Find |vec{x}|,if for a unit vector vec{a},(ve | vedclass

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(C) Given that $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 12$. Using the distributive property of the dot product: $\vec{x} \cdot \vec{x} + \vec{x}

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$\sqrt{13}$

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(C) Given that $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 12$. Using the distributive property of the dot product: $\vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12$. Since the dot product is commutative,$\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}$,so they cancel out: $|\vec{x}|^2 - |\vec{a}|^2 = 12$. Given that $\vec{a}$ is a unit vector,$|\vec{a}| = 1$,so $|\vec{a}|^2 = 1$. Substituting this into the equation: $|\vec{x}|^2 - 1 = 12$. $|\vec{x}|^2 = 13$. Taking the square root on both sides,we get $|\vec{x}| = \sqrt{13}$.

Find $|\vec{x}|$,if for a unit vector $\vec{a}$,$(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 12$.

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