If vec{a} = hat{i} - 2hat{j} + 2hat{k} and ve | vedclass

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(B) The component of $\vec{b}$ perpendicular to $\vec{a}$ is given by the formula $\vec{b}_{\perp} = \vec{b} - 	ext{proj}_{\vec{a}} \vec{b} = \vec{b}

Vedclass · Accepted Answer

$\frac{1}{9}(8\hat{i} - 7\hat{j} - 11\hat{k})$

Vedclass · Answer

(B) The component of $\vec{b}$ perpendicular to $\vec{a}$ is given by the formula $\vec{b}_{\perp} = \vec{b} - 	ext{proj}_{\vec{a}} \vec{b} = \vec{b} - \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} ight) \vec{a}$.
Given $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$.
First, calculate the dot product: $\vec{b} \cdot \vec{a} = (2)(1) + (-3)(-2) + (1)(2) = 2 + 6 + 2 = 10$.
Next, calculate the magnitude squared of $\vec{a}$: $|\vec{a}|^2 = (1)^2 + (-2)^2 + (2)^2 = 1 + 4 + 4 = 9$.
Now, the projection of $\vec{b}$ onto $\vec{a}$ is $\frac{10}{9}(\hat{i} - 2\hat{j} + 2\hat{k})$.
Finally, the perpendicular component is $\vec{b}_{\perp} = (2\hat{i} - 3\hat{j} + \hat{k}) - \frac{10}{9}(\hat{i} - 2\hat{j} + 2\hat{k}) = \frac{1}{9}(18\hat{i} - 27\hat{j} + 9\hat{k} - 10\hat{i} + 20\hat{j} - 20\hat{k}) = \frac{1}{9}(8\hat{i} - 7\hat{j} - 11\hat{k})$.

If $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$, then the component of $\vec{b}$ perpendicular to $\vec{a}$ is

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