Let vec{a}=-hat{i}-hat{j}+hat{k},vec{a} cdot | vedclass

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(B) Given $\vec{a} = -\hat{i} - \hat{j} + \hat{k}$,$\vec{a} \cdot \vec{b} = 1$,and $\vec{a} 	imes \vec{b} = \hat{i} - \hat{j}$.Taking the cross produ

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$3(\hat{i}+\hat{j}+\hat{k})$

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(B) Given $\vec{a} = -\hat{i} - \hat{j} + \hat{k}$,$\vec{a} \cdot \vec{b} = 1$,and $\vec{a} 	imes \vec{b} = \hat{i} - \hat{j}$.Taking the cross product of $\vec{a}$ with $\vec{a} 	imes \vec{b} = \hat{i} - \hat{j}$:$\vec{a} 	imes (\vec{a} 	imes \vec{b}) = \vec{a} 	imes (\hat{i} - \hat{j})$Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$:$(\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -1 & 1 \ 1 & -1 & 0 \end{vmatrix}$Since $\vec{a} \cdot \vec{b} = 1$ and $\vec{a} \cdot \vec{a} = (-1)^2 + (-1)^2 + 1^2 = 3$:$1(\vec{a}) - 3\vec{b} = \hat{i}(0 - (-1)) - \hat{j}(0 - 1) + \hat{k}(1 - (-1))$$\vec{a} - 3\vec{b} = \hat{i} + \hat{j} + 2\hat{k}$We need to find $\vec{a} - 6\vec{b}$.From $\vec{a} - 3\vec{b} = \hat{i} + \hat{j} + 2\hat{k}$,we have $3\vec{b} = \vec{a} - (\hat{i} + \hat{j} + 2\hat{k})$.Substituting $\vec{a} = -\hat{i} - \hat{j} + \hat{k}$:$3\vec{b} = -\hat{i} - \hat{j} + \hat{k} - \hat{i} - \hat{j} - 2\hat{k} = -2\hat{i} - 2\hat{j} - \hat{k}$.Then $6\vec{b} = -4\hat{i} - 4\hat{j} - 2\hat{k}$.Finally,$\vec{a} - 6\vec{b} = (-\hat{i} - \hat{j} + \hat{k}) - (-4\hat{i} - 4\hat{j} - 2\hat{k}) = 3\hat{i} + 3\hat{j} + 3\hat{k} = 3(\hat{i} + \hat{j} + \hat{k})$.

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}$,$\vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$. Then $\vec{a}-6 \vec{b}$ is equal to

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