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(B) The formula for the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is given by $	ext{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot

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$\frac{1}{6}(-\hat{i} + 2\hat{j} - \hat{k})$

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(B) The formula for the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is given by $	ext{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} ight) \vec{b}$.First,calculate the dot product $\vec{a} \cdot \vec{b} = (2)(1) + (3)(-2) + (3)(1) = 2 - 6 + 3 = -1$.Next,calculate the square of the magnitude of $\vec{b}$,which is $|\vec{b}|^2 = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6$.Now,substitute these values into the formula:$	ext{proj}_{\vec{b}} \vec{a} = \frac{-1}{6} (\hat{i} - 2\hat{j} + \hat{k}) = \frac{1}{6} (-\hat{i} + 2\hat{j} - \hat{k})$.

The orthogonal projection vector of $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ on $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$ is

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The projection of vector $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ on vector $\vec{b} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ is . . . . . . .

Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=11 \hat{i}-\hat{j}+\hat{k}$,and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$,then $|\vec{c}|^2$ is equal to:

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