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.