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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Intersection of Lines and Planes

A line and a plane in 3D meet in one of three ways: at a single point, nowhere, or everywhere along the line. One substitution tells you which, and finds the point when there is one.

How It Works

The method. Write the line in parametric form, x=x0+atx = x_0 + at, y=y0+bty = y_0 + bt, z=z0+ctz = z_0 + ct, and substitute into the plane's scalar equation Ax+By+Cz+D=0Ax + By + Cz + D = 0. You get one equation in tt.

One solution. You get a single value of tt, so the line crosses the plane at one point. Put that tt back into the line to get the point, then check it in the plane's equation.

No solution. The tt terms cancel and you are left with something false, such as 5=45 = 4. The line is parallel to the plane and never touches it.

Infinitely many solutions. The tt terms cancel and you are left with something true, such as 6=66 = 6. The line lies in the plane.

Predicting it with the normal. The tt terms cancel exactly when n⃗⋅d⃗=0\vec{n} \cdot \vec{d} = 0, where n⃗=[A,B,C]\vec{n} = [A, B, C] and d⃗\vec{d} is the line's direction. Then the line is perpendicular to the normal, which means parallel to the plane. To tell the two parallel cases apart, test one point of the line in the plane.

Two lines in 3D. Set the parametric equations equal, using a different parameter for each line, tt and ss. That is three equations in two unknowns. Solve two of them, then test the third. If it holds, the lines meet at that point. If it fails and the directions are not parallel, the lines are skew: they are not parallel, yet never meet, which cannot happen in 2D.

Always substitute your point into every equation you started with.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    Find where the line x=1+2tx = 1 + 2t, y=−1+ty = -1 + t, z=3−tz = 3 - t meets the plane 2x+y+z=122x + y + z = 12.
    Show solution
    1. Substitute: 2(1+2t)+(−1+t)+(3−t)=122(1 + 2t) + (-1 + t) + (3 - t) = 12, so 4+4t=124 + 4t = 12 and t=2t = 2.
    2. Point: (1+4,  −1+2,  3−2)=(5,1,1)(1 + 4, \; -1 + 2, \; 3 - 2) = (5, 1, 1).
    3. Check: 10+1+1=1210 + 1 + 1 = 12.

    Answer: (5,1,1)(5, 1, 1)

  2. Medium

    Find the intersection of r⃗=[2,1,0]+t[1,−1,2]\vec{r} = [2, 1, 0] + t[1, -1, 2] and the plane x+3y+z=4x + 3y + z = 4.
    Show solution
    1. Substitute: (2+t)+3(1−t)+2t=4(2 + t) + 3(1 - t) + 2t = 4, so 5+0t=45 + 0t = 4, which is false.
    2. Confirm with the normal: [1,3,1]⋅[1,−1,2]=1−3+2=0[1, 3, 1] \cdot [1, -1, 2] = 1 - 3 + 2 = 0, so the line is parallel to the plane.

    Answer: No intersection: the line is parallel to the plane and off it

  3. Medium

    Find the intersection of r⃗=[1,1,1]+t[2,0,−1]\vec{r} = [1, 1, 1] + t[2, 0, -1] and the plane x+3y+2z=6x + 3y + 2z = 6.
    Show solution
    1. Substitute: (1+2t)+3(1)+2(1−t)=6(1 + 2t) + 3(1) + 2(1 - t) = 6, so 6+0t=66 + 0t = 6, true for every tt.
    2. Check: n⃗⋅d⃗=2+0−2=0\vec{n} \cdot \vec{d} = 2 + 0 - 2 = 0 and the point (1,1,1)(1, 1, 1) gives 1+3+2=61 + 3 + 2 = 6.

    Answer: The line lies in the plane: every point of the line is an intersection

  4. Medium

    Find the intersection of L1:r⃗=[0,−4,4]+t[1,2,−1]L_1: \vec{r} = [0, -4, 4] + t[1, 2, -1] and L2:r⃗=[3,0,5]+s[1,1,1]L_2: \vec{r} = [3, 0, 5] + s[1, 1, 1].
    Show solution
    1. Set components equal: t=3+st = 3 + s, −4+2t=s-4 + 2t = s, 4−t=5+s4 - t = 5 + s.
    2. Substitute the first into the second: −4+6+2s=s-4 + 6 + 2s = s, so s=−2s = -2 and t=1t = 1.
    3. Test the third: 4−1=34 - 1 = 3 and 5−2=35 - 2 = 3. It holds.
    4. Point from L1L_1 at t=1t = 1: (1,−2,3)(1, -2, 3).

    Answer: The lines meet at (1,−2,3)(1, -2, 3)

  5. Test level

    For which kk does the line r⃗=[1,0,2]+t[2,k,−1]\vec{r} = [1, 0, 2] + t[2, k, -1] miss the plane 3x−y+2z=43x - y + 2z = 4? For the other values of kk, where do they meet?
    Show solution
    1. Substitute: 3(1+2t)−kt+2(2−t)=43(1 + 2t) - kt + 2(2 - t) = 4, so 7+(4−k)t=47 + (4 - k)t = 4, giving (4−k)t=−3(4 - k)t = -3.
    2. If k=4k = 4, this reads 0=−30 = -3: no intersection. Check: n⃗⋅d⃗=6−k−2=0\vec{n} \cdot \vec{d} = 6 - k - 2 = 0 at k=4k = 4, and (1,0,2)(1, 0, 2) gives 3+4=7≠43 + 4 = 7 \ne 4, so the line is parallel and off the plane.
    3. If k≠4k \ne 4, t=3k−4t = \frac{3}{k - 4}, one point.

    Answer: It misses the plane when k=4k = 4. Otherwise they meet once, at t=3k−4t = \frac{3}{k - 4}: (1+6k−4,  3kk−4,  2−3k−4)\left(1 + \frac{6}{k - 4}, \; \frac{3k}{k - 4}, \; 2 - \frac{3}{k - 4}\right)

Learn It on Martian Lab

Martian Lab teaches this in the Intersections of Lines & Planes lesson of its MCV4U Lines and Planes unit. Watch it, then take the Lines and Planes practice tests to see if it stuck.

Watch the Intersections of Lines & Planes lesson free

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Free Lines and Planes practice test, with solutions

More Lines and Planes topics: Equations of Planes · Intersection of Three Planes · Equations of Lines in 3D · Distance from a Point to a Line or Plane

More MCV4U topics: Evaluating Limits · Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque · Continuity · Derivative from First Principles · Power, Product and Quotient Rules · Velocity and Acceleration · Increasing, Decreasing and Critical Points · Concavity and the Second Derivative Test · Derivatives of e^x and a^x · Derivatives of Sin, Cos and Tan · Adding and Subtracting Vectors · Cartesian Vectors in 2D and 3D · Force and Velocity Vector Problems

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