Let vec{a}=hat{i}-2hat{j}+3hat{k}, vec{b}=2ha | vedclass

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(D) Given $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, and $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$.First, calculate $\vec

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$12$

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(D) Given $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, and $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$.First, calculate $\vec{v} = \vec{a} 	imes \vec{b}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(-1-6) + \hat{k}(1+4) = -\hat{i} + 7\hat{j} + 5\hat{k}$.Given $\vec{v} \cdot \vec{c} = 11$, so $(-\hat{i} + 7\hat{j} + 5\hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \hat{k}) = 11$.$-\lambda + 7 + 5 = 11 \Rightarrow -\lambda + 12 = 11 \Rightarrow \lambda = 1$.Now, $\vec{c} = \hat{i} + \hat{j} + \hat{k}$. The length of the projection of $\vec{b}$ on $\vec{c}$ is $p = \left| \vec{b} \cdot \frac{\vec{c}}{|\vec{c}|} ight|$.$|\vec{c}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.$p = \left| (2\hat{i} + \hat{j} - \hat{k}) \cdot \frac{(\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}} ight| = \left| \frac{2 + 1 - 1}{\sqrt{3}} ight| = \frac{2}{\sqrt{3}}$.Therefore, $9p^2 = 9 	imes \left( \frac{2}{\sqrt{3}} ight)^2 = 9 	imes \frac{4}{3} = 12$.

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}\times\vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^{2}$ is equal to:

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