QJEEM21Mae7ef: JEE Main 2021 Physics - Class 11 JEEM21Mae7ef Solution: Let , and be three given vectors. If is a vector such that and…

Given, $\overrightarrow a = \widehat i + 2\widehat j - \widehat k$ ,

$\overrightarrow b = \widehat i - \widehat j$ ,

$\overrightarrow c = \widehat i - \widehat j - \widehat k$

$\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a$

$\Rightarrow \overrightarrow r \times \overrightarrow a - \overrightarrow c \times \overrightarrow a = 0$

$\Rightarrow (\overrightarrow r - \overrightarrow c ) \times \overrightarrow a = 0$

$\therefore$ $\overrightarrow r - \overrightarrow c = \lambda \overrightarrow a$

$\Rightarrow \overrightarrow r = \lambda \overrightarrow a + \overrightarrow c$

$\Rightarrow \overrightarrow r \,.\,\overrightarrow b = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b$ (taking dot with $\overrightarrow b$ )

$\Rightarrow 0 = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b$ [ $\because$ $\overrightarrow r \,.\,\overrightarrow b = 0$ ]

$\Rightarrow \lambda (\widehat i + 2\widehat j - \widehat k)\,.\,(\widehat i - \widehat j) + (\widehat i - \widehat j - \widehat k)\,.\,(\widehat i - \widehat j) = 0$

$\Rightarrow \lambda (1 - 2) + 2 = 0$

$\Rightarrow \lambda = 2$

$\therefore$ $\overrightarrow r = 2\overrightarrow a + \overrightarrow c$

$\Rightarrow \overrightarrow r \,.\,\overrightarrow a = 2\overrightarrow a \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a$ [taking dot with ${\overrightarrow a }$ ]

$= 2{\left| {\overrightarrow a } \right|^2} + \overrightarrow a \,.\,\overrightarrow c$

$= 2(1 + 4 + 1) + (1 - 2 + 1)$

$\Rightarrow \overrightarrow r \,.\,\overrightarrow a = 12$