QJEEM22Maea33: JEE Main 2022 Physics - Class 11 JEEM22Maea33 Solution: Let , be a vector which makes equal angles with the coordinate…

${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \Rightarrow {\cos ^2}\alpha = {1 \over 3} \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$

\overrightarrow a = {\lambda \over 3}(\widehat i + \widehat j + \widehat k),\,\lambda > 0

${\lambda \over {\sqrt 3 }}{{(\widehat i + \widehat j + \widehat k)\,.\,(3\widehat i + 4\widehat j)} \over {\sqrt {{3^2} + {4^2}} }} = 7$

$\Rightarrow {\lambda \over {\sqrt 3 }}(3 + 4) = 7 \times 5$

$\therefore$ $\lambda = 5\sqrt 3$

$\overrightarrow a = 5(\widehat i + \widehat j + \widehat k)$

Let $\overrightarrow b = p\widehat i + q\widehat j + r\widehat k$

$\overrightarrow a \,.\,\overrightarrow b = 0$ and $[\overrightarrow a \,\overrightarrow b \,\widehat i] = 0$

$\Rightarrow p + q + r = 0$ ..... (i)

& $$\left|\begin{array}{ccc}p& q& r\\ 1& 1& 1\\ 1& 0& 0\end{array}\right|=0\Rightarrow \begin{array}{c}q=r\\ p=-2r\end{array}$$

$\overrightarrow b = - 2r\widehat i + r\widehat j + r\widehat k$

$\overrightarrow b = r( - 2\widehat i + \widehat j + \widehat k)$

Now $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$

$5\sqrt 3 = \left| r \right|\sqrt b \Rightarrow \left| r \right| = {5 \over {\sqrt 2 }}$

$\Rightarrow$ Projection of $\overrightarrow b$ on $3\widehat i + 4\widehat j = \left| {{{\overrightarrow b \,.\,\left( {3\widehat i + 4\widehat j} \right)} \over {\sqrt {{3^2} + {4^2}} }}} \right|$

$= \left| r \right|{{( - 6 + 4)} \over 5} = \left| {{{ - 2r} \over 5}} \right|$

Projection $= {2 \over 5} \times {5 \over {\sqrt 2 }} = \sqrt 2$

$\therefore$ B is correct.