Question

If the area of the parallelogram whose adjacent sides are $(3\hat{i}-4\hat{j}+\lambda \hat{k})$ and $(2\hat{j}-4\hat{k})$ is $\sqrt{436}$ sq. units and $\lambda \ge 0$, then $\lambda =$

• A. $0$
• B. $4$
• C. $1$
• D. $3$

Answer 1

The correct option is A $0$

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

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -4& \lambda \\ 0& 2& -4\end{array}\right|=(16-2\lambda )\hat{i}+12\hat{j}+6\hat{k}$

Area of parallelogram$=|\overrightarrow{n}|$

$\sqrt{436}=\sqrt{(16-2\lambda {)}^2+{12}^2+36}$

$\Rightarrow 436=144+36+4(8-\lambda )2$

$\Rightarrow 64=(8-\lambda )2$

$\Rightarrow \pm 8=8-\lambda$

$\Rightarrow \lambda =0,16$