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

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(D) Given $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$.First,calculate the cross product $\vec{a} 	imes \vec{b}$

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$15$

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(D) Given $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 1 \ 3 & 2 & 5 \end{vmatrix} = \hat{i}(-15-2) - \hat{j}(10-3) + \hat{k}(4+9) = -17 \hat{i}-7 \hat{j}+13 \hat{k}$.We are given $(\vec{a}-\vec{c}) 	imes \vec{b} = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.Using the distributive property of the cross product,we have $(\vec{a} 	imes \vec{b}) - (\vec{c} 	imes \vec{b}) = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.Since $\vec{c} 	imes \vec{b} = -(\vec{b} 	imes \vec{c})$,we have $(\vec{a} 	imes \vec{b}) + (\vec{b} 	imes \vec{c}) = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.Substituting $\vec{a} 	imes \vec{b}$,we get $(-17 \hat{i}-7 \hat{j}+13 \hat{k}) + (\vec{b} 	imes \vec{c}) = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.Thus,$\vec{d} = \vec{b} 	imes \vec{c} = (-18 \hat{i}-3 \hat{j}+12 \hat{k}) - (-17 \hat{i}-7 \hat{j}+13 \hat{k}) = -\hat{i}+4 \hat{j}-\hat{k}$.Finally,calculate $\vec{a} \cdot \vec{d} = (2 \hat{i}-3 \hat{j}+\hat{k}) \cdot (-\hat{i}+4 \hat{j}-\hat{k}) = (2)(-1) + (-3)(4) + (1)(-1) = -2 - 12 - 1 = -15$.Therefore,$|\vec{a} \cdot \vec{d}| = |-15| = 15$.

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}$ and $\vec{a} \cdot \vec{c}=3$. If $\vec{b} \times \vec{c}=\vec{d}$,then $|\vec{a} \cdot \vec{d}|$ is equal to:

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