Question

Let $\lambda \in R , \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}$. If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$, then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to

• A. 136
• B. 132
• C. 140
• D. 144

Answer 1

Answer: (C)

$ \vec{ a }=\lambda \hat{ i }+2 \hat{ j }-3 \hat{ k } $
$\vec{ b }=\hat{ i }-\lambda \hat{ j }+2 \hat{ k }$
$\Rightarrow(\vec{ b }-\vec{ a }) \times((\vec{ a }+\vec{ b }) \times(\vec{ a } \times \vec{ b }))=8 \hat{ i }-40 \hat{ j }-24 \hat{ k } $
$ \Rightarrow((\vec{ a }-\vec{ b }) \cdot(\vec{ a }+\vec{ b }))(\vec{ a } \times \vec{ b })=8 \hat{ i }-40 j -24 \hat{ k } $
$ \Rightarrow 8(\vec{ a } \times \vec{ b })=8 \hat{ i }-40 \hat{ j }-24 \hat{ k }$
Now, $\vec{a} \times \vec{b}=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2\end{vmatrix}$
$ =(4-3 \lambda) \hat{ i }-(2 \lambda+3) \hat{ j }+\left(-\lambda^2-2\right) \hat{ k } $
$ \Rightarrow \lambda=1 $
$ \therefore \vec{ a }=\hat{ i }+2 \hat{ j }-3 \hat{ k } $
$\vec{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$
$ \Rightarrow \vec{ a }+\vec{ b }=2 \hat{ i }+\hat{ j }-\hat{ k }, \vec{ a }-\vec{ b }=3 \hat{ j }-5 \hat{ k }$
$\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -1 \\ 0 & 3 & -5\end{vmatrix}=2 \hat{i}+10 \hat{ j }+6 \hat{ k }$
$\therefore$ required answer $=4+100+36=140$