The line of intersection of the planes vec{r} | vedclass

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Question

(A) The line of intersection of the planes $\vec{r} \cdot \vec{n_1} = d_1$ and $\vec{r} \cdot \vec{n_2} = d_2$ is parallel to the vector $\vec{v} = \v

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The line of intersection of the planes $\vec{r} \cdot \vec{n_1} = d_1$ and $\vec{r} \cdot \vec{n_2} = d_2$ is parallel to the vector $\vec{v} = \vec{n_1} 	imes \vec{n_2}$.Given $\vec{n_1} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + 4\hat{j} - 2\hat{k}$.Calculating the cross product:$\vec{v} = \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}$.

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

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