Let lambda in R,vec{a} = lambda hat{i} + 2 ha | vedclass

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(A) Given the expression $((\vec{a} + \vec{b}) 	imes (\vec{a} 	imes \vec{b})) 	imes (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$.Usin

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$140$

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(A) Given the expression $((\vec{a} + \vec{b}) 	imes (\vec{a} 	imes \vec{b})) 	imes (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$.Using the vector triple product identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,let $\vec{u} = \vec{a} + \vec{b}$,$\vec{v} = \vec{a} 	imes \vec{b}$,and $\vec{w} = \vec{a} - \vec{b}$.Since $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = (\lambda^2 + 4 + 9) - (1 + \lambda^2 + 4) = 8$,the expression simplifies to $8(\vec{a} 	imes \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$.Thus,$\vec{a} 	imes \vec{b} = \hat{i} - 5 \hat{j} - 3 \hat{k}$.Calculating $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \lambda & 2 & -3 \ 1 & -\lambda & 2 \end{vmatrix} = (4 - 3\lambda)\hat{i} - (2\lambda + 3)\hat{j} + (-\lambda^2 - 2)\hat{k}$.Comparing components: $4 - 3\lambda = 1 \Rightarrow \lambda = 1$. Check: $-(2(1) + 3) = -5$ and $-(1^2 + 2) = -3$. This matches.For $\lambda = 1$,$\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.Then $\vec{a} + \vec{b} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{a} - \vec{b} = 3\hat{j} - 5\hat{k}$.$(\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 0 & 3 & -5 \end{vmatrix} = (-5 + 3)\hat{i} - (-10 - 0)\hat{j} + (6 - 0)\hat{k} = -2\hat{i} + 10\hat{j} + 6\hat{k}$.Since $\lambda = 1$,we need $|1(-2\hat{i} + 10\hat{j} + 6\hat{k})|^2 = (-2)^2 + 10^2 + 6^2 = 4 + 100 + 36 = 140$.

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