Question

Find the value of $\lambda$ so that the points $P,Q,R$ and $S$ on the sides $OA,OB,OC$ and $AB$ respectively of a regular tetrahedron $OABC$ are coplanar. It is given that $\frac{OP}{OA}=\frac{1}{3},\frac{OQ}{OB}=\frac{1}{2}$ and $\frac{OS}{AB}=\lambda$.

• A. $\lambda =\frac{1}{2}$
• B. $\lambda =-1$
• C. $\lambda =0$
• D. For no value of $\lambda$

Answer 1

The correct option is B $\lambda =-1$

Let $\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b}$ and $\overrightarrow{OC}=\overrightarrow{c}$,
then, $\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}$ and $\overrightarrow{OP}=\frac{1}{3}\overrightarrow{a},\overrightarrow{OQ}=\frac{1}{3}\overrightarrow{b},\overrightarrow{OR}=\frac{1}{3}\overrightarrow{c}$

Since, $P,Q,R$ and $S$ are coplanar, then
$\overrightarrow{PS}=\alpha \overrightarrow{PQ}+\beta \overrightarrow{PR}$

($\overrightarrow{PS}$ can be written as a linear combination of $\overrightarrow{PQ}$ and $\overrightarrow{PR}$)
$\Rightarrow \overrightarrow{PS}=\alpha (\overrightarrow{OQ}-\overrightarrow{OP})+\beta (\overrightarrow{OR}-\overrightarrow{OP})$

i.e.$\overrightarrow{OS}-\overrightarrow{OP}=-(\alpha +\beta )\frac{\overrightarrow{a}}{3}+\frac{\alpha }{2}\overrightarrow{b}+\frac{\beta }{3}\overrightarrow{c}$

$\Rightarrow \overrightarrow{OS}=(1-\alpha -\beta )\frac{\overrightarrow{a}}{3}+\frac{\alpha }{2}\overrightarrow{b}+\frac{\beta }{3}\overrightarrow{c}$
$\dots \left[1\right]$
Given $\overrightarrow{OS}=\lambda \overrightarrow{AB}=\lambda (\overrightarrow{b}-\overrightarrow{a})\dots \left[2\right]$
From $\left[1\right]$ and $\left[2\right]$,
$\beta =0,\frac{1-\alpha }{3}=-\lambda$ and $\frac{\alpha }{2}=\lambda$
$\Rightarrow 2\lambda =1+3\lambda$
or, $\lambda =-1$