Let overrightarrow{c} be a vector perpendicul | vedclass

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(B) Since $\overrightarrow{c}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b},$ it must be parallel to their cross product $\ov

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(B) Since $\overrightarrow{c}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b},$ it must be parallel to their cross product $\overrightarrow{a}	imes\overrightarrow{b}.$ Thus,we can write $\overrightarrow{c}=\lambda(\overrightarrow{a}	imes\overrightarrow{b})$ for some scalar $\lambda.$First,calculate $\overrightarrow{a}	imes\overrightarrow{b}$:$\overrightarrow{a}	imes\overrightarrow{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1 - (-2)) - \hat{j}(1 - (-1)) + \hat{k}(2 - 1) = 3\hat{i}-2\hat{j}+\hat{k}.$Given $\overrightarrow{c}\cdot(\hat{i}+\hat{j}+3\hat{k})=8,$ substitute $\overrightarrow{c}=\lambda(3\hat{i}-2\hat{j}+\hat{k})$:$\lambda(3\hat{i}-2\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+3\hat{k})=8$$\lambda(3(1) + (-2)(1) + (1)(3)) = 8$$\lambda(3 - 2 + 3) = 8 \Rightarrow 4\lambda = 8 \Rightarrow \lambda = 2.$Now,calculate $\overrightarrow{c}\cdot(\overrightarrow{a}	imes\overrightarrow{b})$:$\overrightarrow{c}\cdot(\overrightarrow{a}	imes\overrightarrow{b}) = \lambda(\overrightarrow{a}	imes\overrightarrow{b})\cdot(\overrightarrow{a}	imes\overrightarrow{b}) = \lambda|\overrightarrow{a}	imes\overrightarrow{b}|^2.$$|\overrightarrow{a}	imes\overrightarrow{b}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14.$Therefore,$\overrightarrow{c}\cdot(\overrightarrow{a}	imes\overrightarrow{b}) = 2 	imes 14 = 28.$

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