Let vec{a}=hat{i}-3 hat{j}+7 hat{k},vec{b}=2 | vedclass

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(D) Given $(\vec{a}+2 \vec{b}) 	imes \vec{c}=3(\vec{c} 	imes \vec{a})$.Since $\vec{c} 	imes \vec{a} = -(\vec{a} 	imes \vec{c})$,we have $(\vec{a}+

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$30$

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(D) Given $(\vec{a}+2 \vec{b}) 	imes \vec{c}=3(\vec{c} 	imes \vec{a})$.Since $\vec{c} 	imes \vec{a} = -(\vec{a} 	imes \vec{c})$,we have $(\vec{a}+2 \vec{b}) 	imes \vec{c} = -3(\vec{a} 	imes \vec{c})$.$(\vec{a}+2 \vec{b}) 	imes \vec{c} + 3(\vec{a} 	imes \vec{c}) = 0$.$(\vec{a} + 2\vec{b} + 3\vec{a}) 	imes \vec{c} = 0$.$(4\vec{a} + 2\vec{b}) 	imes \vec{c} = 0$.This implies $\vec{c}$ is parallel to $(4\vec{a} + 2\vec{b})$.Let $\vec{c} = \lambda(4\vec{a} + 2\vec{b})$.$4\vec{a} + 2\vec{b} = 4(\hat{i}-3\hat{j}+7\hat{k}) + 2(2\hat{i}-\hat{j}+\hat{k}) = (4+4)\hat{i} + (-12-2)\hat{j} + (28+2)\hat{k} = 8\hat{i} - 14\hat{j} + 30\hat{k}$.So,$\vec{c} = \lambda(8\hat{i} - 14\hat{j} + 30\hat{k})$.Given $\vec{a} \cdot \vec{c} = 130$.$(\hat{i}-3\hat{j}+7\hat{k}) \cdot \lambda(8\hat{i}-14\hat{j}+30\hat{k}) = 130$.$\lambda(8 + 42 + 210) = 130$.$260\lambda = 130 \implies \lambda = \frac{1}{2}$.Thus,$\vec{c} = 4\hat{i} - 7\hat{j} + 15\hat{k}$.Finally,$\vec{b} \cdot \vec{c} = (2\hat{i}-\hat{j}+\hat{k}) \cdot (4\hat{i}-7\hat{j}+15\hat{k}) = 8 + 7 + 15 = 30$.

Let $\vec{a}=\hat{i}-3 \hat{j}+7 \hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$. If $\vec{a} \cdot \vec{c}=130$,then $\vec{b} \cdot \vec{c}$ is equal to ....................

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