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Date	May 2019	Marks available	6	Reference code	19M.2.SL.TZ2.S_7
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	S_7	Adapted from	N/A

Question

The vector equation of line $L$ is given by r $= \left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 4 \\ 5 \\ { - 1} \end{array}} \right)$ .

Point P is the point on $L$ that is closest to the origin. Find the coordinates of P.

Markscheme

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

METHOD 1 (Distance between the origin and P)

correct position vector for OP (A1)

eg $\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { - 1 + 4t} \\ {3 + 5t} \\ {8 - t} \end{array}} \right)$ , ${\text{P}} = \left( { - 1 + 4t{\text{,}}\,\,3 + 5t{\text{,}}\,\,8 - t} \right)$

correct expression for OP or OP² (seen anywhere) A1

eg $\sqrt {{{\left( { - 1 + 4t} \right)}^2} + {{\left( {3 + 5t} \right)}^2} + {{\left( {8 - t} \right)}^2}}$ , ${\left( { - 1 + 4x} \right)^2} + {\left( {3 + 5x} \right)^2} + {\left( {8 - x} \right)^2}$

valid attempt to find the minimum of OP (M1)

eg $d' = 0$ , root on sketch of $d^{'}$ , min indicated on sketch of $d$

$t = - \frac{1}{{14}}{\text{,}}\,\, - 0.0714285$ (A1)

substitute their value of $t$ into $L$ (only award if there is working to find $t$ ) (M1)

eg one correct coordinate, $- 1 + 4\left( { - \frac{1}{{14}}} \right)$

$\left( { - 1.28571{\text{,}}\,\,2.64285{\text{,}}\,\,8.07142} \right)$

$\left( { - \frac{9}{7}{\text{,}}\,\,\frac{{37}}{{14}}{\text{,}}\,\,\frac{{113}}{{14}}} \right)\,\,\, = \left( { - 1.29{\text{,}}\,\,2.64{\text{,}}\,\,8.07} \right)$ A1 N2

METHOD 2 (Perpendicular vectors)

recognizing that closest implies perpendicular (M1)

eg $\overrightarrow {{\text{OP}}} \bot \,L$ (may be seen on sketch), $a \bullet b = 0$

valid approach involving $\overrightarrow {{\text{OP}}}$ (M1)

eg $\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { - 1 + 4t} \\ {3 + 5t} \\ {8 - t} \end{array}} \right){\text{,}}\,\,\left( {\begin{array}{*{20}{c}} 4 \\ 5 \\ { - 1} \end{array}} \right) \bullet \overrightarrow {{\text{OP}}} {\text{,}}\,\,\left( {\begin{array}{*{20}{c}} 4 \\ 5 \\ { - 1} \end{array}} \right) \bot \overrightarrow {{\text{OP}}}$

correct scalar product A1

eg $4\left( { - 1 + 4t} \right) + 5\left( {3 + 5t} \right) - 1\left( {8 - t} \right)$ , $- 4 + 16t + 15 + 25t - 8 + t = 0$ , $42t + 3$

$t = - \frac{1}{{14}}{\text{,}}\,\, - 0.0714285$ (A1)

substitute their value of $t$ into $L$ or $\overrightarrow {{\text{OP}}}$ (only award if scalar product used to find $t$ ) (M1)

eg one correct coordinate, $- 1 + 4\left( { - \frac{1}{{14}}} \right)$

$\left( { - 1.28571{\text{,}}\,\,2.64285{\text{,}}\,\,8.07142} \right)$

$\left( { - \frac{9}{7}{\text{,}}\,\,\frac{{37}}{{14}}{\text{,}}\,\,\frac{{113}}{{14}}} \right)\,\,\, = \left( { - 1.29{\text{,}}\,\,2.64{\text{,}}\,\,8.07} \right)$ A1 N2

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.11—Vector equation of a line in 2d and 3d

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Topic 3—Geometry and trigonometry » AHL 3.13—Scalar and vector products

Topic 3—Geometry and trigonometry

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