Let O be the origin and overrightarrow OA =2 hat i +2 hat j +hat k , overrightarrow OB =hat i -2 hat j +2 hat k and overrightarrow OC =1/2(overrightarrow OB -λ overrightarrow OA ) for some λ 0 . If overrightarrow OB × overrightarrow OC =9/2 , then which of the following statements is(are) TRUE?

Projection of overrightarrow OC on overrightarrow OA is -3/2

Area of the triangle OAB is 9/2

Area of the triangle ABC is 9/2

The acute angle between the diagonals of the parallelogram with adjacent sides overrightarrow OA and overrightarrow OC is π/3

Mathematics · Vector Algebra

JEE Advanced 2021 — Paper 2 — Question 21

Let O be the origin and $\overrightarrow{\mathrm{OA}}=2 \hat{i}+2 \hat{j}+\hat{k}, \overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}-2 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{OC}}=\frac{1}{2}(\overrightarrow{\mathrm{OB}}-\lambda \overrightarrow{\mathrm{OA}})$ for some $\lambda>0$ . If $|\overrightarrow{\mathrm{OB}} \times \overrightarrow{\mathrm{OC}}|=\frac{9}{2}$ , then which of the following statements is(are) TRUE?

Option A:
Projection of $\overrightarrow{\mathrm{OC}}$ on $\overrightarrow{\mathrm{OA}}$ is $-\frac{3}{2}$
Correct
Option B:
Area of the triangle OAB is $\frac{9}{2}$
Correct
Option C:
Area of the triangle ABC is $\frac{9}{2}$
Correct
Option D:
The acute angle between the diagonals of the parallelogram with adjacent sides $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OC}}$ is $\frac{\pi}{3}$

Answer: A, B, C

Step-by-step solution

$\overrightarrow{O A} \cdot \overrightarrow{O B}=0 $$\Rightarrow \overrightarrow{O A} \perp \overrightarrow{O B}$

$\overrightarrow{O C}=\frac{1}{2}((1-2 \lambda) \hat{i}+(-2-2 \lambda) \hat{j}+(2-\lambda) \hat{k})$

$|\overrightarrow{\mathrm{OB}} \times \overrightarrow{\mathrm{OC}}|=\frac{9|\lambda|}{2}=\frac{9}{2}$

$\Rightarrow \lambda= \pm 1$ , as $\lambda>0, \lambda=1$

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

(A) Projection $\overrightarrow{O C}$ on $\overrightarrow{O A}$

\overrightarrow{\mathrm{OC}} \cdot \widehat{O A}=-\frac{3}{2}

(B) Area of $\triangle \mathrm{OAB}=9 / 2$

(D) $\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=\frac{3 \hat{\mathrm{i}}+3 \hat{k}}{2}, \overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}=\frac{5 \hat{i}+8 \hat{j}+\hat{k}}{2}$

(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}})(\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}})=|\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}||\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}| \cos \theta

$\cos \theta=\frac{1}{\sqrt{5}}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra
Topic: Projection & component of a vector along another vector.