JEE Advanced 2021 Paper 2 Q05 Mathematics Vectors & 3D Geometry Scalar & Vector Products Medium

JEE Advanced 2021 Paper 2 · Q05 · Scalar & Vector Products

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

  1. A. Projection of $\vec{OC}$ on $\vec{OA}$ is $-\dfrac{3}{2}$
  2. B. Area of the triangle $OAB$ is $\dfrac{9}{2}$
  3. C. Area of the triangle $ABC$ is $\dfrac{9}{2}$
  4. D. The acute angle between the diagonals of the parallelogram with adjacent sides $\vec{OA}$ and $\vec{OC}$ is $\dfrac{\pi}{3}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

Compute $\vec{OA}\cdot\vec{OB} = 2-4+2 = 0$, so $\vec{OA}\perp\vec{OB}$; also $|\vec{OA}|=|\vec{OB}|=3$. Then $\vec{OC} = \tfrac{1}{2}\vec{OB} - \tfrac{\lambda}{2}\vec{OA}$, and $\vec{OB}\times\vec{OC} = -\tfrac{\lambda}{2}(\vec{OB}\times\vec{OA})$, with $|\vec{OB}\times\vec{OA}| = 9$. So $\dfrac{\lambda}{2}\cdot 9 = \dfrac{9}{2} \Rightarrow \lambda = 1$. (A) $\vec{OC}\cdot\hat{OA} = -\tfrac{1}{2}\cdot 3 = -\dfrac{3}{2}$. TRUE. (B) Area $\triangle OAB = \tfrac{1}{2}|\vec{OA}\times\vec{OB}| = \tfrac{1}{2}\cdot 9 = \dfrac{9}{2}$. TRUE. (C) $\vec{AB} = \vec{OB}-\vec{OA}$, $\vec{AC} = \vec{OC}-\vec{OA}$. Compute $|\vec{AB}\times\vec{AC}| = 9$, so area $= \dfrac{9}{2}$. TRUE. (D) Diagonals are $\vec{OA}+\vec{OC}$ and $\vec{OA}-\vec{OC}$. $\cos\theta = \dfrac{1}{5}$, which is not $\dfrac{1}{2}$. FALSE.

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