1.6 - Lines meeting planes

1.6 - Lines meeting planes

A line can cross a plane once, miss it completely or lie entirely within it. You will use one substitution to distinguish these cases, find intersections, and construct the perpendicular foot from a point to a plane.

Parallel, contained or intersecting

A line's direction can be parallel to a plane even when the line lies inside that plane. How can we tell the two cases apart? Check both the direction and one point, rather than only the direction.

Let the line be r=a+td\mathbf r=\mathbf a+t\mathbf d, where d≠0\mathbf d\ne\mathbf0, and the plane be r⋅n=p\mathbf r\cdot\mathbf n=p, where n≠0\mathbf n\ne\mathbf0. Substitution gives

a⋅n+t(d⋅n)=p.\mathbf a\cdot\mathbf n+t(\mathbf d\cdot\mathbf n)=p.

This is one linear equation for tt. Its coefficient tells you whether there can be a unique intersection.

Key idea

If d⋅n≠0\mathbf d\cdot\mathbf n\ne0, there is one intersection. If d⋅n=0\mathbf d\cdot\mathbf n=0, the line is contained in the plane when a⋅n=p\mathbf a\cdot\mathbf n=p, and parallel with no intersection when a⋅n≠p\mathbf a\cdot\mathbf n\ne p.

A zero coefficient means movement along the line cannot change its scalar product with the normal. Every point of the line therefore stays at the same normal level.

Worked example 1: a line that crosses

Question. Classify the line r=(1,1,0)+t(1,0,0)\mathbf r=(1,1,0)+t(1,0,0) relative to x+2y−z=4x+2y-z=4.

Plan. Substitute its coordinates into the plane and inspect the parameter equation.

(1+t)+2(1)−0=43+t=4t=1.\begin{aligned} (1+t)+2(1)-0&=4\\ 3+t&=4\\ t&=1. \end{aligned}

Answer. There is exactly one intersection, at (2,1,0)(2,1,0).

Check. The direction has scalar product (1,0,0)⋅(1,2,−1)=1≠0(1,0,0)\cdot(1,2,-1)=1\ne0 with the normal, so it cannot stay at one plane level. The point (2,1,0)(2,1,0) gives 2+2=42+2=4.

Change only the direction to (1,0,1)(1,0,1). This removes the parameter from the plane equation. We must then use the base point to distinguish two different outcomes.

Worked example 2: zero coefficient, two possible outcomes

Question. Classify the lines l:r=(1,1,0)+t(1,0,1)l:\mathbf r=(1,1,0)+t(1,0,1) and m:r=(2,1,0)+s(1,0,1)m:\mathbf r=(2,1,0)+s(1,0,1) relative to x+2y−z=4x+2y-z=4.

Plan. Substitute each base point and direction; decide whether the resulting constant statement is true.

For ll, substitution gives

(1+t)+2−t=4⟹3=4.(1+t)+2-t=4\quad\Longrightarrow\quad3=4.

There is no solution, so ll is parallel to and outside the plane. For mm,

(2+s)+2−s=4⟹4=4.(2+s)+2-s=4\quad\Longrightarrow\quad4=4.

Every real ss works, so mm lies in the plane.

Answer. ll is parallel with no intersection; mm is contained in the plane.

Check. Both directions have scalar product 1−1=01-1=0 with the normal. The point (1,1,0)(1,1,0) is outside the plane, whereas (2,1,0)(2,1,0) lies on it, confirming the two classifications independently.

An identity is not a unique solution such as t=0t=0: it allows every parameter. A contradiction allows none.

Watch out

d⋅n=0\mathbf d\cdot\mathbf n=0 alone does not prove that a line misses a plane. Check a point before choosing between containment and parallel separation.

Classify another line without trying to divide by zero.

Finding an intersection

When the parameter coefficient is nonzero, solve for the parameter and put it back into the line. The parameter is a label for the point; it is not itself a coordinate or a distance unless the line has been specially defined that way.

Line-plane intersection parameter

For r=a+td\mathbf r=\mathbf a+t\mathbf d and r⋅n=p\mathbf r\cdot\mathbf n=p,

t=p−a⋅nd⋅nif d⋅n≠0.t=\frac{p-\mathbf a\cdot\mathbf n}{\mathbf d\cdot\mathbf n} \quad\text{if }\mathbf d\cdot\mathbf n\ne0.

Substitute this value into the line to obtain the intersection point.

Not in the formula booklet: learn it.

You can equally substitute the three coordinates directly. The formula is simply that linear equation rearranged.

Worked example 3: obtaining the point

Question. Find where r=(1,0,1)+t(1,2,1)\mathbf r=(1,0,1)+t(1,2,1) meets 2x−y+z=52x-y+z=5.

Plan. Substitute the line into the plane, solve for tt, then calculate all three coordinates.

2(1+t)−2t+(1+t)=53+t=5t=2.\begin{aligned} 2(1+t)-2t+(1+t)&=5\\ 3+t&=5\\ t&=2. \end{aligned}

Thus r=(1,0,1)+2(1,2,1)=(3,4,3)\mathbf r=(1,0,1)+2(1,2,1)=(3,4,3).

Answer. The intersection point is (3,4,3)(3,4,3).

Check. It satisfies the plane because 2(3)−4+3=52(3)-4+3=5. Its displacement from the base point is (2,4,2)=2(1,2,1)(2,4,2)=2(1,2,1), so it also lies on the line.

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