Let O be the origin and → OA = 2 ^ i + 2 ^ j + ^ k , → OB = ^ i - 2...

\overset{\to}{OA} = 2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}

\overset{\to}{OB} = \overset{^}{i} - 2\overset{^}{j} + 2\overset{^}{k}

$\overset{\to}{OC} = \dfrac{1}{2}\left(\overset{\to}{OB} - \lambda\overset{\to}{OA}\right); \lambda > 0$

\left|\overset{\to}{OB} \times \overset{\to}{OC}\right| = \dfrac{9}{2}$</p><p>$⇒ \left|\overset{\to}{OB} \times \dfrac{1}{2}\left(\overset{\to}{OB} - \lambda\overset{\to}{OA}\right)\right| = \dfrac{9}{2}$</p><p>$⇒ \dfrac{1}{2}\left|\overset{\to}{OB} \times \left(\overset{\to}{OB} - \lambda\overset{\to}{OA}\right)\right| = \dfrac{9}{2}$</p><p>$⇒ \dfrac{1}{2}\left|\lambda\left(\overset{\to}{OA} \times \overset{\to}{OB}\right)\right| = \dfrac{9}{2}$</p><p>$⇒ \left|\dfrac{\lambda}{2}\left(\overset{\to}{OA} \times \overset{\to}{OB}\right)\right| = \dfrac{9}{2} . . . \left(1\right)$</p><p>now,$\overset{\to}{OA} \times \overset{\to}{OB} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & 2 & 1 \\ 1 & - 2 & 2\end{cases}\right|$</p><p>$= 6\overset{^}{i} - 3\overset{^}{j} - 6\overset{^}{k}$</p><p>$⇒ \left|\overset{\to}{OA} \times \overset{\to}{OB}\right| = \sqrt{36 + 9 + 36} = 9 . . . \left(2\right)$</p><p>Put in (1) \left|\dfrac{\lambda}{2} \times 9\right| = \dfrac{9}{2}$

$⇒ \left|\lambda\right| = 1$

$⇒ \lambda = \pm 1\left(∵ \lambda > 0\right)$

$⇒ \lambda = 1$

$⇒ \overset{\to}{OC} = \dfrac{1}{2}\left(\overset{\to}{OB} - \overset{\to}{OA}\right)$

\overset{\to}{OC} = \dfrac{1}{2}\left(- \overset{^}{i} - 4\overset{^}{j} + \overset{^}{k}\right)

(A)

Projection of $\overset{\to}{OC}$ on $\overset{\to}{OA}$

$= \dfrac{\overset{\to}{OC} ⋅ \overset{\to}{OA}}{\left|\overset{\to}{OA}\right|}$

= \dfrac{1}{2}\dfrac{\left(- \overset{^}{i} - 4\overset{^}{j} + \overset{^}{k}\right) ⋅ \left(2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}\right)}{\sqrt{4 + 4 + 1}}

$= \dfrac{1}{2}\dfrac{\left(- 2 - 8 + 1\right)}{3}$

$= \dfrac{- 9}{6} = - \dfrac{3}{2}$

(B) Area of $\delta OAB = \dfrac{1}{2}\left|\overset{\to}{OA} \times \overset{\to}{OB}\right|$

$= \dfrac{9}{2}$ sq unit

(C)

Area of

$ABE = \dfrac{1}{2} | \overset{\to}{AB} \times \overset{\to}{AC} |$

\overset{\to}{AB} = - \overset{^}{i} - 4\overset{^}{j} + \overset{^}{k}

\overset{\to}{AC} = - \dfrac{5}{2}\overset{^}{i} - 4\overset{^}{j} - \overset{^}{k}2

\overset{\to}{AB} \times \overset{\to}{AC} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 1 & - 4 & 1 \\ - \dfrac{5}{2} & - 4 & - \dfrac{1}{2}\end{cases}\right|

= 6\overset{^}{i} - 3\overset{^}{j} - 6\overset{^}{k}

$| \overset{\to}{AB} \times \overset{\to}{AC} | = \sqrt{36 + 9 + 36} = 9$

Area of $∆ ABC = \dfrac{1}{2}\left|\overset{\to}{AB} \times \overset{\to}{AC}\right|$

$= \dfrac{9}{2}sq . unit$

(D)

Apply angle between diagonals $\overset{\to}{OD}$ & $\overset{\to}{CA}$

$cos\theta = \dfrac{\left(\overset{\to}{a} + \overset{\to}{c}\right) ⋅ \left(\overset{\to}{a} - \overset{\to}{c}\right)}{\left|\overset{\to}{a} + \overset{\to}{c}\right|\left|\overset{\to}{a} - \overset{\to}{c}\right|}$

\overset{\to}{a} = 2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}

\overset{\to}{c} = \dfrac{1}{2}\left(- \overset{^}{i} + 4\overset{^}{j} + \overset{^}{k}\right)

\overset{\to}{a} + \overset{\to}{c} = \dfrac{3}{2}\overset{^}{i} + \dfrac{3}{2}\overset{^}{k}

\left|\overset{\to}{a} + \overset{\to}{c}\right| = \sqrt{\dfrac{9}{4} + \dfrac{9}{4}} = \dfrac{3\sqrt{2}}{2}$</p><p>$\overset{\to}{a} - \overset{\to}{c} = \dfrac{5}{2}\overset{^}{i} + 4\overset{^}{j} + \dfrac{1}{2}\overset{^}{k}$</p><p>\left|\overset{\to}{a} - \overset{\to}{c}\right| = \sqrt{\dfrac{25}{4} + 16 + \dfrac{1}{4}} = \sqrt{\dfrac{90}{4}} = \dfrac{3\sqrt{10}}{2}$

$cos\theta = \dfrac{\dfrac{15}{4} + \dfrac{3}{4}}{\left(\dfrac{3\sqrt{2}}{2}\right)\left(\dfrac{3\sqrt{10}}{2}\right)}$

$= \dfrac{\dfrac{9}{2}}{\dfrac{9}{4} \times \sqrt{20}} = \dfrac{1}{\sqrt{5}} \ne \dfrac{\pi}{3}$