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Answer: $\left | \overrightarrow{a} \right |^{2}\left | \overrightarrow{b} \right |^{2}-\left ( \overrightarrow{a}.\overrightarrow{b} \right )^{2}$

Given: $\left ( \overrightarrow{a}\times \overrightarrow{b} \right )^{2}$

Hint: Using $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta$

Explanation: We know,

$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta$

Squaring on both sides

\begin{aligned} &(\vec{a} \times \vec{b})^{2}=(|\vec{a}||\vec{b}| \sin \theta)^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\ &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \\ &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \end{aligned}………………….. (*)

We know $\left | \overrightarrow{a}. \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$

On squaring both sides

$\left ( \overrightarrow{a}.\overrightarrow{b} \right )^{2}=\left | \overrightarrow{a} \right |^{2}\left | \overrightarrow{b} \right |^{2}\cos \theta$

Put in (*)

\begin{aligned} &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\ &(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \end{aligned}

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