Let overrightarrow{a} = 2hat{i} + hat{j} + ha | vedclass

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(C) Given $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b} - \vec{c}|^2$.Expanding both sides using the property $|\vec{u} + \vec{v}|^2 = |\vec{u

Vedclass · Accepted Answer

only $(A)$ is correct

Vedclass · Answer

(C) Given $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b} - \vec{c}|^2$.Expanding both sides using the property $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$:$|\vec{a} + \vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} + \vec{b}) \cdot \vec{c} = |\vec{a} + \vec{b}|^2 + |\vec{c}|^2 - 2(\vec{a} + \vec{b}) \cdot \vec{c}$.This simplifies to $4(\vec{a} + \vec{b}) \cdot \vec{c} = 0$.Since $\vec{b} \cdot \vec{c} = 0$,we have $4(\vec{a} \cdot \vec{c}) = 0$,which implies $\vec{a} \cdot \vec{c} = 0$.Statement $(B)$ claims $\vec{a}$ and $\vec{c}$ are parallel,but $\vec{a} \cdot \vec{c} = 0$ implies they are perpendicular (since $\vec{c} 
eq 0$). Thus,$(B)$ is incorrect.For statement $(A)$,consider $|\overrightarrow{a} + \lambda \overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + \lambda^2 |\overrightarrow{c}|^2 + 2\lambda(\vec{a} \cdot \vec{c})$.Since $\vec{a} \cdot \vec{c} = 0$,this becomes $|\overrightarrow{a}|^2 + \lambda^2 |\overrightarrow{c}|^2$.Since $\lambda^2 |\overrightarrow{c}|^2 \geq 0$ for all $\lambda \in R$,it follows that $|\overrightarrow{a} + \lambda \overrightarrow{c}|^2 \geq |\overrightarrow{a}|^2$,which means $|\overrightarrow{a} + \lambda \overrightarrow{c}| \geq |\overrightarrow{a}|$.Thus,$(A)$ is correct.

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