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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Distance from a Point to a Line or Plane

The distance from a point to a plane or a line always means the shortest distance, measured along a perpendicular. Projections and cross products from the vectors units turn that into a formula.

How It Works

Point to plane. Take any point QQ on the plane and draw QP⃗\vec{QP} to the point PP. The perpendicular distance is the part of QP⃗\vec{QP} that lies along the normal, the scalar projection onto n⃗\vec{n}: d=∣QP⃗⋅n⃗∣∣n⃗∣d = \frac{|\vec{QP} \cdot \vec{n}|}{|\vec{n}|}.

The formula. For P(x0,y0,z0)P(x_0, y_0, z_0) and the plane Ax+By+Cz+D=0Ax + By + Cz + D = 0, that projection simplifies to d=∣Ax0+By0+Cz0+D∣A2+B2+C2d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}. Substitute the point into the left side of the scalar equation, take the absolute value, and divide by the length of the normal.

Point to a line in 2D. The same idea, one dimension down: for Ax+By+C=0Ax + By + C = 0, d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.

Point to a line in 3D. A 3D line has no single normal, so use the cross product. With QQ on the line and direction d⃗\vec{d}, d=∣QP⃗×d⃗∣∣d⃗∣d = \frac{|\vec{QP} \times \vec{d}|}{|\vec{d}|}. Why: ∣QP⃗×d⃗∣|\vec{QP} \times \vec{d}| is the area of the parallelogram on those two vectors, and dividing by the base ∣d⃗∣|\vec{d}| leaves the height.

Parallel planes. Pick any point on one plane and find its distance to the other. Write both with the same normal first.

The closest point. To find the foot of the perpendicular on a plane, run a line from PP along n⃗\vec{n} and intersect it with the plane. The distance from PP to that point checks the formula.

A distance is never negative; the absolute value is not optional.

5 Worked Examples

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

  1. Easy

    Find the distance from P(1,−2,3)P(1, -2, 3) to the plane 2x−y+2z−4=02x - y + 2z - 4 = 0.
    Show solution
    1. Substitute: 2(1)−(−2)+2(3)−4=62(1) - (-2) + 2(3) - 4 = 6.
    2. ∣n⃗∣=4+1+4=3|\vec{n}| = \sqrt{4 + 1 + 4} = 3.

    Answer: d=63=2d = \frac{6}{3} = 2

  2. Easy

    Find the distance from P(3,1)P(3, 1) to the line 4x−3y+6=04x - 3y + 6 = 0.
    Show solution
    1. Substitute: 4(3)−3(1)+6=154(3) - 3(1) + 6 = 15.
    2. 16+9=5\sqrt{16 + 9} = 5.

    Answer: d=155=3d = \frac{15}{5} = 3

  3. Medium

    Find the distance between the parallel planes x+2y−2z=3x + 2y - 2z = 3 and x+2y−2z=12x + 2y - 2z = 12.
    Show solution
    1. A point on the first: (3,0,0)(3, 0, 0).
    2. Distance to x+2y−2z−12=0x + 2y - 2z - 12 = 0: ∣3−12∣1+4+4=93\frac{|3 - 12|}{\sqrt{1 + 4 + 4}} = \frac{9}{3}.

    Answer: 33

  4. Medium

    Find the distance from P(2,3,−1)P(2, 3, -1) to the line r⃗=[1,0,2]+t[1,2,2]\vec{r} = [1, 0, 2] + t[1, 2, 2].
    Show solution
    1. Q(1,0,2)Q(1, 0, 2) is on the line, so QP⃗=[1,3,−3]\vec{QP} = [1, 3, -3].
    2. QP⃗×d⃗=[3(2)−(−3)(2),  (−3)(1)−1(2),  1(2)−3(1)]=[12,−5,−1]\vec{QP} \times \vec{d} = [3(2) - (-3)(2), \; (-3)(1) - 1(2), \; 1(2) - 3(1)] = [12, -5, -1].
    3. ∣QP⃗×d⃗∣=144+25+1=170|\vec{QP} \times \vec{d}| = \sqrt{144 + 25 + 1} = \sqrt{170} and ∣d⃗∣=3|\vec{d}| = 3.

    Answer: d=1703≈4.35d = \frac{\sqrt{170}}{3} \approx 4.35

  5. Test level

    Find the distance from P(5,1,3)P(5, 1, 3) to the plane 2x−y+2z+3=02x - y + 2z + 3 = 0, and the point of the plane closest to PP.
    Show solution
    1. Formula: ∣10−1+6+3∣3=183=6\frac{|10 - 1 + 6 + 3|}{3} = \frac{18}{3} = 6.
    2. The line through PP along n⃗=[2,−1,2]\vec{n} = [2, -1, 2]: x=5+2tx = 5 + 2t, y=1−ty = 1 - t, z=3+2tz = 3 + 2t.
    3. Substitute into the plane: 2(5+2t)−(1−t)+2(3+2t)+3=18+9t=02(5 + 2t) - (1 - t) + 2(3 + 2t) + 3 = 18 + 9t = 0, so t=−2t = -2.
    4. Closest point: (1,3,−1)(1, 3, -1). Check: 2−3−2+3=02 - 3 - 2 + 3 = 0, and its distance from PP is ∣−2∣×3=6|-2| \times 3 = 6.

    Answer: The distance is 66, and the closest point is (1,3,−1)(1, 3, -1)

Learn It on Martian Lab

Martian Lab teaches this in the Applications of the Dot & Cross Products lesson of its MCV4U Cartesian Vectors unit. Watch it, then take the Lines and Planes practice tests to see if it stuck.

Watch the Applications of the Dot & Cross Products lesson free

Keep Going

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 · Intersection of Lines and Planes

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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