JEE Advanced 2021 · Paper 2, Mathematics question 5

Find $\lambda$. $\overrightarrow{OA}\cdot\overrightarrow{OB}=2-4+2=0$ and $|\overrightarrow{OA}|=|\overrightarrow{OB}|=3$. So
$$|\overrightarrow{OB}\times\overrightarrow{OC}|=\frac\lambda2|\overrightarrow{OB}\times\overrightarrow{OA}|=\frac{9\lambda}2=\frac92\implies\lambda=1.$$
Then $\overrightarrow{OC}=\frac12(-1,-4,1)$.

Check each option.

• (A) $\dfrac{\overrightarrow{OC}\cdot\overrightarrow{OA}}{|\overrightarrow{OA}|}=\dfrac{\frac12(-2-8+1)}3=-\dfrac32$. True.
• (B) Area of $OAB=\frac12\cdot3\cdot3=\frac92$. True.
• (C) $\overrightarrow{AB}=2\overrightarrow{OC}$ and $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$, so $|\overrightarrow{AB}\times\overrightarrow{AC}|=2|\overrightarrow{OC}\times\overrightarrow{OA}|=9$. The area is $\frac92$. True.
• (D) The diagonals are $(1.5,0,1.5)$ and $(2.5,4,0.5)$, with $\cos\theta=\dfrac{4.5}{2.12\times4.74}\approx0.447$. That is about $63.4^\circ$, not $60^\circ$. False.

(A), (B), (C)