If the vector vec{b} = 3hat{j} + 4hat{k} is w | vedclass

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(B) Given $\vec{b} = 3\hat{j} + 4\hat{k}$ and $\vec{a} = \hat{i} + \hat{j}$.$\vec{b_1}$ is parallel to $\vec{a}$,so $\vec{b_1} = 	ext{proj}_{\vec{a}}

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$6\hat{i} - 6\hat{j} + \frac{9}{2}\hat{k}$

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(B) Given $\vec{b} = 3\hat{j} + 4\hat{k}$ and $\vec{a} = \hat{i} + \hat{j}$.$\vec{b_1}$ is parallel to $\vec{a}$,so $\vec{b_1} = 	ext{proj}_{\vec{a}} \vec{b} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} ight) \vec{a}$.$\vec{b} \cdot \vec{a} = (3\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j}) = 3$.$|\vec{a}|^2 = |\hat{i} + \hat{j}|^2 = 1^2 + 1^2 = 2$.Thus,$\vec{b_1} = \frac{3}{2}(\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}$.Since $\vec{b} = \vec{b_1} + \vec{b_2}$,we have $\vec{b_2} = \vec{b} - \vec{b_1} = (3\hat{j} + 4\hat{k}) - (\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}) = -\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} + 4\hat{k}$.Now,calculate the cross product $\vec{b_1} 	imes \vec{b_2}$:$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3/2 & 3/2 & 0 \ -3/2 & 3/2 & 4 \end{vmatrix}$.$= \hat{i} \left( (3/2)(4) - (0)(3/2) ight) - \hat{j} \left( (3/2)(4) - (0)(-3/2) ight) + \hat{k} \left( (3/2)(3/2) - (3/2)(-3/2) ight)$.$= \hat{i}(6) - \hat{j}(6) + \hat{k}(9/4 + 9/4) = 6\hat{i} - 6\hat{j} + \frac{9}{2}\hat{k}$.

If the vector $\vec{b} = 3\hat{j} + 4\hat{k}$ is written as the sum of a vector $\vec{b_1}$,parallel to $\vec{a} = \hat{i} + \hat{j}$ and a vector $\vec{b_2}$,perpendicular to $\vec{a}$,then $\vec{b_1} \times \vec{b_2}$ is equal to

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