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### Answers (1)

Answer:

L.H.S = R.H.S

Hint:

To solve this we use formula

Given:

$|\vec{a} \times \vec{b}|^{2}=\left|\begin{array}{ll} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right|$

Solution:

RHS \\\\\begin{aligned} &\left|\begin{array}{ll} \left|\vec{a}^{2}\right| & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right| \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{a}) \end{aligned}

$=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}]$

\begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2}-(|\vec{a}||\vec{b}| \cos \theta)^{2} \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \end{aligned}

\begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\\\ &=|\vec{a} \times \vec{b}|^{2} \quad[\because \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta] \end{aligned}

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