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 $\lambda :1$ $\left(\lambda >0\right)$. If $O$ is the origin and $\overrightarrow{OB}·\overrightarrow{OP}-3{\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2=6$, then $\lambda$ is equal to:

Answer 1

Step 1: Finding the dot product of $\overrightarrow{OB}$ and $\overrightarrow{OP}$

Given that the position vectors of the point $A$ as $\left(\hat{i}+\hat{j}+\hat{k}\right)$ and $B$ as $\left(2\hat{i}+\hat{j}+3\hat{k}\right)$. $P$ be the point which divides the line segment$AB$.

Also the equation$\overrightarrow{OB}·\overrightarrow{OP}-3{\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2=6$.

Let us find $\overrightarrow{OP}$using the section formula.

$\Rightarrow \overrightarrow{OP}=\frac{2\lambda +1}{\lambda +1}\hat{i}+\frac{\lambda +1}{\lambda +1}\hat{j}+\frac{3\lambda +1}{\lambda +1}\hat{k}$

Now find the dot product of $\overrightarrow{OB}$ and$\overrightarrow{OP}$.

$$\begin{gathered} \overrightarrow{OB}·\overrightarrow{OP}=\frac{4\lambda +2+\lambda +1+9\lambda +3}{\lambda +1} \\ \overrightarrow{OB}·\overrightarrow{OP}=\frac{14\lambda +6}{\lambda +1} \end{gathered}$$

Step 2: Finding the value of ${\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2$

Finding ${\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2$

$\begin{array}{rcl}{\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2& =& {\left|\overrightarrow{OA}\right|}^2{\overrightarrow{\left|OP\right|}}^2-{\left(\overrightarrow{OA}.\overrightarrow{OP}\right)}^2\\ & =& 3\left(\frac{{\left(2\lambda +1\right)}^2+{\left(\lambda +1\right)}^2+{\left(3\lambda +1\right)}^2}{{\left(\lambda +1\right)}^2}\right)-{\left(\frac{2\lambda +1+\lambda +1+3\lambda +1}{\lambda +1}\right)}^2\\ {\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2& =& \frac{6{\lambda }^2}{{\left(\lambda +1\right)}^2}\end{array}$

Step 3: Finding $\lambda$

Substituting the value of $\overrightarrow{OB}·\overrightarrow{OP}$ and ${\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2$ in the equation$\overrightarrow{OB}·\overrightarrow{OP}-3{\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2=6$.

$$\begin{gathered} \Rightarrow \overrightarrow{OB}·\overrightarrow{OP}-3{\left|\overrightarrow{OA}\times \overrightarrow{OP}\right|}^2=6 \\ \Rightarrow \frac{14\lambda +6}{\lambda +1}-3\frac{6{\lambda }^2}{{\left(\lambda +1\right)}^2}=6 \\ \Rightarrow 10{\lambda }^2-8\lambda =0 \\ \Rightarrow \lambda =0,0.8 \end{gathered}$$

Therefore, the value of $\lambda$ is equal to $0.8$.