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

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(A) Given $\vec{a} 	imes \vec{b} = \vec{d} 	imes \vec{b}$,we can write $(\vec{a} - \vec{d}) 	imes \vec{b} = 0$.This implies that $\vec{a} - \vec{d}

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$6$

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(A) Given $\vec{a} 	imes \vec{b} = \vec{d} 	imes \vec{b}$,we can write $(\vec{a} - \vec{d}) 	imes \vec{b} = 0$.This implies that $\vec{a} - \vec{d}$ is parallel to $\vec{b}$,so $\vec{a} - \vec{d} = \lambda \vec{b}$ for some scalar $\lambda$.Thus,$\vec{d} = \vec{a} - \lambda \vec{b}$.Taking the dot product with $\vec{c}$ on both sides,we get $\vec{d} \cdot \vec{c} = \vec{a} \cdot \vec{c} - \lambda (\vec{b} \cdot \vec{c})$.Given $\vec{d} \cdot \vec{c} = 8$,$\vec{a} \cdot \vec{c} = (2)(1) + (3)(1) + (1)(1) = 6$,and $\vec{b} \cdot \vec{c} = (1)(1) + (-1)(1) + (1)(1) = 1$.Substituting these values: $8 = 6 - \lambda(1) \Rightarrow \lambda = -2$.Therefore,$\vec{d} = \vec{a} + 2\vec{b} = (2\hat{i} + 3\hat{j} + \hat{k}) + 2(\hat{i} - \hat{j} + \hat{k}) = 4\hat{i} + \hat{j} + 3\hat{k}$.Finally,$\vec{d} \cdot \vec{b} = (4\hat{i} + \hat{j} + 3\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = (4)(1) + (1)(-1) + (3)(1) = 4 - 1 + 3 = 6$.

If $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,$\vec{c} = \hat{i} + \hat{j} + \hat{k}$ and let $\vec{d}$ be such that $\vec{a} \times \vec{b} = \vec{d} \times \vec{b}$ and $\vec{d} \cdot \vec{c} = 8$,then the value of $\vec{d} \cdot \vec{b}$ is:

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