Let vec{u}=2 hat{i}-hat{j}+hat{k} and vec{v}= | vedclass

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(D) Given vectors are $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$.First,calculate the cross product $\vec{u} 	imes \vec{v}

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$\sqrt{17}$

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(D) Given vectors are $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$.First,calculate the cross product $\vec{u} 	imes \vec{v}$:$\vec{u} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 0 & -3 & 2 \end{vmatrix}$$= \hat{i}(-2 - (-3)) - \hat{j}(4 - 0) + \hat{k}(-6 - 0)$$= \hat{i}(1) - 4 \hat{j} - 6 \hat{k} = \hat{i} - 4 \hat{j} - 6 \hat{k}$.Since $\vec{w}$ is a unit vector in the $XY$-plane,it can be represented as $\vec{w} = \cos 	heta \hat{i} + \sin 	heta \hat{j}$.Now,calculate the dot product $|(\vec{u} 	imes \vec{v}) \cdot \vec{w}|$:$|(\hat{i} - 4 \hat{j} - 6 \hat{k}) \cdot (\cos 	heta \hat{i} + \sin 	heta \hat{j})| = |\cos 	heta - 4 \sin 	heta|$.The maximum value of an expression of the form $a \cos 	heta + b \sin 	heta$ is $\sqrt{a^2 + b^2}$.Here,$a = 1$ and $b = -4$.Maximum value $= \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$.

Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $XY$-plane. Then,the maximum value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is:

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