Let vec{a}=2 hat{i}+5 hat{j}-hat{k},vec{b}=2 | vedclass

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(B) Given $\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$.Let $\vec{v} = \vec{a} + \vec{b} + \hat{i} = (2+2+1

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$5$

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(B) Given $\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$.Let $\vec{v} = \vec{a} + \vec{b} + \hat{i} = (2+2+1)\hat{i} + (5-2)\hat{j} + (-1+2)\hat{k} = 5\hat{i} + 3\hat{j} + \hat{k}$.Let $\vec{p} = \vec{c} + \hat{i}$.The given equation is $\vec{p} 	imes \vec{v} = \vec{a} 	imes \vec{p}$.This implies $\vec{p} 	imes \vec{v} + \vec{p} 	imes \vec{a} = \vec{0}$,so $\vec{p} 	imes (\vec{v} + \vec{a}) = \vec{0}$.Thus,$\vec{p} = \lambda(\vec{v} + \vec{a})$ for some scalar $\lambda$.$\vec{v} + \vec{a} = (5\hat{i} + 3\hat{j} + \hat{k}) + (2\hat{i} + 5\hat{j} - \hat{k}) = 7\hat{i} + 8\hat{j}$.So,$\vec{c} + \hat{i} = \lambda(7\hat{i} + 8\hat{j}) \Rightarrow \vec{c} = 7\lambda\hat{i} + 8\lambda\hat{j} - \hat{i} = (7\lambda - 1)\hat{i} + 8\lambda\hat{j}$.Given $\vec{a} \cdot \vec{c} = -29$,we have $(2\hat{i} + 5\hat{j} - \hat{k}) \cdot ((7\lambda - 1)\hat{i} + 8\lambda\hat{j}) = -29$.$2(7\lambda - 1) + 5(8\lambda) = -29 \Rightarrow 14\lambda - 2 + 40\lambda = -29 \Rightarrow 54\lambda = -27 \Rightarrow \lambda = -\frac{1}{2}$.Now,$\vec{c} = (7(-\frac{1}{2}) - 1)\hat{i} + 8(-\frac{1}{2})\hat{j} = -\frac{9}{2}\hat{i} - 4\hat{j}$.We need to find $\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k}) = (-\frac{9}{2}\hat{i} - 4\hat{j}) \cdot (-2\hat{i} + \hat{j} + \hat{k}) = (-\frac{9}{2})(-2) + (-4)(1) + (0)(1) = 9 - 4 = 5$.

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