JEE Main 2014 — Vector Algebra Question with Solution

Given $\overrightarrow{x}=3\hat{i}-6\hat{j}-\hat{k}, \overrightarrow{y}=\hat{i}+4\hat{j}-3\hat{k}$ and $\overrightarrow{z}=3\hat{i}-4\hat{j}-12\hat{k}.$

$\Rightarrow \overrightarrow{x}\times \overrightarrow{y}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -6& -1\\ 1& 4& -3\end{array}\right|$

$=\hat{i}\left(18+4\right)-\hat{j}\left(-9+1\right)+\hat{k}\left(12+6\right)=22\hat{i}+8\hat{j}+18\hat{k}=2\left(11\hat{i}+4\hat{j}+9\hat{k}\right)$

Now, we know that the projection of a vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is defined as $\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$

Hence, the magnitude of projection of $\overrightarrow{x}\times \overrightarrow{y}$ on $\overrightarrow{z}$ is $\frac{\left(\overrightarrow{x}\times \overrightarrow{y}\right)·\overrightarrow{z}}{\left|\overrightarrow{z}\right|}$

$=\frac{\left|2\left(11i+4j+9k\right)·\left(3\hat{i}-4\hat{j}-12\hat{k}\right)\right|}{\left|3\hat{i}-4\hat{j}-12\hat{k}\right|}$

$=\frac{\left|2\left(33-16-108\right)\right|}{\sqrt{3^2+4^2+12^2}}$

$=\frac{2\times 91}{\sqrt{169}}=\frac{2\times 91}{13}=14.$