane \(\prod \) containing the vectors a and b and passing through the point (1, 1, −1).
(iv) The plane \(\prod \) intersects the x-y plane in the line l. Find the area of the finite triangular region enclosed by l, the x-axis and the y-axis.[11]
Given two vectors p and q,
(i) show that p\( \cdot \)p = \(|\)p\({|^2}\);
(ii) hence, or otherwise, show that \(|\)p + q\({|^2}\) = \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\);
(iii) deduce that \(|\)p + q\(|\) ≤ \(|\)p\(|\) + \(|\)q\(|\).
▶️Answer/Explanation
Markscheme
(i) use of a\( \cdot \)b = \(|\)a\(|\)\(|\)b\(|\cos \theta \) (M1)
a\( \cdot \)b = –1 (A1)
\(|\)a\(|\) = 7, \(|\)b\(|\) = 5 (A1)
\(\cos \theta = – \frac{1}{{35}}\) A1
(ii) the required cross product is
\(\left| {\begin{array}{*{20}{c}}
i&j&k \\
6&3&2 \\
0&{ – 3}&4
\end{array}} \right| = \) 18i – 24j -18k M1A1
(iii) using r\( \cdot \)n = p\( \cdot \)n the equation of the plane is (M1)
\(18x – 24y – 18z = 12\,\,\,\,\,(3x – 4y – 3z = 2)\) A1
(iv) recognizing that z = 0 (M1)
x-intercept \( = \frac{2}{3}\), y-intercept \( = – \frac{1}{2}\) (A1)
area \( = \left( {\frac{2}{3}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right) = \frac{1}{6}\) A1
[11 marks]
(i) p\( \cdot \)p = \(|\)p\(|\)\(|\)p\(|\cos 0\) M1A1
= \(|\)p\({|^2}\) AG
(ii) consider the LHS, and use of result from part (i)
\(|\)p + q\({|^2}\) = (p + q)\( \cdot \)(p + q) M1
= p\( \cdot \)p + p\( \cdot \)q + q\( \cdot \)p + q\( \cdot \)q (A1)
= p\( \cdot \)p + 2p\( \cdot \)q + q\( \cdot \)q A1
= \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\) AG
(iii) EITHER
use of p\( \cdot \)q \( \leqslant \) \(|\)p\(|\)\(|\)q\(|\) M1
so 0 \( \leqslant \) \(|\)p + q\({|^2}\) = \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\) \( \leqslant \) \(|\)p\({|^2}\) + 2 \(|\)p\(|\)\(|\)q\(|\) + \(|\)q\({|^2}\) A1
take square root (of these positive quantities) to establish A1
\(|\)p + q\(|\) \( \leqslant \) \(|\)p\(|\) + \(|\)q\(|\) AG
OR
M1M1
Note: Award M1 for correct diagram and M1 for correct labelling of vectors including arrows.
since the sum of any two sides of a triangle is greater than the third side,
\(|\)p\(|\) + \(|\)q\(|\) > \(|\)p + q\(|\) A1
when p and q are collinear \(|\)p\(|\) + \(|\)q\(|\) = \(|\)p + q\(|\)
\( \Rightarrow |\)p + q\(|\) \( \leqslant \) \(|\)p\(|\) + \(|\)q\(|\) AG
[8 marks]
Question
The points A(1, 2, 1) , B(−3, 1, 4) , C(5, −1, 2) and D(5, 3, 7) are the vertices of a tetrahedron.
a.Find the vectors \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \).[2]
b.Find the Cartesian equation of the plane \(\prod \) that contains the face ABC.[4]
▶️Answer/Explanation
Markscheme
a\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
{ – 1} \\
3
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
4 \\
{ – 3} \\
1
\end{array}} \right)\) A1A1
Note: Accept row vectors.
[2 marks]
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
{\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
{ – 4}&{ – 1}&3 \\
4&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
8 \\
{16} \\
{16}
\end{array}} \right)\) M1A1
normal \({\boldsymbol{n}} = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) so \({\boldsymbol{r}} \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
1
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) (M1)
\(x + 2y + 2z = 7\) A1
Note: If attempt to solve by a system of equations:
Award A1 for 3 correct equations, A1 for eliminating a variable and A2 for the correct answer.
[4 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]
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]
c.Find the Cartesian equation of the plane containing the triangle ABC.[3]
d.Find a vector equation of (AB).[2]
e.The point D on (AB) is such that \(\overrightarrow {{\text{OD}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) where O is the origin.
(i) Find the coordinates of D.
(ii) Show that D does not lie between A and B.[5]
▶️Answer/Explanation
Markscheme
(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
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]
Find the coordinates of M, the mid-point of [OB].[1]
Show that an equation of the plane \({\mathit{\Pi }}\), containing the square OABC, is \(y + \sqrt 2 z = 0\).[3]
Find a vector equation of the line \(L\), through M, perpendicular to the plane \({\mathit{\Pi }}\).[3]
Find the coordinates of D, the point of intersection of the line \(L\) with the plane whose equation is \(y = 0\).[3]
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
\(\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6\) (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]
Examiners report
[N/A]
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Question
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°.
a.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 \({\Pi _1}\).[4]
(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
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 vectors a \( = \) i \( – {\text{ }}3\)j \( – {\text{ }}2\)k, b \( = – {\text{ }}3\)j \( + {\text{ }}2\)k.
a.Find a \( \times \) b.[2]
b.Hence find the Cartesian equation of the plane containing the vectors a and b, and passing through the point \((1,{\text{ }}0,{\text{ }} – 1)\).[3]
▶️Answer/Explanation
Markscheme
a \( \times \) b \( = – 12\)i \( – {\text{ }}2\)j \( – {\text{ }}3\)k (M1)A1
[2 marks]
METHOD 1
\( – 12x – 2y – 3z = d\) M1
\( – 12 \times 1 – 2 \times 0 – 3( – 1) = d\) (M1)
\( \Rightarrow d = – 9\) A1
\( – 12x – 2y – 3z = – 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)\)
METHOD 2
\(\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 12} \\ { – 2} \\ { – 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ { – 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 12} \\ { – 2} \\ { – 3} \end{array}} \right)\) M1A1
\( – 12x – 2y – 3z = – 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)\) A1
[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 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
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]