Let a , b , c be three vectors such that | a | = 31 , 4 | b | = | c | = 2 and 2 ( a × b ) = 3 ( c × a ) . If the angle between b and c is 2 3 , then ( a × c a · b ) 2 is equal to _______ . [2023]

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Solution

Q.
Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $| \vec{a} |$ $= \sqrt{31},$ $4$ $| \vec{b} |$ $=$ $| \vec{c} |$ $= 2$ and $2 (\vec{a} \times \vec{b}) = 3 (\vec{c} \times \vec{a}) .$ If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 π}{3},$ then ${(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}})}^{2}$ is equal to _______ . [2023]

Ans.
(3)

$Given, 2 (\vec{a} \times \vec{b}) = 3 (\vec{c} \times \vec{a})$

$\Rightarrow \vec{a} \times (2 \vec{b} + 3 \vec{c}) = 0 \Rightarrow \vec{a} \times λ (2 \vec{b} + 3 \vec{c})$

$\Rightarrow$ $| \vec{a} |^{2}$ $= λ^{2}$ $| 2 \vec{b} + 3 \vec{c} |^{2}$ $\Rightarrow | \vec{a} |^{2} = λ^{2} (4 | \vec{b} |^{2} + 9 | \vec{c} |^{2} + 12 \vec{b} \cdot \vec{c})$

$= λ^{2} (4 \times \frac{1}{4} + 9 \times 4 + 12 (\frac{1}{2}) (2) \cos \frac{2 π}{3})$

$= λ^{2} (1 + 36 + 12 (- \frac{1}{2}))$ $= λ^{2} (1 + 36 - 6)$

$\Rightarrow 31 = 31 λ^{2} \Rightarrow λ = \pm 1$

$\Rightarrow \vec{a} = \pm (2 \vec{b} + 3 \vec{c})$

$Now,$ $\frac{| \vec{a} \times \vec{c} |}{| \vec{a} \cdot \vec{b} |} = \frac{2 | \vec{b} \times \vec{c} |}{2 \vec{b} \cdot \vec{b} + 3 \vec{c} \cdot \vec{b}}$

$And$ $| \vec{b} \times \vec{c} |^{2} = | \vec{b} |^{2} | \vec{c} |^{2} - {(\vec{b} \cdot \vec{c})}^{2} = \frac{1}{4} \times 4 - \frac{1}{4} \times 4 \cos^{2} \frac{2 π}{3} = \frac{3}{4}$

$\Rightarrow \frac{| \vec{a} \times \vec{c} |}{| \vec{a} \cdot \vec{b} |} = \frac{2 \times \frac{\sqrt{3}}{2}}{2 \cdot \frac{1}{4} - \frac{3}{2}} = - \sqrt{3}$ $∴$ ${(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}})}^{2} = 3$

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