Let → α = 4 ^ i + 3 ^ j + 5 ^ k and → β = ^ i + 2 ^ j - 4 ^ k . Let...

Given that \overset{\to}{\alpha} = 4\overset{^}{i} + 3\overset{^}{j} + 5\overset{^}{k} and\overset{\to}{\beta} = \overset{^}{i} + 2\overset{^}{j} - 4\overset{^}{k} and \overset{\to}{\beta} = \overset{\to}{{\beta}_{1}} + \overset{\to}{{\beta}_{2}}$</p><p>$⇒ \overset{\to}{{\beta}_{2}} = \overset{\to}{\beta} - \overset{\to}{{\beta}_{1}} . . . . . . . . . . \left(1\right)

since $\overset{\to}{\beta}1$ is parallel to $\overset{\to}{\alpha}$, $\overset{\to}{{\beta}_{1}} = t\overset{\to}{\alpha}$

⇒ \overset{\to}{{\beta}_{1}} = t \left(4\overset{^}{i} + 3\overset{^}{j} + 5\overset{^}{k}\right) = 4t\overset{^}{i} + 3t\overset{^}{j} + 5t\overset{^}{k} . . . . . . . . . \left(2\right)</p><p>Substituting thevalues of\overset{\to}{{\beta}_{1}} and \overset{\to}{\alpha} in ( 1 )$, we get$

\overset{\to}{{\beta}_{2}} = \overset{^}{i} + 2\overset{^}{j} - 4\overset{^}{k} -\left(4t\overset{^}{i} + 3t\overset{^}{j} + 5t\overset{^}{k}\right)= \left(1 - 4t\right)\overset{^}{i} + \left(2 - 3t\right)\overset{^}{j} + \left(- 4 - 5t\right)\overset{^}{k} . . . . . . . . . . . . \left(3\right)</p><p>since\overset{\to}{{\beta}_{2}}$is perpendicular to$\overset{\to}{\alpha}$, so$\overset{\to}{{\beta}_{2}} \cdot \overset{\to}{\alpha} = 0$</p><p>$⇒ \left[$\left(1 - 4t\right)$\overset{^}{i} + \left(2 - 3t\right)\overset{^}{j} + \left(- 4 - 5t\right)\overset{^}{k}\right] \cdot \left[4\overset{^}{i} + 3\overset{^}{j} + 5\overset{^}{k}\right]$= 0$

$⇒ 4\left(1 - 4t\right)$ + 3\left(2 - 3t\right)$+ 5\left(- 4 - 5t\right)$ = 0$</p><p>$⇒ 4 - 16t + 6 - 9t - 20 - 25t = 0$</p><p>$⇒ - 50t = 10⇒ t = $\dfrac{- 1}{5}</p><p>From \left(2\right)$ and \left(3\right), we get$\overset{\to}{{\beta}_{1}} =$\dfrac{- 1}{5}\left(4\overset{^}{i} + 3\overset{^}{j} + 5\overset{^}{k}\right)$</p><p>$\overset{\to}{{\beta}_{2}} = \dfrac{9}{5}\overset{^}{i} + \dfrac{13}{5}\overset{^}{j} - 3\overset{^}{k} = \dfrac{1}{5}\left(9\overset{^}{i} + 13\overset{^}{j} - 15\overset{^}{k}\right)$</p><p>So, the$value of5\overset{\to}{\beta}2 \cdot \left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) will be,$</p><p>$= 5 \times \dfrac{1}{5} (9\overset{^}{i} + 13\overset{^}{j} - 15\overset{^}{k} ) \cdot \left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) = 9 + 13 - 15 = 7$</p>$