The line of intersection of the planes vec r | vedclass

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Question

(C) The direction vector $\vec v$ of the line of intersection is given by the cross product of the normals $\vec n_1 = 3\hat i - \hat j + \hat k$ and

Vedclass · Accepted Answer

$\frac{x - 6/13}{2} = \frac{y - 5/13}{-7} = \frac{z}{-13}$

Vedclass · Answer

(C) The direction vector $\vec v$ of the line of intersection is given by the cross product of the normals $\vec n_1 = 3\hat i - \hat j + \hat k$ and $\vec n_2 = \hat i + 4\hat j - 2\hat k$.$\vec v = \vec n_1 	imes \vec n_2 = \begin{vmatrix} \hat i & \hat j & \hat k \ 3 & -1 & 1 \ 1 & 4 & -2 \end{vmatrix} = \hat i(2 - 4) - \hat j(-6 - 1) + \hat k(12 + 1) = -2\hat i + 7\hat j + 13\hat k$.To find a point on the line,set $z = 0$ in the plane equations:$3x - y = 1$ and $x + 4y = 2$.Multiplying the first by $4$: $12x - 4y = 4$. Adding to the second: $13x = 6 \Rightarrow x = 6/13$.Substituting $x$: $6/13 + 4y = 2 \Rightarrow 4y = 2 - 6/13 = 20/13 \Rightarrow y = 5/13$.The point is $(6/13, 5/13, 0)$.The line equation is $\frac{x - 6/13}{-2} = \frac{y - 5/13}{7} = \frac{z}{13}$,which is equivalent to $\frac{x - 6/13}{2} = \frac{y - 5/13}{-7} = \frac{z}{-13}$.

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:

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