JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q14

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MathematicsSingle correct+4 / -1Question 14 of 90

Let $\vec a=\alpha\hat i+2\hat j-\hat k$ and $\vec b=-2\hat i+\alpha\hat j+\hat k$ , where $\alpha\in\mathbf R$ . If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec a$ and $\vec b$ is $\sqrt{15\left(\alpha^2+4\right)}$ , then the value of $2\left|\vec a\right|^2+\left(\vec a\cdot\vec b\right)\left|\vec b\right|^2$ is equal to :

Answer and worked solution

Official answer: D

Cross product. $\vec a\times\vec b=(2+\alpha,\,2-\alpha,\,\alpha^2+4)$ , so $|\vec a\times\vec b|^2=2(\alpha^2+4)+(\alpha^2+4)^2$ .

Equate with $15(\alpha^2+4)$ : $\alpha^2=9$ .

Value. $|\vec a|^2=14$ , $\vec a\cdot\vec b=-1$ , $|\vec b|^2=14$ : $28-14=14$ .

(D)

Topics:Cross product Dot product

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JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 14