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 $\vec{ OB } \cdot \vec{ OP }-3|\vec{ OA } \times \vec{ OP }|^{2}=6,$ then $\lambda$ is equal to ___

Answer 1

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} $
$=\frac{14 \lambda+6}{\lambda+1}$
$\overline{OA} \times\overline{OP} =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 1&1&1\\ \frac{2\lambda+1}{\lambda+1}&1&\frac{3\lambda+1}{\lambda+1}\end{vmatrix}$
$=\frac{2 \lambda+1}{\lambda+1} \hat{i}+\frac{-\lambda}{\lambda+1} \hat{j}+\frac{-\lambda}{\lambda+1} \hat{k}$
$|\overline{O A} \times \overline{O P}|^{2}=\frac{(2 \lambda+1)^{2}+\lambda^{2}+\lambda^{2}}{(\lambda+1)^{2}}$
$=\frac{6 \lambda^{2}+1}{(\lambda+1)^{2}}$
$\Rightarrow \frac{14 \lambda+6}{\lambda+1}-3 \times \frac{\left(6 \lambda^{2}+1\right)}{(\lambda+1)^{2}}=6$
$\Rightarrow 10 \lambda^{2}-8 \lambda=0$
$\Rightarrow \lambda=0, \frac{8}{10}=0.8$
$\Rightarrow \lambda=0.8$