Let bar{a} = 3 hat{i} + 2 hat{j} + x hat{k} a | vedclass

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(B) First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & x \ 1

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$r \geq 5 \sqrt{\frac{3}{2}}$

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(B) First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & x \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(2 + x) - \hat{j}(3 - x) + \hat{k}(-3 - 2) = (2 + x) \hat{i} + (x - 3) \hat{j} - 5 \hat{k}$.Now,find the magnitude $r = |\vec{a} 	imes \vec{b}| = \sqrt{(2 + x)^2 + (x - 3)^2 + (-5)^2}$.Expand the squares: $r = \sqrt{(4 + 4x + x^2) + (x^2 - 6x + 9) + 25} = \sqrt{2x^2 - 2x + 38}$.Complete the square for the expression inside the square root:$2x^2 - 2x + 38 = 2(x^2 - x) + 38 = 2(x^2 - x + \frac{1}{4} - \frac{1}{4}) + 38 = 2(x - \frac{1}{2})^2 - \frac{1}{2} + 38 = 2(x - \frac{1}{2})^2 + \frac{75}{2}$.Since $(x - \frac{1}{2})^2 \geq 0$,the minimum value of $r^2$ is $\frac{75}{2}$.Thus,$r \geq \sqrt{\frac{75}{2}} = \sqrt{\frac{25 	imes 3}{2}} = 5 \sqrt{\frac{3}{2}}$.Therefore,the condition for $r$ is $r \geq 5 \sqrt{\frac{3}{2}}$.

Let $\bar{a} = 3 \hat{i} + 2 \hat{j} + x \hat{k}$ and $\bar{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\bar{a} \times \bar{b}| = r$ is possible,if

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