Mathematics - Vector Question with Solution | TestHub

Given, $\overrightarrow{a}=\alpha \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=2\hat{i}+\hat{j}-\alpha \hat{k},\alpha >0$. If the projection of $\overrightarrow{a}\times \overrightarrow{b}$ on the vector $-\hat{i}+2\hat{j}-2\hat{k}$ is $30$, then $\alpha$ is equal to

Now $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & 1& -1\\ 2& 1& -\alpha \end{array}\right|=\left(1-\alpha \right)\hat{i}+\left({\alpha }^2-2\right)\hat{j}+\left(\alpha -2\right)\hat{k}$

Projection of $\overrightarrow{a}\times \overrightarrow{b}$ on $-\hat{i}+2\hat{j}-2\hat{k}$

$=\left|\frac{\left(\overrightarrow{a}\times \overrightarrow{b}\right)·\left(-\hat{i}+2\hat{j}-2\hat{k}\right)}{\sqrt{1^2+2^2+2^2}}\right|$

$=\frac{\left(\left(1-\alpha \right)\hat{i}+\left({\alpha }^2-2\right)\hat{j}+\left(\alpha -2\right)\hat{k}\right).\left(-\hat{i}+2\hat{j}-2\hat{k}\right)}{3}$

$=\frac{-1+\alpha +2{\alpha }^2-4-2\alpha +4}{3}$

$=\frac{2{\alpha }^2-\alpha -1}{3}$

Now given length of projection is $30$

So $\frac{2{\alpha }^2-\alpha -1}{3}=30$

$\Rightarrow 2{\alpha }^2-\alpha -91=0$

$\Rightarrow \alpha =7,-\frac{13}{2}$