The diagram shows a cuboid $OABCA'B'C'D'$ in which the lengths of $OA$,$OC$ and $OD'$ are $3a$,$2a$

Question

The diagram shows a cuboid $OABCA'B'C'D'$ in which the lengths of $OA$,$OC$ and $OD'$ are $3a$,$2a$ and $a$ respectively, where $a
e 0$. The point $O$ is taken as origin, with unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ in the direction of $\overrightarrow{OA}$, $\overrightarrow{OC}$, $\overrightarrow{OD'}$ respectively. The points $P$ on $C'B'$ and $Q$ on $OD'$ are such that $C'P=\frac{1}{3}C'B'$ and $OQ=\frac{1}{3}OD'$ respectively. Find the position vectors of $P$ and $Q$. Hence find the vector equation of line $PQ$. Determine whether the lines $PQ$ and $AC'$ are skew lines. Find the shortest distance from $M$, the midpoint of $AB'$ to the line $PQ$. Use vector product to find the area of triangle $PQM$.

Accepted Answer

OP = ai + 2aj + ak; OQ = (1/3)ak; l_PQ: r = (a, 2a, a) + λ(3, 6, 2); Skew lines; Shortest distance = (5√41/14)a units; Area = (5a²√41)/12 units²

ACJC Tutorial 15 Q9

ACJC Tutorial 15 Q9

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