Let a vector → a be coplanar with vectors → b = 2 ^ i + ^ j + ^ k...

Given, a vector $\overset{\to}{a}$ is coplanar with vectors \overset{\to}{b} = 2\overset{^}{i} + \overset{^}{j} + \overset{^}{k} and\overset{\to}{c} = \overset{^}{i} - \overset{^}{j} + \overset{^}{k}$and$\overset{\to}{a}$is perpendicular to$\overset{\to}{d} = 3\overset{^}{i} + 2\overset{^}{j} + 6\overset{^}{k},$and$| \overset{\to}{a} | = \sqrt{10} .$</p><p>If a vector$\overset{\to}{a}$is coplanar with the vectors$\overset{\to}{b} \text{ and } \overset{\to}{c}$then it can be expressed as the linear combination of the vectors.</p><p>Thus,$\overset{\to}{a} = \lambda\overset{\to}{b} + \mu\overset{\to}{c}$</p><p>$⇒ \overset{\to}{a} = \lambda\left(2\overset{^}{i} + \overset{^}{j} + k\right) + \mu\left(\overset{^}{i} - \overset{^}{j} + k\right)$</p><p>$⇒ \overset{\to}{a} = \overset{^}{i}\left(2\lambda + \mu\right) + \overset{^}{j}\left(\lambda - \mu\right) + \overset{^}{k}\left(\lambda + \mu\right)$</p><p>Also, it is given that$\overset{\to}{a}$is perpendicular to$\overset{\to}{d},$and we know that if two vectors are perpendicular, then$\overset{\to}{a} \cdot \overset{\to}{d} = 0$</p><p>$⇒ 3\left(2\lambda + \mu\right) + 2\left(\lambda - \mu\right) + 6\left(\lambda + \mu\right) = 0$</p><p>$⇒ 14\lambda + 7\mu = 0$</p><p>$⇒ \mu = - 2\lambda$</p><p>$⇒ \overset{\to}{a} = \left(0\right)\overset{^}{i} + 3\lambda\overset{^}{j} + \left(- \lambda\right)\overset{^}{k}$</p><p>$⇒ \overset{\to}{a} = \lambda\left(3\overset{^}{j} - \overset{^}{k}\right)$</p><p>$⇒ | \overset{\to}{a} | = \sqrt{10} | \lambda |$</p><p>Given \left|\overset{\to}{a}\right| = \sqrt{10}$

$⇒ | \lambda | = 1$

$⇒ \lambda = 1$ or $- 1$

Also, if $\overset{\to}{a}, \overset{\to}{b} \text{ and } \overset{\to}{c}$ are coplanar, then $[ \overset{\to}{a}\overset{\to}{b}\overset{\to}{c} ] = 0$

Now, $[ \overset{\to}{a}\overset{\to}{b}\overset{\to}{c} ] + [ \overset{\to}{a}\overset{\to}{b}\overset{\to}{d} ] + [ \overset{\to}{a}\overset{\to}{c}\overset{\to}{d} ]$

$= 0 + [ \overset{\to}{a}\overset{\to}{b}\overset{\to}{d} ] + [ \overset{\to}{a}\overset{\to}{c}\overset{\to}{d} ] = [ \overset{\to}{a}\overset{\to}{b} + \overset{\to}{c}\overset{\to}{d} ]$

On putting the values of the vectors for finding the box product, we get

= \left|\begin{cases}0 & 3\lambda & - \lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6\end{cases}\right|

$= - 3\lambda ( 12 ) - \lambda ( 6 ) = - 42\lambda = - 42$ or $42 .$