Let → a = a _ 1 ^ i + a _ 2 ^ j + a _ 3 ^ k , a _ i > 0, i = 1, 2,...

Given \overset{\to}{a} = {a}_{1}\overset{^}{i} + {a}_{2}\overset{^}{j} + {a}_{3}\overset{^}{k}

Since the vector makes equal angles with the coordinate axes $OX, OY and$OZ$, so<p>$\overset{\to}{a} = \lambda\left(\dfrac{1}{\sqrt{3}}\overset{^}{i} + \dfrac{1}{\sqrt{3}}\overset{^}{j} + \dfrac{1}{\sqrt{3}}\overset{^}{k}\right) = \dfrac{\lambda}{\sqrt{3}}\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)$</p><p>Now projection of$\overset{\to}{a}$on$\overset{\to}{b} = 7$</p><p>$⇒ \dfrac{\lambda}{\sqrt{3}}\dfrac{\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) \cdot \left(3\overset{^}{i} + 4\overset{^}{j}\right)}{5} = 7$</p><p>$⇒ \lambda = 5\sqrt{3}$</p><p>So$\overset{\to}{a} = 5\left(\overset{^}{\overset{^}{i}} + \overset{^}{j} + \overset{^}{k}\right)$</p><p>Let$\overset{\to}{b} = 5\alpha\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) + \beta\left(\overset{^}{i}\right)$</p><p>Since both vectors are perpendicular$\overset{\to}{a} \cdot \overset{\to}{b} = 0$</p><p>$⇒ 25\alpha\left(3\right) + 5\beta = 0$</p><p>$⇒ 15\alpha + \beta = 0 ⇒ \beta = - 15\alpha$</p><p>i.e.$\overset{\to}{b} = 5\alpha\left(- 2\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)$</p><p>\left|\overset{\to}{b}\right| = 5\sqrt{3}$

\overset{\to}{b} = \pm \dfrac{5}{\sqrt{2}}\left(- 2\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)

Projection of $\overset{\to}{b}$ on 3\overset{^}{i} + 4\overset{^}{j} is

\dfrac{\overset{\to}{b} \cdot \left(3\overset{^}{i} + 4\overset{^}{j}\right)}{5} = \pm \dfrac{5}{\sqrt{2}}\left(\dfrac{- 6 + 4}{5}\right) = \pm \sqrt{2}