Planes in Space

Example: Determine the equation of the plane that passes through the origin, O,

and contains the points P(1, 1, 1) and Q(1, -1, 3) .

Strategy: Determine the normal vector to the plane and form the dot product between

this vector and an arbitrary vector in the plane.

Let r_o = OP be a vector from the origin to point P. So OP = i + j + k

Let OQ be a vector from the origin to point Q. So OQ = i ˗ j + 3 k

Since both vectors OP and OQ are vectors in the plane, the cross product yields the

normal vector, n.

n = OP x OQ (vector cross product)

i j k

n = det 1 1 1 where det denotes the 3x3 determinant

1 - 1 3

n = i (3 + 1) - j (3 – 1) + k (-1 -1) = 4 i - 2 j -2 k

Also let r be a vector to an arbitrary point (x, y, z) in the plane.

So r = x i + y j + z k

So r - r_o = i (x - 1) - j (y – 1) + k (z -1)

Note: r - r_o is a vector in the desired plane.

and therefore, n and r - r_o are perpendicular to each other so the dot product is zero.

n ∙ ( r - r_o ) = 0 = 4(x - 1) - 2(y – 1) - 2(z -1)

or 2(x - 1) - (y – 1) - (z -1) = 0

2x - y - z = 0 (result for equation of the plane)

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