Question
A ray of light coming from the point (−1, 3, 2) is travelling in the direction of vector and meets the plane (π : x + 3y + 2z -24= ) .
Find the angle that the ray of light makes with the plane.
Answer/Explanation
Markscheme
The normal vector to the plane is \(\left( {\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}} \right)\) . (A1)
EITHER
\(\theta \) is the angle between the line and the normal to the plane.
\(\cos \theta = \frac{{\left( {\begin{array}{*{20}{c}}
4 \\
1 \\
{ – 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}} \right)}}{{\sqrt {14} \sqrt {21} }} = \frac{3}{{\sqrt {14} \sqrt {21} }} = \left( {\frac{3}{{7\sqrt 6 }}} \right)\) (M1)A1A1
\( \Rightarrow \theta = 79.9^\circ {\text{ }}( = 1.394…)\) A1
The required angle is 10.1° (= 0.176) A1
OR
\(\phi \) is the angle between the line and the plane.
\(\sin \phi = \frac{{\left( {\begin{array}{*{20}{c}}
4 \\
1 \\
{ – 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}} \right)}}{{\sqrt {14} \sqrt {21} }} = \frac{3}{{\sqrt {14} \sqrt {21} }}\) (M1)A1A1
\(\phi \) = 10.1° (= 0.176) A2
[6 marks]
Question
A plane \(\pi \) has vector equation r = (−2i + 3j − 2k) + \(\lambda \)(2i + 3j + 2k) + \(\mu \)(6i − 3j + 2k).
(a) Show that the Cartesian equation of the plane \(\pi \) is 3x + 2y − 6z = 12.
(b) The plane \(\pi \) meets the x, y and z axes at A, B and C respectively. Find the coordinates of A, B and C.
(c) Find the volume of the pyramid OABC.
(d) Find the angle between the plane \(\pi \) and the x-axis.
(e) Hence, or otherwise, find the distance from the origin to the plane \(\pi \).
(f) Using your answers from (c) and (e), find the area of the triangle ABC.
Answer/Explanation
Markscheme
(a) EITHER
normal to plane given by
\(\left| {\begin{array}{*{20}{c}}
i&j&k \\
2&3&2 \\
6&{ – 3}&2
\end{array}} \right|\) M1A1
= 12i + 8j – 24k A1
equation of \(\pi \) is \(3x + 2y – 6z = d\) (M1)
as goes through (–2, 3, –2) so d = 12 M1A1
\(\pi :3x + 2y – 6z = 12\) AG
OR
\(x = – 2 + 2\lambda + 6\mu \)
\(y = 3 + 3\lambda – 3\mu \)
\(z = – 2 + 2\lambda + 2\mu \)
eliminating \(\mu \)
\(x + 2y = 4 + 8\lambda \)
\(2y + 3z = 12\lambda \) M1A1A1
eliminating \(\lambda \)
\(3(x + 2y) – 2(2y + 3z) = 12\) M1A1A1
\(\pi :3x + 2y – 6z = 12\) AG
[6 marks]
(b) therefore A(4, 0, 0), B(0, 6, 0) and C(0, 0, 2) A1A1A1
Note: Award A1A1A0 if position vectors given instead of coordinates.
[3 marks]
(c) area of base \({\text{OAB}} = \frac{1}{2} \times 4 \times 6 = 12\) M1
\(V = \frac{1}{3} \times 12 \times 2 = 8\) M1A1
[3 marks]
(d) \(\left( {\begin{array}{*{20}{c}}
3 \\
2 \\
{ – 6}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
0
\end{array}} \right) = 3 = 7 \times 1 \times \cos \phi \) M1A1
\(\phi = \arccos \frac{3}{7}\)
so \(\theta = 90 – \arccos \frac{3}{7} = 25.4^\circ \,\,\,\,\,\)(accept 0.443 radians) M1A1
[4 marks]
(e) \(d = 4\sin \theta = \frac{{12}}{7}\,\,\,\,\,( = 1.71)\) (M1)A1
[2 marks]
(f) \(8 = \frac{1}{3} \times \frac{{12}}{7} \times {\text{area}} \Rightarrow {\text{area}} = 14\) M1A1
Note: If answer to part (f) is found in an earlier part, award M1A1, regardless of the fact that it has not come from their answers to part (c) and part (e).
[2 marks]
Total [20 marks]
Examiners report
The question was generally well answered, although there were many students who failed to recognise that the volume was most logically found using a base as one of the coordinate planes.
Question
The points P(−1, 2, − 3), Q(−2, 1, 0), R(0, 5, 1) and S form a parallelogram, where S is diagonally opposite Q.
Find the coordinates of S.
The vector product \(\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
{ – 13} \\
7 \\
m
\end{array}} \right)\). Find the value of m .
Hence calculate the area of parallelogram PQRS.
Find the Cartesian equation of the plane, \({\prod _1}\) , containing the parallelogram PQRS.
Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane \({\prod _1}\) .
Hence find the point on the plane that is closest to the origin.
A second plane, \({\prod _2}\) , has equation x − 2y + z = 3. Calculate the angle between the two planes.
Answer/Explanation
Markscheme
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 1} \\
3
\end{array}} \right)\) , \(\overrightarrow {{\text{SR}}} = \left( {\begin{array}{*{20}{c}}
{0 – x} \\
{5 – y} \\
{1 – z}
\end{array}} \right)\) (M1)
point S = (1, 6, −2) A1
[2 marks]
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 1} \\
3
\end{array}} \right)\)\(\overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
2 \\
4 \\
1
\end{array}} \right)\) A1
\(\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
{ – 13} \\
7 \\
{ – 2}
\end{array}} \right)\)
m = −2 A1
[2 marks]
area of parallelogram PQRS \( = \left| {\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} } \right| = \sqrt {{{( – 13)}^2} + {7^2} + {{( – 2)}^2}} \) M1
\( = \sqrt {222} = 14.9\) A1
[2 marks]
equation of plane is −13x + 7y − 2z = d M1A1
substituting any of the points given gives d = 33
−13x + 7y − 2z = 33 A1
[3 marks]
equation of line is \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
0 \\
0 \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
{ – 13} \\
7 \\
{ – 2}
\end{array}} \right)\) A1
Note: To get the A1 must have \(\boldsymbol{r} =\) or equivalent.
[1 mark]
\(169\lambda + 49\lambda + 4\lambda = 33\) M1
\(\lambda = \frac{{33}}{{222}}{\text{ }}( = 0.149…)\) A1
closest point is \(\left( { – \frac{{143}}{{74}},\frac{{77}}{{74}}, – \frac{{11}}{{37}}} \right){\text{ }}\left( { = ( – 1.93,{\text{ 1.04, – 0.297)}}} \right)\) A1
[3 marks]
angle between planes is the same as the angle between the normals (R1)
\(\cos \theta = \frac{{ – 13 \times 1 + 7 \times – 2 – 2 \times 1}}{{\sqrt {222} \times \sqrt 6 }}\) M1A1
\(\theta = 143^\circ \) (accept \(\theta = 37.4^\circ \) or 2.49 radians or 0.652 radians) A1
[4 marks]
Examiners report
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = …
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = …
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = …
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = …
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form \(\boldsymbol{r} = \) …
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = …
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = …
Question
The planes \(2x + 3y – z = 5\) and \(x – y + 2z = k\) intersect in the line \(5x + 1 = 9 – 5y = – 5z\) .
Find the value of k .
Answer/Explanation
Markscheme
point on line is \(x = \frac{{ – 1 – 5\lambda }}{5}{\text{, }}y = \frac{{9 + 5\lambda }}{5}{\text{, }}z = \lambda \) or similar M1A1
Note: Accept use of point on the line or elimination of one of the variables using the equations of the planes
\(\frac{{ – 1 – 5\lambda }}{5} – \frac{{9 + 5\lambda }}{5} + 2\lambda = k\) M1A1
Note: Award M1A1 if coordinates of point and equation of a plane is used to obtain linear equation in k or equations of the line are used in combination with equation obtained by elimination to get linear equation in k.
\(k = – 2\) A1
[5 marks]
Examiners report
Many different attempts were seen, sometimes with success. Unfortunately many candidates wasted time with aimless substitutions showing little understanding of the problem.
Question
Find the values of k for which the following system of equations has no solutions and the value of k for the system to have an infinite number of solutions.
\[x – 3y + z = 3\]
\[x + 5y – 2z = 1\]
\[16y – 6z = k\]
Given that the system of equations can be solved, find the solutions in the form of a vector equation of a line, r = a + λb , where the components of b are integers.
The plane \( \div \) is parallel to both the line in part (b) and the line \(\frac{{x – 4}}{3} = \frac{{y – 6}}{{ – 2}} = \frac{{z – 2}}{0}\).
Given that \( \div \) contains the point (1, 2, 0) , show that the Cartesian equation of ÷ is 16x + 24y − 11z = 64 .
The z-axis meets the plane \( \div \) at the point P. Find the coordinates of P.
Find the angle between the line \(\frac{{x – 2}}{3} = \frac{{y + 5}}{4} = \frac{z}{2}\) and the plane \( \div \) .
Answer/Explanation
Markscheme
in augmented matrix form \(\left| {\begin{array}{*{20}{c}}
1&{ – 3}&1&3 \\
1&5&{ – 2}&1 \\
0&{16}&{ – 6}&k
\end{array}} \right|\)
attempt to find a line of zeros (M1)
\({r_2} – {r_1}\left| {\begin{array}{*{20}{c}}
1&{ – 3}&1&3 \\
0&8&{ – 3}&{ – 2} \\
0&{16}&{ – 6}&k
\end{array}} \right|\) (A1)
\({r_3} – 2{r_2}\left| {\begin{array}{*{20}{c}}
1&{ – 3}&1&3 \\
0&8&{ – 3}&{ – 2} \\
0&{0}&{0}&{k + 4}
\end{array}} \right|\) (A1)
there is an infinite number of solutions when \(k = – 4\) R1
there is no solution when
\(k \ne – 4,{\text{ }}(k \in \mathbb{R})\) R1
Note: Approaches other than using the augmented matrix are acceptable.
[5 marks]
using \(\left| {\begin{array}{*{20}{c}}
1&{ – 3}&1&3 \\
0&8&{ – 3}&{ – 2} \\
0&{0}&{ 0}&{k + 4}
\end{array}} \right|\) and letting \(\boldsymbol{z} = \lambda \) (M1)
\(8y – 3\lambda = – 2\)
\( \Rightarrow y = \frac{{3\lambda – 2}}{8}\) (A1)
\(x – 3y + z = 3\)
\( \Rightarrow x – \left( {\frac{{9\lambda – 6}}{8}} \right) + \lambda = 3\) (M1)
\( \Rightarrow 8x – 9\lambda + 6 + 8\lambda = 24\)
\( \Rightarrow x = \frac{{18 + \lambda }}{8}\) (A1)
\( \Rightarrow \left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\frac{{18}}{8}} \\
{ – \frac{2}{8}} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
{\frac{1}{8}} \\
{\frac{3}{8}} \\
1
\end{array}} \right)\) (M1)(A1)
\(r = \left( {\begin{array}{*{20}{c}}
{\frac{9}{4}} \\
{ – \frac{1}{4}} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
1 \\
3 \\
8
\end{array}} \right)\) A1
Note: Accept equivalent answers.
[7 marks]
recognition that \(\left( {\begin{array}{*{20}{c}}
3 \\
{ – 2} \\
0
\end{array}} \right)\) is parallel to the plane (A1)
direction normal of the plane is given by \(\left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
1&3&8 \\
3&{ – 2}&0
\end{array}} \right|\) (M1)
= 16i + 24j – 11k A1
Cartesian equation of the plane is given by 16x + 24y –11z = d and a point which fits this equation is (1, 2, 0) (M1)
\( \Rightarrow 16 + 48 = d\)
d = 64 A1
hence Cartesian equation of plane is 16x + 24y –11z = 64 AG
Note: Accept alternative methods using dot product.
[5 marks]
the plane crosses the z-axis when x = y = 0 (M1)
coordinates of P are \(\left( {0,\,0,\, – \frac{{64}}{{11}}} \right)\) A1
Note: Award A1 for stating \(z = – \frac{{64}}{{11}}\).
Note: Accept. \(\left( {\begin{array}{*{20}{c}}
0 \\
0 \\
{ – \frac{{64}}{{11}}}
\end{array}} \right)\)
[2 marks]
recognition that the angle between the line and the direction normal is given by:
\(\left( {\begin{array}{*{20}{c}}
3 \\
4 \\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{16} \\
{24} \\
{-11}
\end{array}} \right) = \sqrt {29} \sqrt {953} \cos \theta \) where \(\theta \) is the angle between the line and the normal vector M1A1
\( \Rightarrow 122 = \sqrt {29} \sqrt {953} \cos \theta \) (A1)
\( \Rightarrow \theta = 42.8^\circ {\text{ (0.747 radians)}}\) (A1)
hence the angle between the line and the plane is 90° – 42.8° = 47.2° (0.824 radians) A1
[5 marks]
Note: Accept use of the formula a.b = \(\left| {} \right.\)a\(\left. {} \right|\)\(\left| {} \right.\)b\(\left| {\sin \theta } \right.\) .
Examiners report
Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.
Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.
Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.
Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.
Many candidates were able to start this question, but only a few candidates gained full marks. Many candidates successfully used the augmented matrix in part (a) to find the correct answer. Part (b) was less successful with only a limited number of candidates using the calculator to its full effect here and with many candidates making arithmetic and algebraic errors. This was the hardest part of the question. Many candidates understood the vector techniques necessary to answer parts (c), (d) and (e) but a number made arithmetic and algebraic errors in the working.
Question
Consider the planes \({\pi _1}:x – 2y – 3z = 2{\text{ and }}{\pi _2}:2x – y – z = k\) .
Find the angle between the planes \({\pi _1}\)and \({\pi _2}\) .
The planes \({\pi _1}\) and \({\pi _2}\) intersect in the line \({L_1}\) . Show that the vector equation of
\({L_1}\) is \(r = \left( {\begin{array}{*{20}{c}}
0\\
{2 – 3k}\\
{2k – 2}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right)\)
The line \({L_2}\) has Cartesian equation \(5 – x = y + 3 = 2 – 2z\) . The lines \({L_1}\) and \({L_2}\) intersect at a point X. Find the coordinates of X.
Determine a Cartesian equation of the plane \({\pi _3}\) containing both lines \({L_1}\) and \({L_2}\) .
Let Y be a point on \({L_1}\) and Z be a point on \({L_2}\) such that XY is perpendicular to YZ and the area of the triangle XYZ is 3. Find the perimeter of the triangle XYZ.
Answer/Explanation
Markscheme
Note: Accept alternative notation for vectors (eg \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).
\(\boldsymbol{n} = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 2} \\
{ – 3}
\end{array}} \right)\) and \(\boldsymbol{m} = \left( {\begin{array}{*{20}{c}}
2 \\
{ – 1} \\
{ – 1}
\end{array}} \right)\) (A1)
\(\cos \theta = \frac{{\boldsymbol{n} \cdot \boldsymbol{m}}}{{\left| \boldsymbol{n} \right|\left| \boldsymbol{m} \right|}}\) (M1)
\(\cos \theta = \frac{{2 + 2 + 3}}{{\sqrt {1 + 4 + 9} \sqrt {4 + 1 + 1} }} = \frac{7}{{\sqrt {14} \sqrt 6 }}\) A1
\(\theta = 40.2^\circ \,\,\,\,\,(0.702{\text{ rad}})\) A1
[4 marks]
Note: Accept alternative notation for vectors (eg \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).
METHOD 1
eliminate z from x – 2y – 3z = 2 and 2x – y – z = k
\(5x – y = 3k – 2 \Rightarrow x = \frac{{y – (2 – 3k)}}{5}\) M1A1
eliminate y from x – 2y – 3z = 2 and 2x – y – z = k
\(3x + z = 2k – 2 \Rightarrow x = \frac{{z – (2k – 2)}}{{ – 3}}\) A1
x = t, y = (2 − 3k) + 5t and z = (2k − 2) − 3t A1A1
\(r = \left( {\begin{array}{*{20}{c}}
0\\
{2 – 3k}\\
{2k – 2}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right)\) AG
[5 marks]
METHOD 2
\(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 3}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
{ – 1}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
{ – 5}\\
3
\end{array}} \right) \Rightarrow {\text{direction is }}\left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right)\) M1A1
Let x = 0
\(0 – 2y – 3z = 2{\text{ and }}2 \times 0 – y – z = k\) (M1)
solve simultaneously (M1)
\(y = 2 – 3k{\text{ and }}z = 2k – 2\) A1
therefore r \( = \left( {\begin{array}{*{20}{c}}
0\\
{2 – 3k}\\
{2k – 2}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right)\) AG
[5 marks]
METHOD 3
substitute \(x = t,{\text{ }}y = (2 – 3k) + 5t{\text{ and }}z = (2k – 2) – 3t{\text{ into }}{\pi _1}{\text{ and }}{\pi _2}\) M1
for \({\pi _1}:t – 2(2 – 3k + 5t) – 3(2k – 2 – 3t) = 2\) A1
for \({\pi _2}:2t – (2 – 3k + 5t) – (2k – 2 – 3t) = k\) A1
the planes have a unique line of intersection R2
therefore the line is \(r = \left( {\begin{array}{*{20}{c}}
0\\
{2 – 3k}\\
{2k – 2}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right)\) AG
[5 marks]
Note: Accept alternative notation for vectors (eg \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).
\(5 – t = (2 – 3k + 5t) + 3 = 2 – 2(2k – 2 – 3t)\) M1A1
Note: Award M1A1 if candidates use vector or parametric equations of \({L_2}\)
eg \(\left( {\begin{array}{*{20}{c}}
0\\
{2 – 3k}\\
{2k – 2}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
5\\
{ – 3}\\
1
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
{ – 2}\\
2\\
{ – 1}
\end{array}} \right)\) or \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{t = 5 – 2s}\\
{2 – 3k + 5t = – 3 + 2s}\\
{2k – 2 – 3t = 1 + s}
\end{array}} \right.\)
solve simultaneously M1
\(k = 2,{\text{ }}t = 1{\text{ }}(s = 2)\) A1
intersection point (\(1\), \(1\), \( – 1\)) A1
[5 marks]
Note: Accept alternative notation for vectors (eg \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).
\({\overrightarrow l _2} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 2}\\
1
\end{array}} \right)\) A1
\({\overrightarrow l _1} \times {\overrightarrow l _2} = \left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\
1&5&{ – 3}\\
2&{ – 2}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
{ – 7}\\
{ – 12}
\end{array}} \right)\) (M1)A1
\(\boldsymbol{r} \cdot \left( {\begin{array}{*{20}{c}}
1\\
7\\
{12}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
1\\
{ – 1}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1\\
7\\
{12}
\end{array}} \right)\) (M1)
\(x + 7y + 12z = – 4\) A1
[5 marks]
Note: Accept alternative notation for vectors (eg \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).
Let \(\theta \) be the angle between the lines \({\overrightarrow l _1} = \left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right)\) and \({\overrightarrow l _2} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 2}\\
1
\end{array}} \right)\)
\(\cos \theta = \frac{{\left| {2 – 10 – 3} \right|}}{{\sqrt {35} \sqrt 9 }} \Rightarrow \theta = 0.902334…{\text{ }}51.699…^\circ )\) (M1)
as the triangle XYZ has a right angle at Y,
\({\text{XZ}} = a \Rightarrow {\text{YZ}} = a\sin \theta {\text{ and XY}} = a\cos \theta \) (M1)
\({\text{area = 3}} \Rightarrow \frac{{{a^2}\sin \theta \cos \theta }}{2} = 3\) (M1)
\(a = 3.5122…\) (A1)
perimeter \( = a + a\sin \theta + a\cos \theta = 8.44537… = 8.45\) A1
Note: If candidates attempt to find coordinates of Y and Z award M1 for expression of vector YZ in terms of two parameters, M1 for attempt to use perpendicular condition to determine relation between parameters, M1 for attempt to use the area to find the parameters and A2 for final answer.
[5 marks]
Examiners report
Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of k. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.
Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.
Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of k. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.
Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.
Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of k. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.
Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.
Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of k. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.
Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.
Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of k. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.
Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.
Question
Consider the two planes
\({\pi _1}:4x + 2y – z = 8\)
\({\pi _2}:x + 3y + 3z = 3\).
Find the angle between \({\pi _1}\) and \({\pi _2}\), giving your answer correct to the nearest degree.
Answer/Explanation
Markscheme
\({{{n}}_1} = \left( {\begin{array}{*{20}{c}} 4 \\ 2 \\ { – 1} \end{array}} \right)\;\;\;{\text{and}}\;\;\;{{{n}}_2} = \left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 3 \end{array}} \right)\) (A1)(A1)
use of \(\cos \theta = \frac{{{{{n}}_1} \bullet {{{n}}_2}}}{{\left| {{{{n}}_1}} \right|\left| {{{{n}}_2}} \right|}}\) (M1)
\(\cos \theta = \frac{7}{{\sqrt {21} \sqrt {19} }}\;\;\;\left( { = \frac{7}{{\sqrt {399} }}} \right)\) (A1)(A1)
Note: Award A1 for a correct numerator and A1 for a correct denominator.
\(\theta = 69^\circ \) A1
Note: Award A1 for 111°.
[6 marks]
Examiners report
Reasonably well answered. A large number of candidates did not express their final answer correct to the nearest degree.