Let vec{b}=hat{i}+hat{j}+lambda hat{k}, lambd | vedclass

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(C) Given $\vec{a} 	imes \vec{b} = 13 \hat{i} - \hat{j} - 4 \hat{k}$. Since $(\vec{a} 	imes \vec{b}) \cdot \vec{b} = 0$,we have $(13 \hat{i} - \hat{

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$14$

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(C) Given $\vec{a} 	imes \vec{b} = 13 \hat{i} - \hat{j} - 4 \hat{k}$. Since $(\vec{a} 	imes \vec{b}) \cdot \vec{b} = 0$,we have $(13 \hat{i} - \hat{j} - 4 \hat{k}) \cdot (\hat{i} + \hat{j} + \lambda \hat{k}) = 0$. $13 - 1 - 4\lambda = 0 \Rightarrow 4\lambda = 12 \Rightarrow \lambda = 3$. Thus,$\vec{b} = \hat{i} + \hat{j} + 3 \hat{k}$. Using the vector triple product identity $(\vec{a} 	imes \vec{b}) 	imes \vec{b} = (\vec{a} \cdot \vec{b}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{a}$,we have: $(\vec{a} 	imes \vec{b}) 	imes \vec{b} = (13 \hat{i} - \hat{j} - 4 \hat{k}) 	imes (\hat{i} + \hat{j} + 3 \hat{k}) = \hat{i}(-3+4) - \hat{j}(39+4) + \hat{k}(13+1) = \hat{i} - 43 \hat{j} + 14 \hat{k}$. Given $\vec{a} \cdot \vec{b} = -21$ and $|\vec{b}|^2 = 1^2 + 1^2 + 3^2 = 11$,we have $-21 \vec{b} - 11 \vec{a} = \hat{i} - 43 \hat{j} + 14 \hat{k}$. $-21(\hat{i} + \hat{j} + 3 \hat{k}) - 11 \vec{a} = \hat{i} - 43 \hat{j} + 14 \hat{k} \Rightarrow 11 \vec{a} = -22 \hat{i} + 22 \hat{j} - 77 \hat{k} \Rightarrow \vec{a} = -2 \hat{i} + 2 \hat{j} - 7 \hat{k}$. Now,calculate $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$: $\vec{b}-\vec{a} = (1 - (-2))\hat{i} + (1 - 2)\hat{j} + (3 - (-7))\hat{k} = 3 \hat{i} - \hat{j} + 10 \hat{k}$. $\vec{b}+\vec{a} = (1 - 2)\hat{i} + (1 + 2)\hat{j} + (3 - 7)\hat{k} = -\hat{i} + 3 \hat{j} - 4 \hat{k}$. $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j}) = (3 \hat{i} - \hat{j} + 10 \hat{k}) \cdot (0 \hat{i} - 1 \hat{j} + 1 \hat{k}) = 0 + 1 + 10 = 11$. $(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k}) = (-\hat{i} + 3 \hat{j} - 4 \hat{k}) \cdot (1 \hat{i} + 0 \hat{j} - 1 \hat{k}) = -1 + 0 + 4 = 3$. Sum $= 11 + 3 = 14$.

Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in R$. If $\vec{a}$ is a vector such that $\vec{a} \times \vec{b}=13 \hat{i}-\hat{j}-4 \hat{k}$ and $\vec{a} \cdot \vec{b}+21=0$,then $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$ is equal to

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