# Thread: Line of intersection of two planes

1. ## Line of intersection of two planes

How would you find the line of intersection of the planes:

$\displaystyle \mathbf{x}\cdot\begin{pmatrix}1\\-1\\3\end{pmatrix}=0$ and $\displaystyle \mathbf{x}=\lambda\begin{pmatrix}2\\1\\2\end{pmatr ix}+\mu\begin{pmatrix}3\\1\\-3\end{pmatrix}$

2. We have:

$\displaystyle \mathbf{x}\cdot\begin{pmatrix}1\\-1\\3\end{pmatrix}=0$ (1)

$\displaystyle \mathbf{x}=\lambda\begin{pmatrix}2\\1\\2\end{pmatr ix}+\mu\begin{pmatrix}3\\1\\-3\end{pmatrix}$ (2)

If x lies on the intersection, then we must have that x satisfies both (1) and (2)

So.
$\displaystyle \left(\lambda\begin{pmatrix}2\\1\\2\end{pmatrix}+\ mu\begin{pmatrix}3\\1\\-3\end{pmatrix}\right) \cdot \begin{pmatrix}1\\-1\\3\end{pmatrix} = 0$
$\displaystyle \Rightarrow \lambda\begin{pmatrix}2\\1\\2\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\3\end{pmatrix} +\mu\begin{pmatrix}3\\1\\-3\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\3\end{pmatrix} = 0$
$\displaystyle \Rightarrow 7\lambda + -7\mu = 0 \Rightarrow \lambda = \mu$
Which determines a line in $\displaystyle \mathbb{R}^2$

3. Ref attachment, you will know how to do

To introduce new ***

Cross product could be replaced sometimes, $\displaystyle im(A)^{\perp}=ker(A^T)$

For example:

Plane 2 ($\displaystyle V_2$) equation $\displaystyle \mathbf{x}=\lambda\begin{pmatrix}2\\1\\2\end{pmatr ix}+\mu\begin{pmatrix}3\\1\\-3\end{pmatrix}$

It means $\displaystyle V_{2}=im(\begin{bmatrix}2\\1\\2\end{bmatrix},\begi n{bmatrix}3\\1\\-3\end{bmatrix})$

Normal vector of Plane 2 is $\displaystyle \vec{v}_2=V_{2}^{\perp}=im(A)^{\perp}=ker(A^T)$

By this method , we could skip cross product especially when dimension is more than 3

Finally, line cross original point, equation

$\displaystyle \frac{x}{5}=\frac{y}{2}=\frac{z}{-1}$

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