A non-zero vectors $\overrightarrow{\mathbf{a}}$ is such that its projection along the vectors $\ \frac{\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}}{\sqrt{2}}$ and $\frac{- \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}}{\sqrt{2}}$ and $\overrightarrow{\mathbf{k}}$ are equal, then unit vector along $\overrightarrow{\mathbf{a}}$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Let projection be $x$, then
$\overrightarrow{\mathbf{a}} = \frac{x(\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}})}{\sqrt{2}} + \frac{x( - \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}})}{\sqrt{2}} + x\ \widehat{\mathbf{k}}$
$	herefore\overrightarrow{\mathbf{a}} = \frac{2x\widehat{\mathbf{j}}}{\sqrt{2}} + x\ \widehat{\mathbf{k}}\$
$\Rightarrow \overrightarrow{\mathbf{a}} = \frac{\sqrt{2}}{\sqrt{3}}\widehat{\mathbf{j}} + \frac{\widehat{\mathbf{k}}}{\sqrt{3}}\$

Explanation

Reviewed by Dr. Abubacker

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