Question

The projection of the vector $\overrightarrow{a}=4\hat{i}-3\hat{j}+2\hat{k}$ on the vector making equal angles (acute) with coordinate axes having magnitude $\sqrt{3}$ is

• A. $3$
• B. $\sqrt{3}$
• C. $2\sqrt{3}$
• D. $1$

Answer 1

The correct option is B $\sqrt{3}$

$\overrightarrow{a}=4\hat{i}-3\hat{j}+2\hat{k}$

Let $\overrightarrow{b}=x\hat{i}+y\hat{j}+z\hat{k}$

it masses equal angles with coordinate axes.

$\Rightarrow \frac{x}{\sqrt{x^2+y^2+z^2}}=cos{\theta }_x=cos{\theta }_y=\frac{y}{\sqrt{x^2+y^2+z^2}}=cos{\theta }_z=\frac{z}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow x=y=z$

$\therefore b=x(\hat{i}+\hat{j}+\hat{k})$

$|\overline{b}|=\sqrt{3}$

$\Rightarrow \sqrt{x^2+x^2+x^2}=\sqrt{3}$

$\Rightarrow 3x^2=3$

$\Rightarrow k=\pm 1$

$\therefore b=\hat{i}+\hat{j}+\hat{k}orb=-(\hat{i}+\hat{j}+\hat{k})$

Projection of $\overrightarrow{a}$ or $\overrightarrow{b}\Rightarrow \frac{\overrightarrow{a}.\overrightarrow{b}}{|\overline{b}|}$

$\Rightarrow \frac{x(4-3+2)}{\sqrt{3}}$

$\Rightarrow \pm \frac{\left(3\right)}{\sqrt{3}}=\pm \sqrt{3}$