Question

Let the position vectors of points ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ be $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}+3\hat{k},$ respectively. $A$ point ${}^{\prime }{P}^{\prime }$ divides the line segment $AB$ internally in the ratio $\lambda :1(\lambda >0)$. If $O$ is the origin and $\overrightarrow{OB}\cdot \overrightarrow{OP}-3|\overrightarrow{OA}\times \overrightarrow{OP}{|}^2=6,$ then $\lambda$ is equal to ___

Answer 1

Answer: 0.8

Using section formula we get
$\overline{OP}=\frac{2\lambda +1}{\lambda +1}\hat{i}+\frac{\lambda +1}{\lambda +1}\hat{j}+\frac{3\lambda +1}{\lambda +1}\hat{k}$
$\text{ Now }\overline{OB}\cdot \overline{OP}=\frac{4\lambda +2+\lambda +1+9\lambda +3}{\lambda +1}$
$=\frac{14\lambda +6}{\lambda +1}$
$\overline{OA}\times \overline{OP}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ \frac{2\lambda +1}{\lambda +1}& 1& \frac{3\lambda +1}{\lambda +1}\end{array}\right|$
$=\frac{2\lambda +1}{\lambda +1}\hat{i}+\frac{-\lambda }{\lambda +1}\hat{j}+\frac{-\lambda }{\lambda +1}\hat{k}$
$|\overline{OA}\times \overline{OP}{|}^2=\frac{(2\lambda +1{)}^2+{\lambda }^2+{\lambda }^2}{(\lambda +1{)}^2}$
$=\frac{6{\lambda }^2+1}{(\lambda +1{)}^2}$
$\Rightarrow \frac{14\lambda +6}{\lambda +1}-3\times \frac{\left(6{\lambda }^2+1\right)}{(\lambda +1{)}^2}=6$
$\Rightarrow 10{\lambda }^2-8\lambda =0$
$\Rightarrow \lambda =0,\frac{8}{10}=0.8$
$\Rightarrow \lambda =0.8$