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If $\vec{a}$ and $\vec{b}$ are unit vectors, then what is the value of
$|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}$ ?
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$|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}$ ?
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Verified Answer
The correct answer is:
1
$\quad|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^{2}+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^{2}$
$=(|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}| \cdot \sin \theta)^{2}+(|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}| \cdot \cos \theta)^{2}$
$=(1.1 \cdot \sin \theta)^{2}+(1.1 \cdot \cos \theta)^{2}$
$=\sin ^{2} \theta+\cos ^{2} \theta=1$
$=(|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}| \cdot \sin \theta)^{2}+(|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}| \cdot \cos \theta)^{2}$
$=(1.1 \cdot \sin \theta)^{2}+(1.1 \cdot \cos \theta)^{2}$
$=\sin ^{2} \theta+\cos ^{2} \theta=1$
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