Let $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ ${a_i} > 0$, $i = 1,2,3$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\overrightarrow a$ on the vector $3\widehat i + 4\widehat j$ be 7. Let $\overrightarrow b$ be a vector obtained by rotating $\overrightarrow a$ with 90$^\circ$. If $\overrightarrow a$, $\overrightarrow b$ and x-axis are coplanar, then projection of a vector $\overrightarrow b$ on $3\widehat i + 4\widehat j$ is equal to:

<p>$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \Rightarrow {\cos ^2}\alpha = {1 \over 3} \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$$</p> <p>$$\overrightarrow a = {\lambda \over 3}(\widehat i + \widehat j + \widehat k),\,\lambda > 0$$</p> <p>$${\lambda \over {\sqrt 3 }}{{(\widehat i + \widehat j + \widehat k)\,.\,(3\widehat i + 4\widehat j)} \over {\sqrt {{3^2} + {4^2}} }} = 7$$</p> <p>$\Rightarrow {\lambda \over {\sqrt 3 }}(3 + 4) = 7 \times 5$</p> <p>$\therefore$ $\lambda = 5\sqrt 3$</p> <p>$\overrightarrow a = 5(\widehat i + \widehat j + \widehat k)$</p> <p>Let $\overrightarrow b = p\widehat i + q\widehat j + r\widehat k$</p> <p>$\overrightarrow a \,.\,\overrightarrow b = 0$ and $[\overrightarrow a \,\overrightarrow b \,\widehat i] = 0$</p> <p>$\Rightarrow p + q + r = 0$ ..... (i)</p> <p>& $$\left| {\matrix{ p & q & r \cr 1 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right| = 0 \Rightarrow \matrix{ {q = r} \cr {p = - 2r} \cr } $$</p> <p>$\overrightarrow b = - 2r\widehat i + r\widehat j + r\widehat k$</p> <p>$\overrightarrow b = r( - 2\widehat i + \widehat j + \widehat k)$</p> <p>Now $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$</p> <p>$$5\sqrt 3 = \left| r \right|\sqrt b \Rightarrow \left| r \right| = {5 \over {\sqrt 2 }}$$</p> <p>$\Rightarrow$ Projection of $\overrightarrow b$ on $$3\widehat i + 4\widehat j = \left| {{{\overrightarrow b \,.\,\left( {3\widehat i + 4\widehat j} \right)} \over {\sqrt {{3^2} + {4^2}} }}} \right|$$</p> <p>$= \left| r \right|{{( - 6 + 4)} \over 5} = \left| {{{ - 2r} \over 5}} \right|$</p> <p>Projection $= {2 \over 5} \times {5 \over {\sqrt 2 }} = \sqrt 2$</p> <p>$\therefore$ B is correct.</p>