Let → a = 4 ^ i + 3 ^ j and → b = 3 ^ i - 4 ^ j + 5 ^ k and → c is...

Given:

\overset{\to}{a} = 4\overset{^}{i} + 3\overset{^}{j}

\overset{\to}{b} = 3\overset{^}{i} - 4\overset{^}{j} + 5\overset{^}{k}

So,

\overset{\to}{a} \times \overset{\to}{b} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 4 & 3 & 0 \\ 3 & - 4 & 5\end{cases}\right|

⇒ \overset{\to}{a} \times \overset{\to}{b} = 15\overset{^}{i} - 20\overset{^}{j} - 25\overset{^}{k}

Let \overset{\to}{c} = x\overset{^}{i} + y\overset{^}{j} + z\overset{^}{k}

Then,

$\overset{\to}{c} \cdot ( \overset{\to}{a} \times \overset{\to}{b} ) + 25 = 0$

$⇒ 15x - 20y - 25z + 25 = 0$

$⇒ 3x - 4y - 5z = - 5 . . . \left(1\right)$

Also,

\overset{\to}{c} \cdot \left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) = 4

$⇒ x + y + z = 4 . . . . \left(2\right)$

And projection of $\overset{\to}{c}$ on $\overset{\to}{a}$ is

$\dfrac{\overset{\to}{c} \cdot \overset{\to}{a}}{| \overset{\to}{a} |} = 1$

⇒ \dfrac{\left(x\overset{^}{i} + y\overset{^}{j} + z\overset{^}{k}\right) \cdot \left(4\overset{^}{i} + 3\overset{^}{j}\right)}{\sqrt{16 + 9}} = 1

$⇒ 4x + 3y = 5 . . . . \left(3\right)$

Solving \left(1\right), \left(2\right) \text{ and } \left(3\right)$, we get</p><p>$x = 2, y = - 1, z = 3$</p><p>So,</p><p>$\overset{\to}{c} = 2\overset{^}{i} - \overset{^}{j} + 3\overset{^}{k}$</p><p>Projection of$\overset{\to}{c}$on$\overset{\to}{b}$is</p><p>$\dfrac{\overset{\to}{c} \cdot \overset{\to}{b}}{\left|\overset{\to}{b}\right|} = \dfrac{\left(2\overset{^}{i} - \overset{^}{j} + 3\overset{^}{k}\right) \cdot \left(3\overset{^}{i} - 4\overset{^}{j} + 5\overset{^}{k}\right)}{\sqrt{9 + 16 + 25}} = \dfrac{25}{5\sqrt{2}} = \dfrac{5}{\sqrt{2}}$