Let vec{a}=-hat{i}+2hat{j}+2hat{k}, vec{b}=8h | vedclass

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(B) Given $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$.Let $\vec{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$.Since $\v

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$27$

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(B) Given $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$.Let $\vec{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$.Since $\vec{a}	imes\vec{c}=\vec{b}$, we have:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 2 \ c_{1} & c_{2} & c_{3} \end{vmatrix} = 8\hat{i}+7\hat{j}-3\hat{k}$$(2c_{3}-2c_{2})\hat{i} - (-c_{3}-2c_{1})\hat{j} + (-c_{2}-2c_{1})\hat{k} = 8\hat{i}+7\hat{j}-3\hat{k}$Comparing components:$2c_{3}-2c_{2}=8 \Rightarrow c_{3}-c_{2}=4 \Rightarrow c_{3}=c_{2}+4$$c_{3}+2c_{1}=7$$-c_{2}-2c_{1}=-3 \Rightarrow c_{2}+2c_{1}=3$Given $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$, so $c_{1}+c_{2}+c_{3}=4$.Substituting $c_{3}=c_{2}+4$: $c_{1}+c_{2}+c_{2}+4=4 \Rightarrow c_{1}+2c_{2}=0 \Rightarrow c_{1}=-2c_{2}$.Using $c_{2}+2c_{1}=3$: $c_{2}+2(-2c_{2})=3 \Rightarrow -3c_{2}=3 \Rightarrow c_{2}=-1$.Then $c_{1}=-2(-1)=2$ and $c_{3}=-1+4=3$.So $\vec{c}=2\hat{i}-\hat{j}+3\hat{k}$.Then $\vec{a}+\vec{c} = (-\hat{i}+2\hat{j}+2\hat{k}) + (2\hat{i}-\hat{j}+3\hat{k}) = \hat{i}+\hat{j}+5\hat{k}$.$|\vec{a}+\vec{c}|^{2} = 1^{2}+1^{2}+5^{2} = 1+1+25 = 27$.

Let $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$, $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a}\times\vec{c}=\vec{b}$. If $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$, then $|\vec{a}+\vec{c}|^{2}$ is equal to:

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