Question

OABC is a tetrahedron andOA=a,OB=b and OC=c. The points P and Q are such that OA=AP and 2OB=BQ. The point M is the midpoint of PQ. Find (i) AB, (ii) PQ, (iii) CQ, (iv) QM, (v) MB and (vi) OM in terms of a, b and c.

Answer 1

Answer on Question #81181 – Math – Geometry

Question

OABC is a tetrahedron and $OA = a$, $OB = b$ and $OC = c$. The points $P$ and $Q$ are such that $OA = AP$ and $2OB = BQ$. The point $M$ is the midpoint of $PQ$. Find (i) $AB$, (ii) $PQ$, (iii) $CQ$, (iv) $QM$, (v) $MB$ and (vi) $OM$ in terms of $a, b$ and $c$.

Let $\overrightarrow{OA} = \vec{a}$, $\overrightarrow{OB} = \vec{b}$, $\overrightarrow{OC} = \vec{c}$.

Then

$$\overrightarrow{OP} = 2\vec{a}, \overrightarrow{OQ} = 3\vec{b}$$ $$\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB} => \overrightarrow{AB} = \vec{b} - \vec{a}$$ $$\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ} => \overrightarrow{PQ} = 3\vec{b} - 2\vec{a}$$ $$\overrightarrow{OC} + \overrightarrow{CQ} = \overrightarrow{OQ} => \overrightarrow{CQ} = 3\vec{b} - \vec{c}$$ $$\overrightarrow{QM} = -\frac{1}{2}\overrightarrow{PQ} = \vec{a} - \frac{3}{2}\vec{b}$$ $$\overrightarrow{OM} = \overrightarrow{OQ} + \overrightarrow{QM} = 3\vec{b} + \vec{a} - \frac{3}{2}\vec{b} = \vec{a} + \frac{3}{2}\vec{b}$$ $$\overrightarrow{OM} + \overrightarrow{MB} = \overrightarrow{OB} => \overrightarrow{MB} = \vec{b} - \left(\vec{a} + \frac{3}{2}\vec{b}\right) = -\vec{a} - \frac{1}{2}\vec{b}$$

Answer:

(i) $\overrightarrow{AB} = \vec{b} - \vec{a}$, (ii) $\overrightarrow{PQ} = 3\vec{b} - 2\vec{a}$, (iii) $\overrightarrow{CQ} = 3\vec{b} - \vec{c}$, (iv) $\overrightarrow{QM} = \vec{a} - \frac{3}{2}\vec{b}$,

(v) $\overrightarrow{MB} = -\vec{a} - \frac{1}{2}\vec{b}$, (vi) $\overrightarrow{OM} = \vec{a} + \frac{3}{2}\vec{b}$.

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