JEE MAIN - Mathematics (2022 - 26th June Evening Shift - No. 11)

Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$, $$\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j + \widehat k$$ be three given vectors. Let $$\overrightarrow v $$ be a vector in the plane of $$\overrightarrow a $$ and $$\overrightarrow b $$ whose projection on $$\overrightarrow c $$ is $${2 \over {\sqrt 3 }}$$. If $$\overrightarrow v \,.\,\widehat j = 7$$, then $$\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)$$ is equal to :
6
7
8
9

Explanation

Let $$\overrightarrow v = {\lambda _1}\overrightarrow a + {\lambda _2}\overrightarrow b $$, where $${\lambda _1},\,{\lambda _2} \in R$$.

$$ = ({\lambda _1} + 2{\lambda _2})\widehat i + ({\lambda _1} - 3{\lambda _2})\widehat j + (2{\lambda _1} + {\lambda _2})\widehat k$$

$$\because$$ Projection of $$\overrightarrow v $$ on $$\overrightarrow c $$ is $${2 \over {\sqrt 3 }}$$

$$\therefore$$ $${{{\lambda _1} + 2{\lambda _2} - {\lambda _1} + 3{\lambda _2} + 2{\lambda _1} + {\lambda _2}} \over {\sqrt 3 }} = {2 \over {\sqrt 3 }}$$

$$\therefore$$ $${\lambda _1} + 3{\lambda _2} = 1$$ ..... (i)

and $$\overrightarrow v \,.\,\widehat j = 7 \Rightarrow {\lambda _1} - 3{\lambda _2} = 7$$ ... (ii)

from equation (i) and (ii)

$${\lambda _1} = 4$$, $${\lambda _2} = - 1$$

$$\therefore$$ $$\overrightarrow v = 2\widehat i + 7\widehat j + 7\widehat k$$

$$\therefore$$ $$\overrightarrow v \,.\,(\widehat i + \widehat k) = 2 + 7$$

$$ = 9$$

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