1.6 - The line where two planes meet

1.6 - The line where two planes meet

Two nonparallel planes meet along a line. You will find its direction from the plane normals, find a point that satisfies both equations, and combine the two pieces into a verified vector equation.

Finding the direction

Two walls meet along an edge. Why should crossing their normal vectors find the direction of that edge? Movement along the edge stays in both planes, so its direction is perpendicular to both normals.

For example, the planes x+y+z=4x+y+z=4 and x−y+z=2x-y+z=2 have normals (1,1,1)(1,1,1) and (1,−1,1)(1,-1,1). A cross product supplies precisely the direction we need.

Direction of a plane intersection

If n1×n2≠0\mathbf n_1\times\mathbf n_2\ne\mathbf0, a direction of the intersection line is

d=n1×n2.\mathbf d=\mathbf n_1\times\mathbf n_2.

Any nonzero scalar multiple of d\mathbf d describes the same direction. If the cross product is zero, the planes are parallel or coincident and there is no unique intersection line.

Not in the formula booklet: learn it.

The zero case requires a further check. When the normals are proportional, scale one entire equation to match the other's left-hand side; compare the constants to distinguish coincident planes from parallel separated planes.

Worked example 1: cross the normals

Question. Find a direction vector for the intersection of x+y+z=4x+y+z=4 and x−y+z=2x-y+z=2.

Plan. Cross their normals, then simplify the nonzero result.

n1×n2=(1,1,1)×(1,−1,1)=(2,0,−2)=2(1,0,−1).\begin{aligned} \mathbf n_1\times\mathbf n_2 &=(1,1,1)\times(1,-1,1)\\ &=(2,0,-2)=2(1,0,-1). \end{aligned}

Answer. One direction vector is (1,0,−1)(1,0,-1).

Check. Its scalar products with the two normals are 1+0−1=01+0-1=0 and 1−0−1=01-0-1=0. Algebraically, subtracting the plane equations fixes y=1y=1; varying xx and zz by opposite amounts preserves both equations, agreeing with the direction.

Change only the second plane's normal to a multiple of the first. Crossing the normals now diagnoses why the method cannot produce an intersection direction.

Worked example 2: parallel or coincident planes

Question. Compare x+y+z=4x+y+z=4 with each of 2x+2y+2z=92x+2y+2z=9 and 2x+2y+2z=82x+2y+2z=8.

Plan. Check that the normals are proportional, then compare the constants after matching their left-hand sides.

The normals are (1,1,1)(1,1,1) and (2,2,2)(2,2,2), whose cross product is zero. Doubling the first plane equation gives

2x+2y+2z=8.2x+2y+2z=8.

This contradicts the constant 99 in the first comparison, but exactly matches the constant 88 in the second.

Answer. The plane with constant 99 is parallel and separate; the plane with constant 88 is coincident with the original plane. Neither pair specifies one intersection line.

Check. Every point of x+y+z=4x+y+z=4 gives left-hand side 88 in the doubled equations. None can give 99, while every one gives 88.

Nonparallel normals give two independent linear constraints in three coordinates, leaving one free parameter. That is why the intersection is a line rather than a single point.

Watch out

A zero cross product is not an intersection direction. The zero vector cannot direct a line; classify the planes before proceeding.

Find and verify another common direction.

Finding a point on both planes

A direction does not locate the intersection line. You also need one point satisfying both plane equations. With two nonparallel planes, choose a convenient value for one coordinate and solve for the other two, changing the choice if necessary.

Setting a coordinate to zero often keeps the arithmetic short, but it is a choice, not an extra condition in the problem. A particular coordinate may be fixed along the entire intersection line, so setting it to a different value would miss the line completely.

Worked example 3: a convenient zero coordinate

Question. Find one point common to x+y+z=4x+y+z=4 and x−y+z=2x-y+z=2.

Plan. Set z=0z=0 and solve the two resulting equations for x,yx,y.

x+y=4x−y=22x=6add the equationsx=3,y=1.\begin{aligned} x+y&=4\\ x-y&=2\\ 2x&=6 &&\text{add the equations}\\ x&=3,\qquad y=1. \end{aligned}

Answer. One common point is (3,1,0)(3,1,0).

Check. Substitution into the original equations gives 3+1+0=43+1+0=4 and 3−1+0=23-1+0=2. Both must hold.

Now change the pair of planes so that the intersection has a fixed nonzero value of zz. The usual first choice z=0z=0 will fail, even though the planes still intersect.

Continue this lesson.

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