Question

The value of $\lambda$ so that the points $P,Q,R,S$ on the sides $OA,OB,OC$ and $AB$ of a regular tetrahedron are co-planar when $\frac{OP}{OA}=\frac{1}{3};\frac{OQ}{OB}=\frac{1}{2};\frac{OR}{OC}=\frac{1}{3}$ and $\frac{OS}{AB}=\lambda$ is

• A. $\lambda =\frac{1}{2}$
• B. $\lambda =-1$
• C. $\lambda =0$
• D. $\lambda =2$

Answer 1

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

$\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}$ $\Rightarrow \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}\overrightarrow{OP}=\frac{1}{3}\overrightarrow{a}\overrightarrow{OQ}=\frac{1}{2}\overline{b}$ $\overrightarrow{OR}=\frac{1}{3}\overrightarrow{c}\because P,Q,R,S\text{are coplanar points}\Rightarrow \overrightarrow{PQ},\overrightarrow{PR},\overrightarrow{RS}\text{are coplanar}\Rightarrow \overrightarrow{PS}=\alpha \overrightarrow{PQ}+\beta \overrightarrow{PQ}$ $\Rightarrow \overrightarrow{OS}-\overrightarrow{OP}=\alpha (\frac{1}{2}\overrightarrow{b}-\frac{1}{3}\overrightarrow{a})+\beta (\frac{1}{3}\overrightarrow{c}-\frac{1}{3}\overrightarrow{a})\Rightarrow \overrightarrow{OS}=(1-\alpha -\beta )\frac{\overrightarrow{a}}{3}+\frac{\alpha }{2}\overrightarrow{b}+\frac{\beta }{3}\overrightarrow{c}\text{But}\overrightarrow{OS}=\lambda \overrightarrow{AB}=\lambda (\overrightarrow{b}-\overrightarrow{a})$
$\Rightarrow \lambda (\overrightarrow{b}-\overrightarrow{a})=(1-\alpha -\beta )\frac{\overrightarrow{a}}{3}+\frac{\alpha }{2}\overrightarrow{b}+\frac{\beta }{2}\overrightarrow{c}$ $\Rightarrow \beta =0;\frac{1-\alpha -0}{3}=-\lambda ;\frac{\alpha }{2}=\lambda \therefore \lambda =-1$