The Vector Equation of a Plane

A flat sheet with two directions in it

A plane is a flat surface that runs on without end in every direction along it. A tabletop, a wall and a sheet of glass are pieces of planes.

A line has one direction. A plane has two: on a tabletop you can move left and right, and you can move toward and away from you, and by mixing the two you can reach every point of the top. Any two directions in the plane that are not along the same line will do.

A sloping plane over a square of the floor, with the x-, y- and z-axes drawn from one corner. Two families of parallel lines are ruled across it, one in each of two directions, and every point of the sheet is where a line of one family crosses a line of the other.

One counter, then two

The vector equation of a line is r = a + t d: start at the point with position vector a and take t steps of the direction d. As the number t runs through every value, from negative to positive, the point r runs along the whole line, and nowhere else.

Now add a second direction c with its own number $\mu$μ. From each point of the line, walking $\mu$μ steps of c reaches a whole line parallel to c. Those parallel lines, one through each point of the first line, sit side by side and fill a flat sheet: a plane.

So two numbers are needed to name a point of a plane, just as one number names a point of a line.

One counter: $r=a+\lambda d$r = a + λd with a = (1, 1) and d = (2, 1). Drag the handle $\lambda$λ. Every value of $\lambda$λ lands on the same dashed line, so a single direction and a single counter can only ever sweep out a line.

The equation

The vector equation of a plane is $r=a+\lambda b+\mu c$r = a + λb + μc.

Here a is the position vector of one point in the plane, the point where you reach the plane. The vectors b and c are two directions that lie in the plane. The numbers $\lambda$λ and $\mu$μ, called parameters, say how far to walk along each direction.

Read it as a journey: go from the origin to the point a, then walk $\lambda$λ lots of b, then $\mu$μ lots of c. A negative $\lambda$λ or $\mu$μ walks the opposite way. The letters a, b and c are vectors; $\lambda$λ and $\mu$μ are ordinary numbers.

Each pair of numbers names one point

Take $r=\left(1,0,2\right)+\lambda \left(1,1,0\right)+\mu \left(0,1,1\right)$r = (1, 0, 2) + λ(1, 1, 0) + μ(0, 1, 1).

With $\lambda =1$λ = 1 and $\mu =0$μ = 0: r = (1, 0, 2) + (1, 1, 0) = (2, 1, 2). With $\lambda =1$λ = 1 and $\mu =2$μ = 2: r = (1, 0, 2) + (1, 1, 0) + 2(0, 1, 1) = (1, 0, 2) + (1, 1, 0) + (0, 2, 2) = (2, 3, 4).

With $\lambda =2$λ = 2 and $\mu =1$μ = 1: r = (1, 0, 2) + (2, 2, 0) + (0, 1, 1) = (3, 3, 3). With $\lambda =0$λ = 0 and $\mu =0$μ = 0 you stay at a = (1, 0, 2), and with $\lambda =-1$λ = −1 and $\mu =1$μ = 1 you get (1, 0, 2) − (1, 1, 0) + (0, 1, 1) = (0, 0, 3).

Every pair $\left(\lambda ,\mu \right)$(λ, μ) gives one point of the plane, and every point of the plane comes from exactly one pair.

The plane seen face-on, so that it is the page itself. From the point A, two strides of b and then one stride of c reach the point R, where r = a + 2b + c. Other values of $\lambda$λ and $\mu$μ reach every other point of the page.

Is a point on the plane?

Write the equation one component at a time. For $r=\left(1,0,2\right)+\lambda \left(1,1,0\right)+\mu \left(0,1,1\right)$r = (1, 0, 2) + λ(1, 1, 0) + μ(0, 1, 1), the components are $x=1+\lambda$x = 1 + λ, $y=\lambda +\mu$y = λ + μ and $z=2+\mu$z = 2 + μ.

To test the point (3, 5, 5), find $\lambda$λ from x and $\mu$μ from z, then check y. From x = 3, $1+\lambda =3$1 + λ = 3, so $\lambda =2$λ = 2. From z = 5, $2+\mu =5$2 + μ = 5, so $\mu =3$μ = 3. Then y should be $\lambda +\mu =5$λ + μ = 5, and it is. So (3, 5, 5) lies on the plane, at $\lambda =2$λ = 2 and $\mu =3$μ = 3.

The point (3, 4, 5) gives the same $\lambda =2$λ = 2 and $\mu =3$μ = 3, but then y should be 5, not 4. Two numbers can always be chosen to fit two of the components; the third component is the test.

The two directions must point different ways

If c is a multiple of b, say c = 2b, then $\lambda b+\mu c=\lambda b+2\mu b=\left(\lambda +2\mu \right)b$λb + μc = λb + 2μb = (λ + 2μ)b. The two counters combine into one number, $\lambda +2\mu$λ + 2μ, so $r=a+\left(\lambda +2\mu \right)b$r = a + (λ + 2μ)b, which is the equation of a line. The second direction has added nothing new, and the sweep never leaves that line.

So before using b and c, check that neither is a multiple of the other. (2, 1, 3) and (4, 2, 6) fail: every component is multiplied by 2. (2, 1, 3) and (4, 2, 7) pass: x and y are multiplied by 2, but 3 × 2 = 6, not 7. Two vectors that point different ways span a plane whatever their lengths.

From A, the direction b = (2, 1) and the direction c = 2b = (4, 2) lie along one line. Every combination $\lambda b+\mu c$λb + μc is $\left(\lambda +2\mu \right)b$(λ + 2μ)b, so every point reached stays on that line.

A normal from the two directions

The vector product of the two directions is perpendicular to both, so it is perpendicular to the whole plane: a normal. For b = (1, 1, 0) and c = (0, 1, 1), b × c = (1 × 1 − 0 × 1, 0 × 0 − 1 × 1, 1 × 1 − 1 × 0) = (1, −1, 1).

Check it with dot products: (1, 1, 0) · (1, −1, 1) = 1 − 1 + 0 = 0 and (0, 1, 1) · (1, −1, 1) = 0 − 1 + 1 = 0.

Every point r of the plane has r · n = a · n, and a · n = 1 − 0 + 2 = 3, so the same plane is x − y + z = 3. The points found above agree: (2, 1, 2) gives 2 − 1 + 2 = 3, (2, 3, 4) gives 2 − 3 + 4 = 3, and (3, 3, 3) gives 3 − 3 + 3 = 3.

Two directions a and b, of lengths 3 and 2, lie in the shaded plane at 90° to each other, and a × b stands at right angles to that plane with length 3 × 2 = 6. Swing b round until it lies along a: the parallelogram flattens, a × b becomes zero, and two directions along one line no longer fix a plane.

A plane through three points

Three points A, B and C that are not in a line fix one plane. Use A as the point, and the two edges from it, AB = b − a and AC = c − a, as the directions: $r=a+\lambda \left(b-a\right)+\mu \left(c-a\right)$r = a + λ(b − a) + μ(c − a).

For A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3), AB = (−1, 2, 0) and AC = (−1, 0, 3), so $r=\left(1,0,0\right)+\lambda \left(-1,2,0\right)+\mu \left(-1,0,3\right)$r = (1, 0, 0) + λ(−1, 2, 0) + μ(−1, 0, 3). Check: $\lambda =1$λ = 1 and $\mu =0$μ = 0 give (0, 2, 0), which is B, and $\lambda =0$λ = 0 and $\mu =1$μ = 1 give (0, 0, 3), which is C.

The directions must be vectors along the plane, from one of its points to another. The position vectors b and c run from the origin, which is usually not in the plane at all.

Many equations for one plane

Any point of the plane can replace a: $r=\left(2,1,2\right)+\lambda \left(1,1,0\right)+\mu \left(0,1,1\right)$r = (2, 1, 2) + λ(1, 1, 0) + μ(0, 1, 1) is the same plane, starting from the point found at $\lambda =1$λ = 1 and $\mu =0$μ = 0. Any multiple of b or c, and any two other non-parallel directions in the plane, such as b + c = (1, 2, 1) and b − c = (1, 0, −1), describe it too.

The pairs $\left(\lambda ,\mu \right)$(λ, μ) change from one equation to another, but the set of points does not.

The usual mistakes

Walking b once when $\lambda =2$λ = 2. The number in front says how many strides: 2b is two whole copies of b.

Leaving out $\mu$μ. With $\mu =0$μ = 0 the point stays on the line through a in the direction b; reaching across the plane takes some of c as well.

Taking a, or $\lambda$λ and $\mu$μ, as the directions. a is a position, the place where the plane is reached, and $\lambda$λ and $\mu$μ are numbers that say how far. The directions are b and c.

Missing a shared factor. (1, 2, 3) and (3, 6, 9) look different, but the second is 3 times the first, so they span only a line.