## The intersection of 2 planes is always a line

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

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

???r=r_0+tv???

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

The parametric equations for the line of intersection are provided by

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

where ???a???, ???b??? and also ???c??? space the coefficients from the vector equation ???r=aold i+bold j+cold k???.

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## Example difficulty of how to uncover the heat where two planes intersect, in parametric for

Example

Find the parametric equations because that the line of intersection that the planes.

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

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

We need to uncover the vector equation of the heat of intersection. In bespeak to acquire it, we’ll need to an initial find ???v???, the cross product the the common vectors that the offered planes.

The typical vectors because that the plane are

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

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

The overcome product the the typical vectors is

We likewise need a point on the heat of intersection. To gain it, we’ll usage the equations of the given planes as a system of direct 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 heat of intersection, first find a allude on the line, and also the overcome 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 allude on the heat of intersection is

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

???r_0=2old i-old j+0old k???

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

Now fine plug ???v??? and ???r_0??? right into the vector equation.

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???r=r_0+tv???

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

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

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

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

With the vector equation for the line of intersection in hand, we can uncover the parametric equations for the same line. Corresponding up ???r=aold i+bold j+cold k??? v 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 for the heat of intersection are

???x=2???

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

???z=-3t???

For integrals comprise exponential functions, try using the strength for the substitution.