Build coordinates from a cuboid · source reference

Printed question

The diagram above shows a cuboid with horizontal rectangular base $ABCD$, where $AB=5$ units, $AD=3$ units and $AP=2$ units. The edges $AP,BQ,CR,DS$ are vertical, and $PQRS$ is the top of the cuboid. The point $A$ is taken as the origin and vectors $\mathbf i,\mathbf j,\mathbf k$, each of length 1 unit, are taken along $AB,AD,AP$ respectively.

(i) Show that the line $DQ$ has equation $\mathbf r=3\mathbf j+\lambda(5\mathbf i-3\mathbf j+2\mathbf k)$, where $\lambda$ is a parameter. Find an equation of the line $BS$. [4]

(ii) Hence calculate the acute angle between the lines $DQ,BS$, giving your answer to the nearest $0.1^\circ$. [3]

The point $X$ lies on $RC$ such that $RX=\frac23RC$, and the point $M$ is the midpoint of $BD$. The plane $p$ contains the points $R,B,M$.

(iii) Find, in scalar product form, an equation of the plane $p$. [3]

(iv) Find the exact value of the shortest distance from $X$ to the plane $p$. [3]

[2009/IJC/Prelim/I/13]

Printed working

(i) $\overrightarrow{DQ}=5\mathbf i-3\mathbf j+2\mathbf k$.

Vector equation of line $DQ:\mathbf r=\overrightarrow{OD}+\lambda\overrightarrow{DQ}$, $\lambda\in\mathbb R$, i.e. $\mathbf r=(0,3,0)+\lambda(5,-3,2)$, $\lambda\in\mathbb R$ (Shown).

$\overrightarrow{BS}=-5\mathbf i+3\mathbf j+2\mathbf k$.

Vector equation of line $BS:\mathbf r=(5,0,0)+\mu(-5,3,2)$, $\mu\in\mathbb R$.

(ii) Let the acute angle between the lines $DQ,BS$ be $\theta$.

$\cos\theta=\frac{|\overrightarrow{DQ}\cdot\overrightarrow{BS}|}{|\overrightarrow{DQ}||\overrightarrow{BS}|}=\frac{|(5,-3,2)\cdot(-5,3,2)|}{\sqrt{38}\sqrt{38}}=\frac{15}{19}$.

$\therefore\theta=37.9^\circ$ (correct to 1 d.p.).

(iii) $\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=-5\mathbf i+3\mathbf j$.

$\overrightarrow{BR}=\overrightarrow{BC}+\overrightarrow{CR}=3\mathbf j+2\mathbf k$.

A normal is $(-5,3,0)\times(0,3,2)=(6,10,-15)$.

Equation of plane $\pi:\mathbf r\cdot(6,10,-15)=(5,0,0)\cdot(6,10,-15)=30$.

(iv) $\overrightarrow{BX}=\overrightarrow{BC}+\overrightarrow{CX}=3\mathbf j+\frac23\mathbf k$ (from diagram).

Shortest distance from $X$ to plane $p$

$=\frac{|\overrightarrow{BX}\cdot(6,10,-15)|}{|(6,10,-15)|}=\frac{|(0,3,2/3)\cdot(6,10,-15)|}{\sqrt{36+100+225}}=\frac{20}{19}$ units.

Teacher notes

For vector equation of line, must write “$\mathbf r=\ldots,\lambda\in\mathbb R$”.

Angle between 2 lines is angle between the direction vectors of the 2 lines. For acute angle, put in the modulus sign for $\overrightarrow{DQ}\cdot\overrightarrow{BS}$.

Shortest distance from $X$ to $p=\frac{|\overrightarrow{BX}\cdot\mathbf n|}{|\mathbf n|}=BX\cos\theta=XF$.

Printed answer key

(i) $\mathbf r=(5,0,0)+\mu(-5,3,2)$, $\mu\in\mathbb R$. (ii) $37.9^\circ$. (iii) $\mathbf r\cdot(6,10,-15)=30$. (iv) $\frac{20}{19}$ units.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question booklet PDF pages 4–12 (printed 80–88); solution booklet PDF pages 17–44. Column vectors are typeset as ordered coordinate triples here; the entries, products, intermediate equations and original parameter/domain restrictions are retained.