Question
The vector equation of line \(l\) is given as \(\left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
3 \\
6
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
{ – 1} \\
2 \\
{ – 1}
\end{array}} \right)\) .
Find the Cartesian equation of the plane containing the line \(l\) and the point A(4, − 2, 5) .
▶️Answer/Explanation
Markscheme
EITHER
\(l\) goes through the point (1, 3, 6) , and the plane contains A(4, –2, 5)
the vector containing these two points is on the plane, i.e.
\(\left( {\begin{array}{*{20}{c}}
1 \\
3 \\
6
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
4 \\
{ – 2} \\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 3} \\
5 \\
1
\end{array}} \right)\) (M1)A1
\(\left( {\begin{array}{*{20}{c}}
{ – 1} \\
2 \\
{ – 1}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
{ – 3} \\
5 \\
1
\end{array}} \right) = \left| {\begin{array}{*{20}{c}}
{\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
{ – 1}&2&{ – 1} \\
{ – 3}&5&1
\end{array}} \right| = 7{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}}\) M1A1
\(\left( {\begin{array}{*{20}{c}}
4 \\
{ – 2} \\
5
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
7 \\
4 \\
1
\end{array}} \right) = 25\) (M1)
hence, Cartesian equation of the plane is \(7x + 4y + z = 25\) A1
OR
finding a third point M1
e.g. (0, 5, 5) A1
three points are (1, 3, 6), (4, –2, 5), (0, 5, 5)
equation is \(ax + by + cz = 1\)
system of equations M1
\(a + 3b + 6c = 1\)
\(4a – 2b + 5c = 1\)
\(5b + 5c = 1\)
\(a = \frac{7}{{25}}\) , \(b = \frac{4}{{25}}\) , \(c = \frac{1}{{25}}\) , from GDC M1A1
so \(\frac{7}{{25}}x + \frac{4}{{25}}y + \frac{1}{{25}}z = 1\) A1
or \(7x + 4y + z = 25\)
[6 marks]
Examiners report
There were many successful answers to this question, as would be expected. There seemed to be some students, however, that had not been taught the vector geometry section
Question
(a) Find the coordinates of the point \(A\) on \({l_1}\) and the point \(B\) on \({l_2}\) such that \(\overrightarrow {{\text{AB}}} \) is perpendicular to both \({l_1}\) and \({l_2}\) .
(b) Find \(\left| {{\text{AB}}} \right|\) .
(c) Find the Cartesian equation of the plane \(\prod \) which contains \({l_1}\) and does not intersect \({l_2}\) .
▶️Answer/Explanation
Markscheme
(a) on \({l_1}\) A(\( – 3 + 3\lambda \), \( – 4 + 2\lambda \), \(6 – 2\lambda \)) A1
on \({l_2}\) \({l_2}:r = \left( {\begin{array}{*{20}{c}}
4 \\
{ – 7} \\
3
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
{ – 3} \\
4 \\
{ – 1}
\end{array}} \right)\) (M1)
\( \Rightarrow \) B(\(4 – 3\mu \), \( – 7 + 4\mu \), \( – 3 – \mu \)) A1
\(\overrightarrow {{\text{BA}}} = {\boldsymbol{a}} – {\boldsymbol{b}} = \left( {\begin{array}{*{20}{c}}
{3\lambda + 3\mu – 7} \\
{2\lambda – 4\mu + 3} \\
{ – 2\lambda + \mu + 9}
\end{array}} \right)\) (M1)A1
EITHER
\({\text{BA}} \bot {l_1} \Rightarrow {\text{BA}} \cdot \left( {\begin{array}{*{20}{c}}
3 \\
2 \\
{ – 2}
\end{array}} \right) = 0 \Rightarrow 3\left( {3\lambda + 3\mu – 7} \right) + 2\left( {2\lambda – 4\mu + 3} \right) – 2\left( { – 2\lambda + \mu + 9} \right) = 0\) M1
\( \Rightarrow 17\lambda – \mu = 33\) A1
\({\text{BA}} \bot {l_2} \Rightarrow {\text{BA}} \cdot \left( {\begin{array}{*{20}{c}}
{ – 3} \\
4 \\
{ – 1}
\end{array}} \right) = 0 \Rightarrow – 3\left( {3\lambda + 3\mu – 7} \right) + 4\left( {2\lambda – 4\mu + 3} \right) – \left( { – 2\lambda + \mu + 9} \right) = 0\) M1
\( \Rightarrow \lambda – 26\mu = – 24\) A1
solving both equations above simultaneously gives
\(\lambda = 2\); \(\mu = 1 \Rightarrow \) A(3, 0, 2), B(1, –3, –4) A1A1A1A1
OR
\(\left| {\begin{array}{*{20}{c}}
{\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
3&2&{ – 2} \\
{ – 3}&4&{ – 1}
\end{array}} \right| = 6{\boldsymbol{i}} + 9{\boldsymbol{j}} + 18{\boldsymbol{k}}\) M1A1
so \(\overrightarrow {{\text{AB}}} = p\left( {\begin{array}{*{20}{c}}
2 \\
3 \\
6
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{3\lambda + 3\mu – 7} \\
{2\lambda – 4\mu + 3} \\
{ – 2\lambda + \mu + 9}
\end{array}} \right)\) M1A1
\({3\lambda + 3\mu – 2p = 7}\)
\({2\lambda – 4\mu – 3p = – 3}\)
\({ – 2\lambda + \mu – 6p = – 9}\)
\(\lambda = 2\), \(\mu = 1\), \(p = 1\) A1A1
A(\( – 3 + 6\), \( – 4 + 4\), \(6 – 4\)) \(=\) (\(3\), \(0\), \(2\)) A1
B(\(4 – 3\), \( – 7 + 4\), \( – 3 – 1\)) \(=\) (\(1\), \( – 3\), \( – 4\)) A1
[13 marks]
(b) \({\text{AB}} = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 3} \\
{ – 4}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
3 \\
0 \\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 2} \\
{ – 3} \\
{ – 6}
\end{array}} \right)\) (A1)
\(\left| {{\text{AB}}} \right| = \sqrt {{{\left( { – 2} \right)}^2} + {{\left( { – 3} \right)}^2} + {{\left( { – 6} \right)}^2}} = \sqrt {49} = 7\) M1A1
[3 marks]
(c) from (b) \(2{\boldsymbol{i}} + 3{\boldsymbol{j}} + 6{\boldsymbol{k}}\) is normal to both lines
\({l_1}\) goes through (–3, –4, 6) \( \Rightarrow \left( {\begin{array}{*{20}{c}}
{ – 3} \\
{ – 4} \\
6
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
2 \\
3 \\
6
\end{array}} \right) = 18\) M1A1
hence, the Cartesian equation of the plane through \({l_1}\) , but not \({l_2}\) , is \(2x + 3y + 6z = 18\) A1
[3 marks]
Total [19 marks]
Examiners report
There were a lot of arithmetic errors in the treatment of this question, even though it was apparent that many students did understand the methods involved. In (a) many students failed to realise that \(\overrightarrow {{\text{AB}}} \) should be a multiple of the cross product of the two direction vectors, rather than the cross product itself, and many students failed to give the final answer as coordinates.
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.
a.Find the coordinates of S.[2]
b.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 .[2]
c.Hence calculate the area of parallelogram PQRS.[2]
d.Find the Cartesian equation of the plane, \({\prod _1}\) , containing the parallelogram PQRS.[3]
e.Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane \({\prod _1}\) .[1]
f.Hence find the point on the plane that is closest to the origin.[3]
g.A second plane, \({\prod _2}\) , has equation x − 2y + z = 3. Calculate the angle between the two planes.[4]
▶️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 points P(−1, 2, − 3), Q(−2, 1, 0), R(0, 5, 1) and S form a parallelogram, where S is diagonally opposite Q.
a.Find the coordinates of S.[2]
b.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 .[2]
c.Hence calculate the area of parallelogram PQRS.[2]
d.Find the Cartesian equation of the plane, \({\prod _1}\) , containing the parallelogram PQRS.[3]
e.Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane \({\prod _1}\) .[1]
f.Hence find the point on the plane that is closest to the origin.[3]
g.A second plane, \({\prod _2}\) , has equation x − 2y + z = 3. Calculate the angle between the two planes.[4]
▶️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
Two planes \({\Pi _1}\) and \({\Pi _2}\) have equations \(2x + y + z = 1\) and \(3x + y – z = 2\) respectively.
a.Find the vector equation of L, the line of intersection of \({\Pi _1}\) and \({\Pi _2}\).[6]
b.Show that the plane \({\Pi _3}\) which is perpendicular to \({\Pi _1}\) and contains L, has equation \(x – 2z = 1\).[4]
c.The point P has coordinates (−2, 4, 1) , the point Q lies on \({\Pi _3}\) and PQ is perpendicular to \({\Pi _2}\). Find the coordinates of Q.[6]
▶️Answer/Explanation
Markscheme
(a) METHOD 1
solving simultaneously (gdc) (M1)
\(x = 1 + 2z;{\text{ }}y = – 1 – 5z\) A1A1
\(L:\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 1} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
2 \\
{ – 5} \\
1
\end{array}} \right)\) A1A1A1
Note: \({1^{{\text{st}}}}\) A1 is for r =.
[6 marks]
METHOD 2
direction of line \( = \left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
3&1&{ – 1} \\
2&1&1
\end{array}} \right|\) (last two rows swapped) M1
= 2i − 5j + k A1
putting z = 0, a point on the line satisfies \(2x + y = 1,{\text{ }}3x + y = 2\) M1
i.e. (1, −1, 0) A1
the equation of the line is
\(\left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 1} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
2 \\
{ – 5} \\
1
\end{array}} \right)\) A1A1
Note: Award A0A1 if \(\left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right)\) is missing.
[6 marks]
\(\left( {\begin{array}{*{20}{c}}
2 \\
1 \\
1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
2 \\
{ – 5} \\
1
\end{array}} \right)\) M1
= 6i − 12k A1
hence, n = i − 2k
\({\boldsymbol{n}} \cdot {\boldsymbol{a}} = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
{ – 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
{ – 1} \\
0
\end{array}} \right) = 1\) M1A1
therefore r \( \cdot \) n = a \( \cdot \) n \( \Rightarrow x – 2z = 1\) AG
[4 marks]
METHOD 1
P = (−2, 4, 1), Q = \((x,{\text{ }}y,{\text{ }}z)\)
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{x + 2} \\
{y – 4} \\
{z – 1}
\end{array}} \right)\) A1
\(\overrightarrow {{\text{PQ}}} \) is perpendicular to \(3x + y – z = 2\)
\( \Rightarrow \overrightarrow {{\text{PQ}}} \) is parallel to 3i + j − k R1
\( \Rightarrow x + 2 = 3t;{\text{ }}y – 4 = t;{\text{ }}z – 1 = – t\) A1
\(1 – z = t \Rightarrow x + 2 = 3 – 3z \Rightarrow x + 3z = 1\) A1
solving simultaneously \(x + 3z = 1;{\text{ }}x – 2z = 1\) M1
\(5z = 0 \Rightarrow z = 0;{\text{ }}x = 1,{\text{ }}y = 5\) A1
hence, Q = (1, 5, 0)
[6 marks]
METHOD 2
Line passing through PQ has equation
\({\mathbf{r}} = \begin{array}{*{20}{c}}
{ – 2} \\
4 \\
1
\end{array} + t\begin{array}{*{20}{c}}
3 \\
1 \\
{ – 1}
\end{array}\) M1A1
Meets \({\pi _3}\) when:
\( – 2 + 3t – 2(1 – t) = 1\) M1A1
t = 1 A1
Q has coordinates (1, 5, 0) A1
[6 marks]
Examiners report
Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.
Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.
Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.