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If $(\vec{a} \times \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2}=676$ and $|\vec{b}|=2$, then $|\vec{a}|$ is equal to
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The correct answer is:
13
$$
\begin{array}{l}
\text { Since, }(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^{2}+(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})^{2}=676 \\
\Rightarrow(|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \sin \theta \hat{\mathrm{n}})^{2}+(|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \cos \theta)^{2}=676 \\
\Rightarrow|\overrightarrow{\mathrm{a}}|^{2} \cdot|\overrightarrow{\mathrm{b}}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=676 \\
\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}(2)^{2}=676 \Rightarrow|\overrightarrow{\mathrm{a}}|^{2}=169 \\
\Rightarrow|\overrightarrow{\mathrm{a}}|=13
\end{array}
$$
\begin{array}{l}
\text { Since, }(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^{2}+(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})^{2}=676 \\
\Rightarrow(|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \sin \theta \hat{\mathrm{n}})^{2}+(|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \cos \theta)^{2}=676 \\
\Rightarrow|\overrightarrow{\mathrm{a}}|^{2} \cdot|\overrightarrow{\mathrm{b}}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=676 \\
\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}(2)^{2}=676 \Rightarrow|\overrightarrow{\mathrm{a}}|^{2}=169 \\
\Rightarrow|\overrightarrow{\mathrm{a}}|=13
\end{array}
$$
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