Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\vec{a}| = \sqrt{31}$, $4|\vec{b}| = |\vec{c}| = 2$ and $2\vec{a} imes \vec{b} = 3\vec{c} imes \vec{a}$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$, then $\left(\frac{\vec{a} imes \vec{c}}{\vec{a} \cdot…

Question

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\vec{a}| = \sqrt{31}$, $4|\vec{b}| = |\vec{c}| = 2$ and $2\vec{a} 	imes \vec{b} = 3\vec{c} 	imes \vec{a}$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$, then $\left(\frac{\vec{a} 	imes \vec{c}}{\vec{a} \cdot \vec{b}}ight)^2$ is equal to _____ .

Aicharya · Accepted Answer

Given, $|\vec{a}| = \sqrt{31}$, $4|\vec{b}| = |\vec{c}| = 2$, $2\vec{a} \cdot \vec{b} = 3\vec{c} \cdot \vec{a}$ and angle between $\vec{b}$ and $\vec{c}$ is given as $\frac{2\pi}{3}$. Now solving, $3\vec{c} \cdot \vec{a} + 2\vec{b} \cdot \vec{a} = 0$. Thus, $3\vec{c} \cdot 2\vec{b} \cdot \vec{a} = 0$. This means $3\vec{c} \cdot 2\vec{b}$ and $\vec{a}$ are parallel vectors.

So, let $3 \vec{c} \times 2 \vec{b} = λ \vec{a}$

Now squaring both sides we get,
$9 {|\vec{c}|}^{2} + 4 {|\vec{b}|}^{2} + 12 (\vec{b} \cdot \vec{c}) = λ^{2} {|\vec{a}|}^{2}$

$\Rightarrow 36 + 1 + 12 \times \frac{1}{2} \times 2 (\cos (2 \frac{π}{3})) = λ^{2} (31)$

$\Rightarrow λ^{2} = 1$

$\Rightarrow λ = \pm 1$

Now putting the value of $λ$ in $(3 \vec{c} \times 2 \vec{b}) = λ \vec{a}$ we get,

$3 \vec{c} + 2 \vec{b} = \pm \vec{a} \dots (1)$

Now taking dot product with $\vec{b}$ in above equation we get,

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

$\Rightarrow \vec{a} \cdot \vec{b} = \pm (- \frac{3}{2} + \frac{1}{2}) = \pm (- 1)$

$\Rightarrow {(\vec{a} \cdot \vec{b})}^{2} = 1$

Again taking $3 (\vec{c} \times \vec{a}) = 2 (\vec{a} \times \vec{b})$ and sqauring both side,

$\Rightarrow {(\vec{c} \times \vec{a})}^{2} = \frac{4}{9} {(\vec{a} \times \vec{b})}^{2}$

$\Rightarrow {(\vec{c} \times \vec{a})}^{2} = \frac{4}{9} [{|\vec{a}|}^{2} {|\vec{b}|}^{2} - {(\vec{a} \cdot \vec{b})}^{2}]$

$\Rightarrow {(\vec{c} \times \vec{a})}^{2} = \frac{4}{9} [\frac{31}{4} - (1)]$

$\Rightarrow {(\vec{c} \times \vec{a})}^{2} = \frac{4}{9} \times \frac{27}{4} = 3$

Hence, the value of ${(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}})}^{2} = \frac{3}{1} = 3$ .

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\vec{a}| = \sqrt{31}$, $4|\vec{b}| = |\vec{c}|…

Solution