How to Find a Plane With 3 Points

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The equation of a plane in three-dimensional space can be written in algebraic notation as ax + by + cz = d, where at least one of the real-number constants "a," "b," and "c" must not be zero, and "x", "y" and "z" represent the axes of the three-dimensional plane. If three points are given, you can determine the plane using vector cross products. A vector is a line in space. A cross product is the multiplication of two vectors.

    Get the three points on the plane. Label them "A," "B" and "C." For example, assume these points are A = (3, 1, 1); B = (1, 4, 2); and C = (1, 3, 4).

    Find two different vectors on the plane. In the example, choose vectors AB and AC. Vector AB goes from point-A to point-B, and vector AC goes from point-A to point-C. So subtract each coordinate in point-A from each coordinate in point-B to get vector AB: (-2, 3, 1). Similarly, vector AC is point-C minus point-A, or (-2, 2, 3).

    Compute the cross product of the two vectors to get a new vector, which is normal (or perpendicular or orthogonal) to each of the two vectors and also to the plane. The cross product of two vectors, (a1, a2, a3) and (b1, b2, b3), is given by N = i(a2b3 - a3b2) + j(a3b1 - a1b3) + k(a1b2 - a2b1). In the example, the cross product, N, of AB and AC is i[(3 x 3) - (1 x 2)] + j[(1 x -2) - (-2 x 3)] + k[(-2 x 2) - (3x - 2)], which simplifies to N = 7i + 4j + 2k. Note that “i,” “j” and “k” are used to represent vector coordinates.

    Derive the equation of the plane. The equation of the plane is Ni(x - a1) + Nj(y - a2) + Nk(z - a3) = 0, where (a1, a2, a3) is any point in the plane and (Ni, Nj, Nk) is the normal vector, N. In the example, using point C, which is (1, 3, 4), the equation of the plane is 7(x - 1) + 4(y - 3) + 2(z - 4) = 0, which simplifies to 7x - 7 + 4y - 12 + 2z - 8 = 0, or 7x + 4y + 2z = 27.

    Verify your answer. Substitute the original points to see if they satisfy the equation of the plane. To conclude the example, if you substitute any of the three points, you will see that the equation of the plane is indeed satisfied.


    • See Resources for tips on how to use systems of three simultaneous equations to find the equation of a plane.


About the Author

Based in Ottawa, Canada, Chirantan Basu has been writing since 1995. His work has appeared in various publications and he has performed financial editing at a Wall Street firm. Basu holds a Bachelor of Engineering from Memorial University of Newfoundland, a Master of Business Administration from the University of Ottawa and holds the Canadian Investment Manager designation from the Canadian Securities Institute.

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