Find the vector equation of L, the line of intersection of ${\Pi _1}$ and ${\Pi _2}$.

Show that the plane ${\Pi _3}$ which is perpendicular to ${\Pi _1}$ and contains L, has equation $x - 2z = 1$.

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.

(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]

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.