Vector Algebra Question 13
Question 13 - 30 January - Shift 2
Let $\lambda \in \mathbb{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}$.
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
(1) 140
(2) 132
(3) 144
(4) 136
Show Answer
Answer: (1)
Solution:
Formula: Properties of Scalar product of vectors, Properties of Cross product of vectors
$ \begin{aligned} & \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} \end{aligned} $
Now, $\vec{a} \times \vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2\end{vmatrix} $
$=(4-3 \lambda) \hat{i}-(2 \lambda+3) \hat{j}+(-\lambda^{2}-2) \hat{k}$
$\Rightarrow \lambda=1$
$\therefore \overrightarrow{{}a}=\hat{i}+2 \hat{j}-3 \hat{k}$
$\overrightarrow{{}b}=\hat{i}-\hat{j}+2 \hat{k}$
$\Rightarrow \vec{a}+\vec{b}=2 \hat{i}+\hat{j}-\hat{k}, \vec{a}-\vec{b}=3 \hat{j}-5 \hat{k}$
$\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 0 & 3 & -5\end{vmatrix} =2 \hat{i}+10 \hat{j}+6 \hat{k}$
$\therefore$ Required answer $=4+100+36=140$
