Question

What is the value of $\lambda$ for which $(\lambda \hat{i}+\hat{j}-\hat{k})\times (3\hat{i}-2\hat{j}+4\hat{k})=(2\hat{i}-11\hat{j}-7\hat{k})$?

• A. $2$
• B. $-2$
• C. $1$
• D. $7$

Answer 1

The correct option is A $2$

$(\lambda \hat{i}+\hat{j}-\hat{k})\times (3\hat{i}-2\hat{j}+4\hat{k})=(2\hat{i}-11\hat{j}-7\hat{k})$

$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \lambda & 1& -1\\ 3& -2& 4\end{array}\right|=(2\hat{i}-11\hat{j}-7\hat{k})$

$\Rightarrow (-2\lambda -3)\hat{k}+(-3-4\lambda )\hat{j}+(4-2)\hat{i}=(2\hat{i}-11\hat{j}-7\hat{k})$

So $-2\lambda -3=-7$

$\Rightarrow 2\lambda =4$

So $\lambda =2$
Option A is correct