QJEEM22Ma2965: JEE Main 2022 Physics - Class 11 JEEM22Ma2965 Solution: Let , , be three non-coplanar vectors such that = 4 , = 9 and = …

Given,

$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c$ ..... (i)

$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a$ ..... (ii)

$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b$ .... (iii)

Taking dot products with $\overrightarrow c ,\overrightarrow a ,\overrightarrow b$ we get

$\overrightarrow a \,.\,\overrightarrow b = \overrightarrow b \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow a = 0$

Hence,

(i) $\Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b | = 4\,.\,|\overrightarrow c |$ ..... (iv)

(ii) $\Rightarrow |\overrightarrow b |\,.\,|\overrightarrow c | = 9\,.\,|\overrightarrow a |$ ..... (v)

(iii) $\Rightarrow |\overrightarrow c |\,.\,|\overrightarrow a | = \alpha \,.\,|\overrightarrow b |$ .... (vi)

Multiplying (iv), (v) and (vi)

$\Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b |\,.\,|\overrightarrow c | = 36\alpha$ ..... (vii)

Dividing (vii) by (iv) $\Rightarrow |\overrightarrow c {|^2} = 9\alpha \Rightarrow |\overrightarrow c | = 3\sqrt \alpha$ ..... (viii)

Dividing (vii) by (v) $\Rightarrow |\overrightarrow a {|^2} = 4\alpha \Rightarrow |\overrightarrow a | = 2\sqrt \alpha$

Dividing (viii) by (vi) $\Rightarrow |\overrightarrow b {|^2} = 36 \Rightarrow |\overrightarrow b | = 6$

Now, as given, $3\sqrt \alpha + 2\sqrt \alpha + 6 = {1 \over {36}} \Rightarrow \sqrt \alpha = {{ - 43} \over {36}}$