Question

Let the position vectors of points $A$ and $B$ be $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}+3\hat{k}$, respectively. A point $p$ 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

• A. 0.80
• B. 0.800
• C. 0.8

Answer 1

Answer: (A), (B), (C)

$\overrightarrow{OA}=(1,1,1),\overrightarrow{OB}=(2,1,3)$

Using section formula, we get $\overrightarrow{OP}=\left(\frac{2\lambda +1}{\lambda +1}\right)\hat{i}+\left(\frac{\lambda +1}{\lambda +1}\right)\hat{j}+\left(\frac{3\lambda +1}{\lambda +1}\right)\hat{k}$

Now $\overrightarrow{OB}\cdot \overrightarrow{OP}=\frac{4\lambda +2+\lambda +1+9\lambda +3}{\lambda +1}=\frac{14\lambda +6}{\lambda +1}$

and $\overrightarrow{OA}\times \overrightarrow{OP}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ \frac{2\lambda +1}{\lambda +1}& 1& \frac{3\lambda +1}{\lambda +1}\end{array}\right|$
$=\left(\frac{2\lambda }{\lambda +1}\right)\hat{i}+\left(\frac{-\lambda }{\lambda +1}\right)\hat{j}+\left(\frac{-\lambda }{\lambda +1}\right)\hat{k}$
$\Rightarrow |\overrightarrow{OA}\times \overrightarrow{OP}{|}^2=\frac{6{\lambda }^2}{(\lambda +1{)}^2}$
According to the question, we have
$\overrightarrow{OB}\cdot \overrightarrow{OP}-3{|\overrightarrow{OA}\times \overrightarrow{OP}|}^2=6$
$\Rightarrow \frac{14\lambda +6}{\lambda +1}-3\times \frac{6{\lambda }^2}{(\lambda +1{)}^2}=6$
$\Rightarrow (14\lambda +6)(\lambda +1)-18{\lambda }^2=6(\lambda +1{)}^2$$\Rightarrow 10{\lambda }^2-8\lambda =0$
$\Rightarrow \lambda =0,\frac{8}{10}$
$\Rightarrow \lambda =0.8(\because \lambda >0)$