If vec{a} = -hat{i} + hat{j} + hat{k} and vec | vedclass

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(A) Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$,and $\vec{c} \perp \vec{b}$,$\vec{c}$ must be parallel to $\vec{b} 	imes (\vec{a} 	imes

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$-\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$

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(A) Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$,and $\vec{c} \perp \vec{b}$,$\vec{c}$ must be parallel to $\vec{b} 	imes (\vec{a} 	imes \vec{b})$.Let $\vec{\alpha} = \vec{b} 	imes (\vec{a} 	imes \vec{b})$.Using the vector triple product formula $\vec{b} 	imes (\vec{a} 	imes \vec{b}) = (\vec{b} \cdot \vec{b})\vec{a} - (\vec{b} \cdot \vec{a})\vec{b}$.Calculate $\vec{b} \cdot \vec{b} = 2^2 + 0^2 + 1^2 = 5$.Calculate $\vec{b} \cdot \vec{a} = (2)(-1) + (0)(1) + (1)(1) = -2 + 0 + 1 = -1$.Thus,$\vec{\alpha} = 5(-\hat{i} + \hat{j} + \hat{k}) - (-1)(2\hat{i} + 0\hat{j} + \hat{k}) = -5\hat{i} + 5\hat{j} + 5\hat{k} + 2\hat{i} + \hat{k} = -3\hat{i} + 5\hat{j} + 6\hat{k}$.Since $\vec{c} = \lambda \vec{\alpha}$,we have $\vec{c} = \lambda(-3\hat{i} + 5\hat{j} + 6\hat{k})$.Given $\vec{a} \cdot \vec{c} = 7$,we substitute $\vec{a}$ and $\vec{c}$:$(-\hat{i} + \hat{j} + \hat{k}) \cdot \lambda(-3\hat{i} + 5\hat{j} + 6\hat{k}) = 7$.$\lambda(3 + 5 + 6) = 7 \implies 14\lambda = 7 \implies \lambda = \frac{1}{2}$.Therefore,$\vec{c} = \frac{1}{2}(-3\hat{i} + 5\hat{j} + 6\hat{k}) = -\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$.

If $\vec{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 0\hat{j} + \hat{k}$,find a vector $\vec{c}$ satisfying the following conditions:
$(i)$ $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$.
$(ii)$ $\vec{c}$ is perpendicular to $\vec{b}$.
$(iii)$ $\vec{a} \cdot \vec{c} = 7$.

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If $\vec{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 0\hat{j} + \hat{k}$,find a vector $\vec{c}$ satisfying the following conditions:$(i)$ $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$.$(ii)$ $\vec{c}$ is perpendicular to $\vec{b}$.$(iii)$ $\vec{a} \cdot \vec{c} = 7$.