The points A, B, C have position vectors i + j + 2k , i + 2j + 3k , 3i + k respectively and lie in the plane $\pi$ .

(a)     Find

(i)     the area of the triangle ABC;

(ii)     the shortest distance from C to the line AB;

(iii)     the cartesian equation of the plane $\pi$ .

The line L passes through the origin and is normal to the plane $\pi$ , it intersects $\pi$ at the

point D.

(b)     Find

(i)     the coordinates of the point D;

(ii)     the distance of $\pi$ from the origin.

(a)     (i)     METHOD 1

$\overrightarrow {{\text{AB}}}  = \boldsymbol{b} - \boldsymbol{a} = \left( {\begin{array}{*{20}{c}}   1 \\   2 \\   3 \end{array}} \right) - \left( {\begin{array}{*{20}{c}}   1 \\   1 \\   2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   0 \\   1 \\   1 \end{array}} \right)$     (A1)

$\overrightarrow {{\text{AC}}}  = \boldsymbol{c} - \boldsymbol{a} = \left( {\begin{array}{*{20}{c}}   3 \\   0 \\   1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}}   1 \\   1 \\   2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   2 \\   { - 1} \\   { - 1} \end{array}} \right)$     (A1)

$\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left| {\begin{array}{*{20}{c}}   \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\   0&1&1 \\   2&{ - 1}&{ - 1} \end{array}} \right|$     M1

= i(−1 + 1) − j(0 − 2) + k (0 − 2)     (A1)

= 2j − 2k     A1

Area of triangle ABC}} $= \frac{1}{2}\left| {2{\mathbf{j}} - 2{\mathbf{k}}} \right| = {\text{ }}\frac{1}{2}\sqrt 8$   $( = \sqrt 2 )$   sq. units     M1A1

Note: Allow FT on final A1.

METHOD 2

$\left| {{\text{AB}}} \right| = \sqrt 2 ,{\text{ }}\left| {{\text{BC}}} \right| = \sqrt {12} ,{\text{ }}\left| {{\text{AC}}} \right| = \sqrt 6$     A1A1A1

Using cosine rule, e.g. on ${\hat C}$     M1

$\cos C = \frac{{6 + 12 - 2}}{{2\sqrt {72} }} = \frac{{2\sqrt 2 }}{3}$     A1

$\therefore {\text{Area }}\Delta {\text{ABC}} = \frac{1}{2}ab\sin C$     M1

$= \frac{1}{2}\sqrt {12} \sqrt 6 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right)$

$= 3\sqrt 2 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right){\text{ }}\left( { = \sqrt 2 } \right)$     A1

Note: Allow FT on final A1.

(ii)     ${\text{AB}} = \sqrt 2$     A1

$\sqrt 2  = \frac{1}{2}{\text{AB}} \times h = \frac{1}{2}\sqrt 2  \times h{\text{ , }}h$ equals the shortest distance     (M1)

$\Rightarrow h = 2$     A1

(iii)     METHOD 1

(\pi \) has form $r \cdot \left( {\begin{array}{*{20}{c}}   0 \\   2 \\   { - 2} \end{array}} \right) = d$     (M1)

Since (1, 1, 2) is on the plane

$d = \left( {\begin{array}{*{20}{c}}   1 \\   1 \\   2 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}   0 \\   2 \\   { - 2} \end{array}} \right) = 2 - 4 = -2$     M1A1

Hence $r \cdot \left( {\begin{array}{*{20}{c}}   0 \\   2 \\   { - 2} \end{array}} \right) = - 2$

$2y - 2z = - 2{\text{ (or }}y - z = - 1)$     A1

METHOD 2

$r = \left( {\begin{array}{*{20}{c}}   1 \\   1 \\   2 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}   0 \\   1 \\   1 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}   2 \\   { - 1} \\   { - 1} \end{array}} \right)$     (M1)

$x = 1 + 2\mu$     (i)

$y = 1 + \lambda - \mu$     (ii)

$z = 2 + \lambda - \mu$     (iii)     A1

Note: Award A1 for all three correct, A0 otherwise.

From (i) $\mu = \frac{{x - 1}}{2}$

substitute in (ii) $y = 1 + \lambda - \left( {\frac{{x - 1}}{2}} \right)$

$\Rightarrow \lambda = y - 1 + \left( {\frac{{x - 1}}{2}} \right)$

substitute $\lambda$ and $\mu$ in (iii)     M1

$\Rightarrow z = 2 + y - 1 + \left( {\frac{{x - 1}}{2}} \right) - \left( {\frac{{x - 1}}{2}} \right)$

$\Rightarrow y - z = - 1$     A1

[14 marks]

(b)     (i)     The equation of OD is

$r = \lambda \left( {\begin{array}{*{20}{c}}   0 \\   2 \\   { - 2} \end{array}} \right)$, $\left( {{\text{or }}r = \lambda \left( {\begin{array}{*{20}{c}}   0 \\   1 \\   { - 1} \end{array}} \right)} \right)$     M1

This meets $\pi$ where

$2\lambda + 2\lambda = - 1$     (M1)

$\lambda = - \frac{1}{4}$     A1

Coordinates of D are $\left( {0, - \frac{1}{2},\frac{1}{2}} \right)$     A1

(ii)     $\left| {\overrightarrow {{\text{OD}}} } \right| = \sqrt {0 + {{\left( { - \frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = \frac{1}{{\sqrt 2 }}$     (M1)A1

[6 marks]

Total [20 marks]

It was disappointing to see that a number of candidates did not appear to be well prepared for this question and made no progress at all. There were a number of schools where no candidate made any appreciable progress with the question. A good number of students, however, were successful with part (a) (i). A good number of candidates were also successful with part a (ii) but few realised that the shortest distance was the height of the triangle. Candidates used a variety of methods to answer (a) (iii) but again a reasonable number of correct answers were seen. Candidates also had a reasonable degree of success with part (b), with a respectable number of correct answers seen.