Let vec{A} = 2 hat{i} - 3 hat{j} + 4 hat{k} a | vedclass

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(B) Given vectors are $\vec{A} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = 4 \hat{i} + \hat{j} + 2 \hat{k}$.To find the cross product $\vec{A}

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$2 \sqrt{110}$

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(B) Given vectors are $\vec{A} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = 4 \hat{i} + \hat{j} + 2 \hat{k}$.To find the cross product $\vec{A} 	imes \vec{B}$,we use the determinant method:$\vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 4 \ 4 & 1 & 2 \end{vmatrix}$$= \hat{i} [(-3)(2) - (4)(1)] - \hat{j} [(2)(2) - (4)(4)] + \hat{k} [(2)(1) - (-3)(4)]$$= \hat{i} [-6 - 4] - \hat{j} [4 - 16] + \hat{k} [2 + 12]$$= -10 \hat{i} + 12 \hat{j} + 14 \hat{k}$Now,calculate the magnitude $|\vec{A} 	imes \vec{B}| = \sqrt{(-10)^2 + (12)^2 + (14)^2}$$= \sqrt{100 + 144 + 196}$$= \sqrt{440}$$= \sqrt{4 	imes 110} = 2 \sqrt{110}$.

Let $\vec{A} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = 4 \hat{i} + \hat{j} + 2 \hat{k}$,then $|\vec{A} \times \vec{B}|$ is equal to:

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