(vec{a}+vec{b}) cdot (vec{a}+vec{b}) = |vec{a | vedclass

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(B) Given the equation: $(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$. Expanding the left side using the distributive proper

Vedclass · Accepted Answer

$\vec{a}$ and $\vec{b}$ are perpendicular to each other.

Vedclass · Answer

(B) Given the equation: $(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$. Expanding the left side using the distributive property of the dot product: $\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 + |\vec{b}|^2$. Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2$,and the dot product is commutative $(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a})$,we get: $|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2$. Subtracting $|\vec{a}|^2 + |\vec{b}|^2$ from both sides: $2(\vec{a} \cdot \vec{b}) = 0$. This implies $\vec{a} \cdot \vec{b} = 0$. Since $\vec{a} 
eq \vec{0}$ and $\vec{b} 
eq \vec{0}$,the dot product of two non-zero vectors is zero if and only if they are perpendicular to each other. Thus,the correct option is $B$.

$(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$ if and only if . . . . . . (where $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$).

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