IBDP Maths analysis and approaches Topic: AHL 3.16 vector product of two vectors: IB Style Questions HL Paper 1

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]

b.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\(|\). [8]

 
▶️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]

a.

(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]

b.

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

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}}
  { – 4} \\
  { – 1} \\
  3
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}}
  4 \\
  { – 3} \\
  1
\end{array}} \right)\)     A1A1

Note: Accept row vectors.

 

[2 marks]

a.

\(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left| {\begin{array}{*{20}{c}}
  {\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
  { – 4}&{ – 1}&3 \\
  4&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
  8 \\
  {16} \\
  {16}
\end{array}} \right)\)     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]

b.

Examiners report

Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.

a.

Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.

b.

Question

For non-zero vectors \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\), show that

(i)     if \(\left| {{\boldsymbol{a}} – {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\), then \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) are perpendicular;

(ii)     \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {({\boldsymbol{a}} \cdot {\boldsymbol{b}})^2}\).

[8]
a.

The points A, B and C have position vectors \({\boldsymbol{a}}\), \({\boldsymbol{b}}\) and \({\boldsymbol{c}}\).

(i)     Show that the area of triangle ABC is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\).

(ii)     Hence, show that the shortest distance from B to AC is

\[\frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} – {\boldsymbol{a}}} \right|}}{\text{.}}\]

[7]
b.
▶️Answer/Explanation

Markscheme

(i)     \(\left| {{\boldsymbol{a}} – {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\)

\( \Rightarrow \left( {{\boldsymbol{a}} – {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} – {\boldsymbol{b}}} \right) = \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right)\)     (M1)

\( \Rightarrow {\left| {\boldsymbol{a}} \right|^2} – 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2} + 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2}\)     A1

\( \Rightarrow 4{\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0 \Rightarrow {\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0\)     A1

therefore \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) are perpendicular     R1

Note: Allow use of 2-d components.

 

Note: Do not condone sloppy vector notation, so we must see something to the effect that \({\left| {\boldsymbol{c}} \right|^2} = {\boldsymbol{c}} \cdot {\boldsymbol{c}}\) is clearly being used for the M1.

 

Note: Allow a correct geometric argument, for example that the diagonals of a parallelogram have the same length only if it is a rectangle.

 

(ii)     \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left( {\left| {\boldsymbol{a}} \right|\left| {\boldsymbol{b}} \right|\sin \theta } \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\sin ^2}\theta \)     M1A1

\({\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\cos ^2}\theta \)     M1

\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {1 – {{\cos }^2}\theta } \right)\)     A1
\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {{{\sin }^2}\theta } \right)\)

\( \Rightarrow {\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2}\)     AG

[8 marks]

a.

(i)     area of triangle \( = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|\)     (M1)

\( = \frac{1}{2}\left| {\left( {{\boldsymbol{b}} – {\boldsymbol{a}}} \right) \times \left( {{\boldsymbol{c}} – {\boldsymbol{a}}} \right)} \right|\)     A1

\( = \frac{1}{2}\left| {{\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{b}} \times {\boldsymbol{ – a}} + {\boldsymbol{ – a}} \times {\boldsymbol{c}} + {\boldsymbol{ – a}} \times {\boldsymbol{ – a}}} \right|\)     A1

\({\boldsymbol{b}} \times {\boldsymbol{ – a}} = {\boldsymbol{a}} \times {\boldsymbol{b}}\); \({\boldsymbol{c}} \times {\boldsymbol{a}} = {\boldsymbol{ – a}} \times {\boldsymbol{c}}\); \({\boldsymbol{ – a}} \times {\boldsymbol{ – a}} = 0\)     M1

hence, area of triangle is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\)     AG

 

(ii)     D is the foot of the perpendicular from B to AC

area of triangle \({\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right|\)     A1

therefore

\(\frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|\)     M1

hence, \(\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{{\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|}}{{\left| {\overrightarrow {{\text{AC}}} } \right|}}\)     A1

\( = \frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} – {\boldsymbol{a}}} \right|}}\)     AG

[7 marks]

b.

Examiners report

(i) The majority of candidates were very sloppy in their use of vector notation. Some candidates used Cartesian coordinates, which was acceptable. Part (ii) was well done.

a.

Part (i) was usually well started, but not completed satisfactorily. Many candidates understood the geometry involved in this part.

b.

Question

Consider the points A(1, 2, 3), B(1, 0, 5) and C(2, −1, 4).

Find \(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} \).

[4]
a.

Hence find the area of the triangle ABC.

[2]
b.
▶️Answer/Explanation

Markscheme

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  5
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0 \\
  { – 2} \\
  2
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}}
  2 \\
  { – 1} \\
  4
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  { – 3} \\
  1
\end{array}} \right)\)     A1A1

Note: Award the above marks if the components are seen in the line below.

 

\(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  0&{ – 2}&2 \\
  1&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
  4 \\
  2 \\
  2
\end{array}} \right)\)     (M1)A1

[4 marks]

a.

area \( = \frac{1}{2}\left| {\left( {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right)} \right|\)     (M1)

\( = \frac{1}{2}\sqrt {{4^2} + {2^2} + {2^2}}  = \frac{1}{2}\sqrt {24} {\text{ }}\left( { = \sqrt 6 } \right)\)     A1

Note: Award M0A0 for attempts that do not involve the answer to (a).

[2 marks]

b.

Examiners report

Candidates showed a good understanding of the vector techniques required in this question.

a.

Candidates showed a good understanding of the vector techniques required in this question.

b.

Question

The vertices of a triangle ABC have coordinates given by A(−1, 2, 3), B(4, 1, 1) and C(3, −2, 2).

(i)     Find the lengths of the sides of the triangle.

(ii)     Find \(\cos {\rm{B\hat AC}}\).

[6]
a.

(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]
b.

Find the Cartesian equation of the plane containing the triangle ABC.

[3]
c.

Find a vector equation of (AB).

[2]
d.

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]
e.
▶️Answer/Explanation

Markscheme

(i)     \(\overrightarrow {{\text{AB}}}  = \overrightarrow {{\text{OB}}}  – \overrightarrow {{\text{OA}}}  = \) 5ij – 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.

 

\(\overrightarrow {{\text{AB}}}  = \) |5ij – 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]

a.

\(\overrightarrow {{\text{BC}}}  \times \overrightarrow {{\text{CA}}}  = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  { – 1}&{ – 3}&1 \\
  { – 4}&4&1
\end{array}} \right|\)     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]

b.

attempt at the use of “(ra)\( \cdot \)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]

c.

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]

d.

(i)     \(\overrightarrow {{\text{OD}}}  = \) (–i + 2j + 3k) + t(5ij – 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]

e.

Examiners report

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3×3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

a.

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3×3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

b.

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3×3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

c.

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3×3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

d.

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3×3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

e.

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 \(16x + \alpha y – 3z = \beta \).

Find the vector \(\overrightarrow {{\text{CA}}}  \times \overrightarrow {{\text{CB}}} \).

[4]
a.

Find an exact value for the area of the triangle ABC.

[3]
b.

Show that the Cartesian equation of the plane \({\Pi _1}\), containing the triangle ABC, is \(2x + 3y – 4z = 2\).

[3]
c.

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]
d.

Find the value of \(\alpha \) if all three planes contain \({L_1}\).

[3]
e.

Find conditions on \(\alpha \) and \(\beta \) if the plane \({\Pi _3}\) does not intersect with \({L_1}\).

[2]
f.
▶️Answer/Explanation

Markscheme

\(\overrightarrow {{\rm{CA}}}  = \left( \begin{array}{c}1\\ – 2\\ – 1\end{array} \right)\)     (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]

a.

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]

b.

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]

c.

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]

d.

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]

e.

\(\alpha  =  – 5\)     A1

\(\beta  \ne 16\)     A1

[2 marks]

f.

Examiners report

Part a) proved an easy start, though a few (weaker) candidates still believe \(\overrightarrow {{\text{CA}}} \) to be \(\overrightarrow {{\text{OC}}}  – \overrightarrow {{\text{OA}}} \).

a.

Part b) was an easy 3 marks and incorrect answers were rare.

b.

Part c) was answered well, though reasoning sometimes seemed sparse, especially given that this was a ‘show that’ question.

c.

Part d) proved more challenging, despite being a very standard question. Many candidates gained only 2 marks, either through correctly calculating the direction vector, or by successfully eliminating one of the variables. A number of clear fully correct solutions were seen, though the absence of ‘r  =’ is still prevalent, and candidates might be reminded of the correct form for the vector equation of a line.

d.

Part e) proved a puzzle for most, though an attempt to use row reduction on an augmented matrix seemed to be the choice way for most successful candidates.

e.

Only the very best were able to demonstrate a complete understanding of intersecting planes and thus answer part f) correctly.

f.

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 \(\overrightarrow {{\text{OC}}}  = {{c}}\).

Find \(\overrightarrow {{\text{BR}}} \) in terms of \({{a}}\), \({{b}}\) and \({{c}}\).

[2]
a.

(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]
b.

Show that the line segment [\(CP\)] also includes the point \(G\).

[3]
c.

The coordinates of the points \(A\)\(B\) and \(C\) are \((1,{\text{ }}3,{\text{ }}1)\), \((3,{\text{ }}7,{\text{ }} – 5)\) and \((2,{\text{ }}2,{\text{ }}1)\) respectively.

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]
d.
▶️Answer/Explanation

Markscheme

\(\overrightarrow {{\text{BR}}}  = \overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{AR}}} \;\;\;\left( { = \overrightarrow {{\text{BA}}}  + \frac{1}{2}\overrightarrow {{\text{AC}}} } \right)\)     (M1)

\( = ({{a}} – {{b}}) + \frac{1}{2}({{c}} – {{a}})\)

\( = \frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}\)     A1

[2 marks]

a.

(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]

b.

\(\overrightarrow {{\text{CP}}}  = \frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}\)     (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]

c.

\(\overrightarrow {{\text{OG}}}  = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 7 \\ { – 5} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right)\)     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]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

Question

Consider the vectors a \( = \) i \( – {\text{ }}3\)j \( – {\text{ }}2\)k, b \( =  – {\text{ }}3\)j \( + {\text{ }}2\)k.

Find a \( \times \) b.

[2]
a.

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]
b.
▶️Answer/Explanation

Markscheme

a \( \times \) b \( =  – 12\)i \( – {\text{ }}2\)j \( – {\text{ }}3\)k     (M1)A1

[2 marks]

a.

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]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

The vectors a , b , c satisfy the equation a + b + c = 0 . Show that a \( \times \) b = b \( \times \) c = c \( \times \) a .

▶️Answer/Explanation

Markscheme

taking cross products with a,     M1

a \( \times \) (a + b + c) = a \( \times \) 0 = 0     A1

using the algebraic properties of vectors and the fact that a \( \times \) a = 0 ,     M1

a \( \times \) b + a \( \times \) c = 0     A1

a \( \times \) b = c \( \times \) a     AG

taking cross products with b,     M1

b \( \times \) (a + b + c) = 0

b \( \times \) a + b \( \times \) c = 0     A1

a \( \times \) b = b \( \times \) c     AG

this completes the proof

[6 marks]

Examiners report

[N/A]

Question

Consider the vectors \(a= \begin{pmatrix} 2\\ 0\\ 3 \end{pmatrix}\) , \(b= \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}\) , \(c= \begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) .

  1. Find \(a × b\) and \(b × a\)
  2. Show that \(a × b\) is perpendicular to both \(a\) and \(b\)
  3. Find \((a . b)c\)
  4. Find \((a × b). c\) and \(a .(b × c)\)
▶️Answer/Explanation

Ans:

  1. \(a × b\) = \(c= \begin{pmatrix} -6\\ -3\\ -4\end{pmatrix}\) and \(b × a\) = \( \begin{pmatrix} 6\\ 3\\ -4\end{pmatrix}\)
  2. \((a × b)⋅ a\) = −12 + 0 +12 = 0, \( (a × b)⋅ b \) = −6 − 6 +12 = 0
  3. \((a ⋅ b)c\) = 11 \(\begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) = \(\begin{pmatrix} 44\\ 11\\ 0\end{pmatrix}\)
  4. \((a × b)⋅ c\) = \(\begin{pmatrix} -6\\ -3\\ -4\end{pmatrix}.\)\(\begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) = −24 − 3 = −27 , \(a ⋅(b × c)\) = \(\begin{pmatrix} 2\\ 0\\ 3\end{pmatrix}.\)\(\begin{pmatrix} -3\\ 12\\ -7\end{pmatrix}\)= -6-21 = -27

Question

Consider the vectors \(a= \begin{pmatrix} 2\\ 0\\ 3 \end{pmatrix}\) , \(b= \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}\) , \(c= \begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) .

  1. Find \(a × b\) and \(b × a\)
  2. Show that \(a × b\) is perpendicular to both \(a\) and \(b\)
  3. Find \((a . b)c\)
  4. Find \((a × b). c\) and \(a .(b × c)\)
▶️Answer/Explanation

Ans:

  1. \(a × b\) = \(c= \begin{pmatrix} -6\\ -3\\ -4\end{pmatrix}\) and \(b × a\) = \( \begin{pmatrix} 6\\ 3\\ -4\end{pmatrix}\)
  2. \((a × b)⋅ a\) = −12 + 0 +12 = 0, \( (a × b)⋅ b \) = −6 − 6 +12 = 0
  3. \((a ⋅ b)c\) = 11 \(\begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) = \(\begin{pmatrix} 44\\ 11\\ 0\end{pmatrix}\)
  4. \((a × b)⋅ c\) = \(\begin{pmatrix} -6\\ -3\\ -4\end{pmatrix}.\)\(\begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) = −24 − 3 = −27 , \(a ⋅(b × c)\) = \(\begin{pmatrix} 2\\ 0\\ 3\end{pmatrix}.\)\(\begin{pmatrix} -3\\ 12\\ -7\end{pmatrix}\)= -6-21 = -27

Question

Consider the vectors \(a= \begin{pmatrix} 2\\ 0\\ 3 \end{pmatrix}\) , \(b= \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}\) , \(c= \begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) .

  1. Find \(a × b\) and \(b × a\)
  2. Show that \(a × b\) is perpendicular to both \(a\) and \(b\)
  3. Find \((a . b)c\)
  4. Find \((a × b). c\) and \(a .(b × c)\)
▶️Answer/Explanation

Ans:

  1. \(a × b\) = \(c= \begin{pmatrix} -6\\ -3\\ -4\end{pmatrix}\) and \(b × a\) = \( \begin{pmatrix} 6\\ 3\\ -4\end{pmatrix}\)
  2. \((a × b)⋅ a\) = −12 + 0 +12 = 0, \( (a × b)⋅ b \) = −6 − 6 +12 = 0
  3. \((a ⋅ b)c\) = 11 \(\begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) = \(\begin{pmatrix} 44\\ 11\\ 0\end{pmatrix}\)
  4. \((a × b)⋅ c\) = \(\begin{pmatrix} -6\\ -3\\ -4\end{pmatrix}.\)\(\begin{pmatrix} 4\\ 1\\ 0\end{pmatrix}\) = −24 − 3 = −27 , \(a ⋅(b × c)\) = \(\begin{pmatrix} 2\\ 0\\ 3\end{pmatrix}.\)\(\begin{pmatrix} -3\\ 12\\ -7\end{pmatrix}\)= -6-21 = -27

Question

Consider the points \(A(1,2,3), B(3,0,5), C(4,3,5)\) and \(D(5,7,3)\). Find

  1. \(AB × AC\)
  2. the cartesian equation of the plane \(Π\) defined by the triangle \(ABC\)
  3. the area of the triangle \(ABC\)
  4. the volume of the tetrahedron \(ABCD\), given \(\frac{1}{6}|AD .( AB × AC)|\)
  5. the distance of the point \(D\) from the plane \(Π\), by using the results \((c)\) and \((d)\).
▶️Answer/Explanation

Ans:

  1. \(AB × AC = \begin{pmatrix} 2\\ -2\\ 2\end{pmatrix} × \begin{pmatrix} 3\\ 1\\ 2\end{pmatrix} = \begin{pmatrix} -6\\ 2\\ 8\end{pmatrix}\)
  2. We can use the parallel vector \(n= \begin{pmatrix} -3\\ 1\\ 4\end{pmatrix}\). The equation is \(− 3x + y + 4z =11 \)
  3. Area of triangle \(ABC = \frac{1}{2} |AB × AC| = \frac{\sqrt{36+ 4+ 64}}{2} = \frac{\sqrt{104}}{2} = \sqrt{26}\)
  4. Volume of tetrahedron \(ABCD = \frac{1}{6} | AD.(AB × AC)| = \frac{1}{6}|\begin{pmatrix} 4\\ 5\\ 0\end{pmatrix}.\begin{pmatrix} -6\\ 2\\ 8\end{pmatrix} |=\frac{7}{3}\)
  5. \(Volume = \frac{1}{3}(base)×(height) ⇔ \frac{7}{3} = \frac{1}{3} \sqrt{26} ×(height) ⇔ height= \frac{7}{\sqrt{26}}\)

Question

Consider the lines
\(L_1:r=\begin{pmatrix} 1\\ 2\\ 3\end{pmatrix}+\lambda \begin{pmatrix} 2\\ 3\\ 1\end{pmatrix}\)  \(L_2:r=\begin{pmatrix} 1\\ 1\\ 1\end{pmatrix}+\mu \begin{pmatrix} 4\\ 7\\ 4\end{pmatrix}\)
and the planes
\(Π_1: 2x + 3y + z = 7\)    \(Π_2: 4x + 7y + 4z = 19\) Find

  1. The angle between the lines \(L_1\) and \(L_2\)
  2. The angle between the planes \(Π_1\) and \(Π_2\)
  3. The angle between the line \(L_1\) and the plane \(Π_2\)
  4. The angle between the \(y\)-axis and the plane \(Π_2\)
  5. The point of intersection of the lines \(L_1\) and \(L_2\)
  6. The point of intersection of the line \(L_1\) and plane \(Π_1\)
  7. The line of intersection of the planes \(Π_1\) and \(Π_2\)
▶️Answer/Explanation

Ans:

\(\overrightarrow{u}= \begin{pmatrix} 2\\ 3\\ 1\end{pmatrix}\)    \(\overrightarrow{v}= \begin{pmatrix} 4\\ 7\\ 4\end{pmatrix}\)
\(\overrightarrow{u}.\overrightarrow{v}=8+21+4=33\)
\(|\overrightarrow{u}|=\sqrt{14},|\overrightarrow{v}|=9\)

\((a)\) and \((b)\)    \(w>v = \frac{\overrightarrow{u}.\overrightarrow{v}}{|\overrightarrow{u}||\overrightarrow{v}|}= \frac{33}{9\sqrt{14}} ⇒ v=11.5\)° or \(0.2 rad\)

\((c)\) \(sin v = \frac{33}{9\sqrt{14}} ⇒ v=78.5\)° or \(1.37 rad\)

\((d)\) the \(y\)-axis has direction vector \(\begin{pmatrix} 0\\ 1\\ 0\end{pmatrix} sin \theta = \frac{7}{\sqrt{81}} ⇒ \theta= 51.06 \)°

\((e)\) \(\overrightarrow{r_1}=\overrightarrow{r_2} ⇒ \begin{Bmatrix}1+2\lambda=1+4\mu \\ 2+3\lambda=1+7\mu \\ 3+\lambda=1+4\mu \end{Bmatrix} ⇒ \begin{Bmatrix}2\lambda-4\mu=0 \\ 3\lambda-7\mu=-1 \\ \lambda-4\mu =-2\end{Bmatrix} ⇒ \lambda=2,\mu=1\)
They intersect at \(P(5,8,5)\).

\((f)\) \(L_1:\) \(x=1+2\lambda,y=2+3\lambda,z=3+\lambda\) lies in \(Π_1:2(1+2\lambda)+3(2+3\lambda)+(3+\lambda)=7 ⇒ 14\lambda=-4 ⇒ \lambda=-\frac{2}{7}\)
They intersect at \(Q(\frac{3}{7},\frac{8}{7},\frac{19}{7})\).

\((f)\) By Gauss elimination (Rref) \(\begin{pmatrix} 1 & 0 & -\frac{5}{2} & -4\\ 0 & 1 & 2 & 5 \end{pmatrix}\)
The line of intersection is \(r=\begin{pmatrix} -4\\ 5\\ 0\end{pmatrix}+\lambda \begin{pmatrix} \frac{5}{2}\\ -2\\ 1\end{pmatrix}\).

Question

The position vectors of points \(P\) and \(Q\) are: \(p = 3i + 2j + k\)  \(q = i + 3j – 2k\)

  1. Find the vector product \(p × q\).
  2. Using your answer to part (a), or otherwise, find the area of the parallelogram with two sides \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\)
▶️Answer/Explanation

Ans:

  1. The vector product, \(p × q = –7i + 7j + 7k\)                                                           
  2. Area of parallelogram \(= |p × q|=\sqrt{147}\) or \(7\sqrt{3}\) or 12.1 units2          

Question

  1. Find a vector perpendicular to the two vectors: \(\overrightarrow{OP}=\overrightarrow{i}-3\overrightarrow{j}+2\overrightarrow{k}\)   \(\overrightarrow{OQ}=-2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}\)
  2. If \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) are position vectors for the points \(P\) and \(Q\), use your answer to part (a), or otherwise, to find the area of the triangle \(OPQ\).
▶️Answer/Explanation

Ans:

  1. A perpendicular vector can be found from the vector product \(\overrightarrow{OP}×\overrightarrow{OQ} =\overrightarrow{i}-3\overrightarrow{j}-5\overrightarrow{k} \)   
  2. Area △\(OPQ\) = \(\frac{1}{2}|\overrightarrow{OP}||\overrightarrow{OQ}| sin \theta\) , where \(\theta\) is the angle between \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) = \(\frac{1}{2} |\overrightarrow{OP}×\overrightarrow{OQ}| = \frac{\sqrt{35}}{2}\)   

Question

Let \(a= \begin{pmatrix} 2\\ 1\\ 0\end{pmatrix}, b= \begin{pmatrix} -1\\ p\\ 6\end{pmatrix}\) and \(c=\begin{pmatrix} 2\\ -4\\ 3\end{pmatrix}\)

  1. Find \(a × b\)
  2. Find the value of \(p\), given that \(a × b\) is parallel to \(c\)
▶️Answer/Explanation

Ans:

  1. \(a × b= \begin{matrix} \begin{vmatrix} i& j & k\\ 2& 1 & 0\\ -1& p & 6 \end{vmatrix}\\ & & \\ & & \end{matrix}\)                      
    \(=6i-12j+(2p+1)k\)                       
  2. METHOD 1
    \(a × b\) parallel to \(c\) ⇒
    \(\begin{pmatrix} 6\\ -12\\ 2p+1\end{pmatrix}=k\begin{pmatrix} 2\\ -4\\ 3\end{pmatrix} \)                
    \(k=3\)                     
    \(p=4\)                 
    METHOD 2
    \(b\) is perpendicular to \(c\) ⇒
    \(\begin{pmatrix} -1\\ p\\ 6\end{pmatrix}.\begin{pmatrix} 2\\ -4\\ 3\end{pmatrix}=0\)               
    \(-2-4p+18=0\)                     
    ⇒ \(p=4\)                                            

Question

For the vectors \(a = 2i + j – 2k, b = 2i –j – k\) and \(c = i + 2j + 2k\), show that:

  1. \(a × b = –3i – 2j – 4k\)
  2. \((a × b) × c = –(b • c)a \)
▶️Answer/Explanation

Ans:

  1. \(a × b = \begin{vmatrix} i& j& k\\ 2 & 1 & -2\\ 2 & -1 & -1 \end{vmatrix} = (–1 – 2)i – (–2 + 4)j + (–2 – 2)k\)      
    \(= –3i – 2j – 4k\)                                                                                              
  2. \((a × b) × c = \begin{vmatrix} i& j& k\\ -3 & -2 & -4\\ 1 & 2 & 2 \end{vmatrix} = (–4 + 8)i – (–6 + 4)j + (–6 + 2)k \)
    \(= 4i + 2j – 4k \)                                                                                                  
    \(b . c = 2 – 2 – 2 = – 2,\)                                                                                  
    and so – \((b . c)a = 2a = 4i + 2j – 4k = (a × b) × c\)                 

Question

The acute angle between the vectors 3i − 4j − 5k and 5i − 4j + 3k is denoted by θ.

Find cos θ.

▶️Answer/Explanation

Markscheme

cos θ = \(\frac{{\left( {3i – 4j – 5k} \right) \bullet \left( {5i – 4j + 3k} \right)}}{{\left| {3i – 4j – 5k} \right|\left| {5i – 4j + 3k} \right|}}\)      (M1)

\( = \frac{{16}}{{\sqrt {50} \sqrt {50} }}\)     A1A1

Note: A1 for correct numerator and A1 for correct denominator.

\( = \frac{8}{{25}}\left( { = \frac{{16}}{{50}} = 0.32} \right)\)      A1

[4 marks]

Examiners report

[N/A]
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