Let vec{a}, vec{b} and vec{c} be three unit v | vedclass

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(A) Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.Squaring both sides: $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{

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$\left(-\frac{3}{2}, 3 \vec{a} 	imes \vec{b}ight)$

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(A) Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.Squaring both sides: $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = 0$.$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$1+1+1+2\lambda = 0 \Rightarrow 3+2\lambda = 0 \Rightarrow \lambda = -\frac{3}{2}$.Now,$\vec{a}+\vec{b}+\vec{c}=\vec{0} \Rightarrow \vec{a}+\vec{b}=-\vec{c}$.Taking cross product with $\vec{a}$: $\vec{a} 	imes (\vec{a}+\vec{b}) = \vec{a} 	imes (-\vec{c}) \Rightarrow \vec{a} 	imes \vec{b} = \vec{c} 	imes \vec{a}$.Similarly,taking cross product with $\vec{b}$: $(\vec{a}+\vec{b}) 	imes \vec{b} = (-\vec{c}) 	imes \vec{b} \Rightarrow \vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c}$.Thus,$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$.Therefore,$\vec{d} = \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b} = 3(\vec{a} 	imes \vec{b})$.The ordered pair is $\left(-\frac{3}{2}, 3 \vec{a} 	imes \vec{b}ight)$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

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