A vector parallel to the line of intersection | vedclass

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Question

(A) The line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is perpendi

Vedclass · Accepted Answer

$-2 \hat{i}+7 \hat{j}+13 \hat{k}$

Vedclass · Answer

(A) The line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is perpendicular to each of the normal vectors $\bar{n}_1 = 3 \hat{i}-\hat{j}+\hat{k}$ and $\bar{n}_2 = \hat{i}+4 \hat{j}-2 \hat{k}$.Therefore,the line is parallel to the vector $\bar{n}_1 	imes \bar{n}_2$.Calculating the cross product:$\bar{n}_1 	imes \bar{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ 1 & 4 & -2 \end{vmatrix}$$= \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1))$$= \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1)$$= -2 \hat{i} + 7 \hat{j} + 13 \hat{k}$.

$A$ vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is

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