Mar 12, 2018

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

Explanation:

Since ${\vec{b}}_{1}$ and ${\vec{b}}_{2}$ are parallel and perpendicular to $\vec{a}$, respectively, we have ${\vec{b}}_{2} \cdot \vec{a} = 0$ and

$| {\vec{b}}_{1} | = \frac{{\vec{b}}_{1} \cdot \vec{a}}{|} \vec{a} | = \frac{\vec{b} \cdot \vec{a}}{|} \vec{a} |$

So that

${\vec{b}}_{1} = \frac{\vec{b} \cdot \vec{a}}{|} \vec{a} | \frac{\vec{a}}{|} \vec{a} | = \frac{\left(\vec{b} \cdot \vec{a}\right) \vec{a}}{|} \vec{a} {|}^{2}$

and

${\vec{b}}_{2} = \vec{b} - {\vec{b}}_{1} = \vec{b} - \frac{\left(\vec{b} \cdot \vec{a}\right) \vec{a}}{|} \vec{a} {|}^{2}$

So

${\vec{b}}_{1} \times {\vec{b}}_{2} = {\vec{b}}_{1} \times \left(\vec{b} - {\vec{b}}_{1}\right) = {\vec{b}}_{1} \times \vec{b} = \frac{\left(\vec{b} \cdot \vec{a}\right) \left(\vec{a} \times \vec{b}\right)}{|} \vec{a} {|}^{2}$

Now, given $\vec{b} = 3 \hat{j} + 4 \hat{k}$ and $\vec{a} = \hat{i} + \hat{j}$, we have

$\vec{b} \cdot \vec{a} = 3$
$\vec{a} \times \vec{b} = \left(\hat{i} + \hat{j}\right) \times \left(3 \hat{j} + 4 \hat{k}\right) = 3 \left(\hat{i} \times \hat{j}\right) + 4 \left(\hat{i} \times \hat{k}\right) + 4 \hat{j} \times \hat{k} = 3 \hat{k} - 4 \hat{j} + 4 \hat{i}$
$| \vec{a} {|}^{2} = 2$

So that

${\vec{b}}_{1} \times {\vec{b}}_{2} = \frac{3 \left(4 \hat{i} - 4 \hat{j} + 3 \hat{k}\right)}{2} = 6 \hat{i} - 6 \hat{j} + \frac{9}{2} \hat{k}$