Let vec{a} = 3hat{i} + 2hat{j} + xhat{k} and | vedclass

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(A) 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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(A) First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & x \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(2 + x) - \hat{j}(3 - x) + \hat{k}(-3 - 2) = (x + 2)\hat{i} + (x - 3)\hat{j} - 5\hat{k}$.Next,find the magnitude $r = |\vec{a} 	imes \vec{b}|$:$r^2 = (x + 2)^2 + (x - 3)^2 + (-5)^2$$r^2 = (x^2 + 4x + 4) + (x^2 - 6x + 9) + 25$$r^2 = 2x^2 - 2x + 38 = 2(x^2 - x + 19)$.Complete the square for the expression inside the parenthesis:$r^2 = 2\left((x - \frac{1}{2})^2 + 19 - \frac{1}{4}ight) = 2\left((x - \frac{1}{2})^2 + \frac{75}{4}ight) = 2(x - \frac{1}{2})^2 + \frac{75}{2}$.Since $(x - \frac{1}{2})^2 \geq 0$,the minimum value of $r^2$ is $\frac{75}{2}$.Therefore,$r^2 \geq \frac{75}{2} \implies r \geq \sqrt{\frac{75}{2}} = \sqrt{\frac{25 	imes 3}{2}} = 5\sqrt{\frac{3}{2}}$.Thus,the condition is $r \geq 5\sqrt{\frac{3}{2}}$.

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

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