Let O be the origin and {OA}=2{i}+2{j}+{k},\ {OB}={i}-2{j}+2{k} and {OC}=1/2({OB}-λ{OA}) for some λ>0. If |{OB}×{OC}|=9/2, then which of the following statements is (are) TRUE? [JEE Adv 2021 P2]

Projection of $\overrightarrow{OC}$ on $\overline{OA}$ is $-\frac{3}{2}$
Area of the triangle $OAB$ is $\frac{9}{2}$
Area of the triangle $ABC$ is $\frac{9}{2}$
The acute angle between the diagonals of the parallelogram with adjacent sides $\overline{OA}$ and $\overline{OC}$ is $\frac{\pi}{3}$

Solution:

\overline{OB}\times\overline{OC}=\frac{1}{2}\overline{OB}\times(\overline{OB}-\lambda\overline{OA})

=\frac{\lambda}{2}(\overline{OA}\times\overline{OB})

....(i)

Since,

\overline{OA}\cdot\overline{OB}=0

therfore

\overline{OA}\perp^{r}\overline{OB}

\therefore |\overline{OB}\times\overline{OC}|=\left|\frac{\lambda}{2}(\overline{OA}\times\overline{OB})\right|

from(i)

=\frac{|\lambda|}{2}|\overline{OA}||\overline{OB}|\sin\frac{\pi}{2}

\Rightarrow \frac{9}{2}=\frac{|\lambda|}{2}\times 3\times 3

\Rightarrow \lambda=1(

given

\lambda>0)

So,

\overrightarrow{OC}=\frac{\overrightarrow{OB}-\overrightarrow{OA}}{2}

=\frac{1}{2}(-\hat{i}-4\hat{j}+\hat{k})

(a) Projection

\overline{OC}

\overline{OA}

=\frac{\overline{OC}\cdot\overline{OA}}{|\overline{OA}|}=\frac{\frac{1}{2}(-2-8+1)}{3}=-\frac{3}{2}

(b) Area of

\triangle DAB

=\frac{1}{2}|\overline{OA}\times\overline{OB}|=\frac{9}{2}

(b) Area of

\triangle DAB

=\frac{1}{2}|\overline{OA}\times\overline{OB}|=\frac{9}{2}

\triangle ABC

=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&-4&1\\-\frac{5}{2}&-4&-\frac{1}{2}\end{vmatrix}

=\frac{1}{2}|6\hat{i}-3\hat{j}-6\hat{k}|=\frac{9}{2}

Let acute angle between diagonals is

\theta

then

\cos\theta=\frac{(\overline{OA}+\overline{OC})\cdot(\overline{OA}-\overline{OC})}{(\overline{OA}+\overline{OC})(\overline{OA}-\overline{OC})}=\frac{18}{3\sqrt{2}\sqrt{90}}

\therefore \theta\neq\frac{\pi}{3}

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