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 $\lambda >0,$ then $\lambda$ is

Answer 1

Area of parallelogram $=|\overrightarrow{a}\times \overrightarrow{b}|$
Area of parallelogram $=|(3\hat{i}-4\hat{j}+\lambda \hat{k})\times (2\hat{j}-4\hat{k})|$
$\Rightarrow |(16-2\lambda )\hat{i}+12\hat{j}+6\hat{k}|=\sqrt{436}$
$\Rightarrow \sqrt{(16-2\lambda {)}^2+{12}^2+6^2}=\sqrt{436}$
Squaring both sides
$\Rightarrow (16-2\lambda {)}^2+144+36=436$
$\Rightarrow (16-2\lambda {)}^2=256$
$\Rightarrow (16-2\lambda )=\pm 16$
$\Rightarrow \lambda =0,16$