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

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(B) Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}+\hat{k}$, $\vec{c}=\hat{i}+3\hat{j}-2\hat{k}$, and $\vec{d}=2\hat{i

Vedclass · Accepted Answer

$-63$

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(B) Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}+\hat{k}$, $\vec{c}=\hat{i}+3\hat{j}-2\hat{k}$, and $\vec{d}=2\hat{i}+\hat{j}-\hat{k}$.First, calculate $l = \vec{b} \cdot \vec{c} = (1)(1) + (-2)(3) + (1)(-2) = 1 - 6 - 2 = -7$.Next, calculate $m = \vec{b} \cdot \vec{a} = (1)(1) + (-2)(1) + (1)(1) = 1 - 2 + 1 = 0$.We need to evaluate the scalar triple product $[(m\vec{b}+l\vec{a}) \quad \vec{b} \quad \vec{d}]$.Substituting $m=0$ and $l=-7$, the expression becomes $[(0\vec{b} + (-7)\vec{a}) \quad \vec{b} \quad \vec{d}] = [-7\vec{a} \quad \vec{b} \quad \vec{d}]$.By the properties of the scalar triple product, this is equal to $-7 [\vec{a} \quad \vec{b} \quad \vec{d}]$.Now, calculate the scalar triple product $[\vec{a} \quad \vec{b} \quad \vec{d}] = \vec{a} \cdot (\vec{b} 	imes \vec{d})$.$\vec{b} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(-1-2) + \hat{k}(1+4) = \hat{i} + 3\hat{j} + 5\hat{k}$.Then, $\vec{a} \cdot (\vec{b} 	imes \vec{d}) = (\hat{i}+\hat{j}+\hat{k}) \cdot (\hat{i} + 3\hat{j} + 5\hat{k}) = 1(1) + 1(3) + 1(5) = 1 + 3 + 5 = 9$.Finally, the value is $-7 	imes 9 = -63$.

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