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

The Correct Option is C

Let \( \mathbf{a} = \lambda i + 2j - 3k \) 
and \( \mathbf{b} = i - \lambda j + 2k \). We are given the equation: \[ \left( (\mathbf{a} + \mathbf{b}) \times (\mathbf{a} \times \mathbf{b}) \right) \left( (\mathbf{a} - \mathbf{b}) \right) = 8i - 40j - 24k \] Now, solve for the value of \( \lambda \): \[ (\mathbf{a} - \mathbf{b}) \times (\mathbf{a} \times \mathbf{b}) = 8i - 40j - 24k \] \[ 8 (\mathbf{a} \times \mathbf{b}) = 8i - 40j - 24k \] Now, calculate \( \mathbf{a} \times \mathbf{b} \): \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} i & j & k  \lambda & 2 & -3  1 & -\lambda & 2 \end{vmatrix} = (4 - 3\lambda) i - (2\lambda + 3) j + (-\lambda^2 - 2) k\)

Since \( \mathbf{a} \times \mathbf{b} = 8i - 40j - 24k \), we solve for \( \lambda = 1 \). Thus, we have: \[ \mathbf{a} + \mathbf{b} = 2i + 3j - 5k \] Then, \((\mathbf{a} + \mathbf{b}) \times (\mathbf{a} - \mathbf{b}) = \begin{vmatrix} i & j & k  2 & 3 & -5  1 & -1 & 2 \end{vmatrix} = 2i + 10j + 6k\) 

Hence, the required answer is \( 4 + 100 + 36 = 140 \).