# Parametric equations for the intersection of planes

## The intersection of two planes is always a line

If two planes intersect each other, the intersection will always be a line.

The vector equation for the line of intersection is given by

???r=r_0+tv???

where ???r_0??? is a point on the line and ???v??? is the vector result of the cross product of the normal vectors of the two planes.

The parametric equations for the line of intersection are given by

???x=a???, ???y=b???, and ???z=c???

where ???a???, ???b??? and ???c??? are the coefficients from the vector equation ???r=a\bold i+b\bold j+c\bold k???.

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## Finding the parametric equations that represent the line of intersection of two planes

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## Example problem of how to find the line where two planes intersect, in parametric for

**Example**

Find the parametric equations for the line of intersection of the planes.

???2x+y-z=3???

???x-y+z=3???

We need to find the vector equation of the line of intersection. In order to get it, we’ll need to first find ???v???, the cross product of the normal vectors of the given planes.

The normal vectors for the planes are

For the plane ???2x+y-z=3???, the normal vector is ???a\langle2,1,-1\rangle???

For the plane ???x-y+z=3???, the normal vector is ???b\langle1,-1,1\rangle???

The cross product of the normal vectors is

???v=|a\times b|=\begin{vmatrix}\bold i&\bold j&\bold k\\2&1&-1\\1&-1&1\end{vmatrix}???

???v=|a\times b|=\bold i\begin{vmatrix}1&-1\\-1&1\end{vmatrix}-\bold j\begin{vmatrix}2&-1\\1&1\end{vmatrix}+\bold k\begin{vmatrix}2&1\\1&-1\end{vmatrix}???

???v=|a\times b|=\left[(1)(1)-(-1)(-1)\right]\bold i-\left[(2)(1)-(-1)(1)\right]\bold j+\left[(2)(-1)-(1)(1)\right]\bold k???

???v=|a\times b|=(1-1)\bold i-(2+1)\bold j+(-2-1)\bold k???

???v=|a\times b|=0\bold i-3\bold j-3\bold k???

???v=|a\times b|=\langle0,-3,-3\rangle???

We also need a point on the line of intersection. To get it, we’ll use the equations of the given planes as a system of linear equations. If we set ???z=0??? in both equations, we get

???2x+y-z=3???

???2x+y-0=3???

???2x+y=3???

and

???x-y+z=3???

???x-y+0=3???

???x-y=3???

To find the line of intersection, first find a point on the line, and the cross product of the normal vectors

Now we’ll add the equations together.

???(2x+x)+(y-y)=3+3???

???3x+0=6???

???x=2???

Plugging ???x=2??? back into ???x-y=3???, we get

???2-y=3???

???-y=1???

???y=-1???

Putting these values together, the point on the line of intersection is

???(2,-1,0)???

???r_0=2\bold i-\bold j+0\bold k???

???r_0=\langle 2,-1,0\rangle???

Now we’ll plug ???v??? and ???r_0??? into the vector equation.

???r=r_0+tv???

???r=(2\bold i-\bold j+0\bold k)+t(0\bold i-3\bold j-3\bold k)???

???r=2\bold i-\bold j+0\bold k+0\bold it-3\bold jt-3\bold kt???

???r=2\bold i-\bold j-3\bold jt-3\bold kt???

???r=(2)\bold i+(-1-3t)\bold j+(-3t)\bold k???

With the vector equation for the line of intersection in hand, we can find the parametric equations for the same line. Matching up ???r=a\bold i+b\bold j+c\bold k??? with our vector equation ???r=(2)\bold i+(-1-3t)\bold j+(-3t)\bold k???, we can say that

???a=2???

???b=-1-3t???

???c=-3t???

Therefore, the parametric equations for the line of intersection are

???x=2???

???y=-1-3t???

???z=-3t???

For integrals containing exponential functions, try using the power for the substitution.