## The intersection of two planes is always a line

If 2 planes crossing each other, the intersection will constantly be a line.

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The vector equation because that the heat of intersection is given by

???r=r_0+tv???

where ???r_0??? is a point on the line and ???v??? is the vector an outcome of the overcome product that the common vectors that the two planes.

The parametric equations because that the line of intersection are offered by

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

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

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## Finding the parametric equations that represent the line of intersection of 2 planes

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

**Example**

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

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

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

We need to uncover the vector equation that the line of intersection. In stimulate to get it, we’ll require to an initial find ???v???, the cross product that the normal vectors that the provided planes.

The normal vectors because that the plane are

For the airplane ???2x+y-z=3???, the typical vector is ???alangle2,1,-1 angle???

For the plane ???x-y+z=3???, the typical vector is ???blangle1,-1,1 angle???

The overcome product that the common vectors is

We also need a suggest on the heat of intersection. To obtain it, we’ll use the equations that the offered planes together a system of linear equations. If we set ???z=0??? in both equations, us 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 uncover the line of intersection, very first find a allude on the line, and the overcome product that the common vectors

Now we’ll add the equations together.

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

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

???x=2???

Plugging ???x=2??? earlier 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=2old i-old j+0old k???

???r_0=langle 2,-1,0 angle???

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

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???r=r_0+tv???

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

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

???r=2old i-old j-3old jt-3old kt???

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

With the vector equation because that the heat of intersection in hand, we can find the parametric equations because that the exact same line. Matching up ???r=aold i+bold j+cold k??? with our vector equation ???r=(2)old i+(-1-3t)old j+(-3t)old k???, we have the right to say that

???a=2???

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

???c=-3t???

Therefore, the parametric equations because that the heat of intersection are

???x=2???

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

???z=-3t???

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