17N.1.AHL.TZ0.H_2

pestleMathematicsAAHLPaper 117N· ahl-3-14-vector-equation-of-linesource ↗

The points A and B are given by ${\text{A}}(0,{\text{ }}3,{\text{ }} - 6)$ and ${\text{B}}(6,{\text{ }} - 5,{\text{ }}11)$ .

The plane Π is defined by the equation $4x - 3y + 2z = 20$ .

Find a vector equation of the line L passing through the points A and B.

[3]

Find the coordinates of the point of intersection of the line L with the plane Π.

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)$ (A1)

r = $\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)$ or r = $\left( {\begin{array}{*{20}{c}} 6 \\ { - 5} \\ {11} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)$ M1A1

Note: Award M1A0 if r = is not seen (or equivalent).

[3 marks]

substitute line L in $\Pi :4(6\lambda ) - 3(3 - 8\lambda ) + 2( - 6 + 17\lambda ) = 20$ M1

$82\lambda = 41$

$\lambda = \frac{1}{2}$ (A1)

r = $\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)$

so coordinate is $\left( {3,{\text{ }} - 1,{\text{ }}\frac{5}{2}} \right)$ A1

Note: Accept coordinate expressed as position vector $\left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)$ .

[3 marks]

Examiners’ report

[N/A]