Let vec alpha = 3hat i + hat j and vec beta = | vedclass

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(A) Given $\vec{\alpha} = 3\hat{i} + \hat{j}$ and $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}.$Since $\vec{\beta}_1$ is parallel to $\vec{\alpha},$ w

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$\frac{1}{2}(-3\hat i + 9\hat j + 5\hat k)$

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(A) Given $\vec{\alpha} = 3\hat{i} + \hat{j}$ and $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}.$Since $\vec{\beta}_1$ is parallel to $\vec{\alpha},$ we have $\vec{\beta}_1 = \lambda \vec{\alpha} = \lambda(3\hat{i} + \hat{j}).$Given $\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2,$ we have $\vec{\beta}_2 = \vec{\beta}_1 - \vec{\beta} = \lambda(3\hat{i} + \hat{j}) - (2\hat{i} - \hat{j} + 3\hat{k}) = (3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k}.$Since $\vec{\beta}_2$ is perpendicular to $\vec{\alpha},$ $\vec{\beta}_2 \cdot \vec{\alpha} = 0.$$((3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k}) \cdot (3\hat{i} + \hat{j}) = 0.$$3(3\lambda - 2) + 1(\lambda + 1) = 0 \implies 9\lambda - 6 + \lambda + 1 = 0 \implies 10\lambda = 5 \implies \lambda = \frac{1}{2}.$Thus,$\vec{\beta}_1 = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}$ and $\vec{\beta}_2 = (\frac{3}{2} - 2)\hat{i} + (\frac{1}{2} + 1)\hat{j} - 3\hat{k} = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}.$Now,$\vec{\beta}_1 	imes \vec{\beta}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \frac{3}{2} & \frac{1}{2} & 0 \ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix}.$$= \hat{i}(-\frac{3}{2} - 0) - \hat{j}(-\frac{9}{2} - 0) + \hat{k}(\frac{9}{4} - (-\frac{1}{4})) = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{10}{4}\hat{k} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k}.$$= \frac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k}).$

Let $\vec \alpha = 3\hat i + \hat j$ and $\vec \beta = 2\hat i - \hat j + 3\hat k.$ If $\vec \beta = \vec \beta _1 - \vec \beta _2,$ where $\vec \beta _1$ is parallel to $\vec \alpha$ and $\vec \beta _2$ is perpendicular to $\vec \alpha,$ then $\vec \beta _1 \times \vec \beta _2$ is equal to

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