In \( \mathbb{R}^2 \), the scalar (Cartesian) equation of a line was derived using the notion of a normal vector, \( \vec{n}\), and a vector in the plane, \( \overrightarrow{P_0P}\).

This notion can be extended to three dimensions to derive the scalar equation of a plane.

Let \( \vec{n} = (A, B, C) \) be a normal vector to a plane that contains the fixed point \( P_0~(x_0, y_0, z_0) \).

Let \( P~(x, y, z) \) be any other arbitrary point in the plane. Then by the definition of the dot product,

is the scalar equation of a plane in \(\mathbb{R}^3\).

In particular, all points in the plane must satisfy this equation and conversely, any point which satisfies this equation must lie on the plane.

Observe, as we did for the scalar equation of a line in \(\mathbb{R}^2\), that the components of the normal vector \((A,B,C)\) form the coefficients of the scalar equation (and vice versa).

Find the scalar equation of a plane with normal \( (1, -2, 5) \) and containing the point \( P_0~(3, 4, -1) \).

Solution

Let \( P~(x, y, z) \) be any point in the plane. Then,

is the scalar equation of the plane.

Alternative Method

Since \( \vec{n} = (1, -2, 5) \) is the normal, the scalar equation of the plane is of the form \( x - 2y + 5z + D = 0 \), with the constant, \( D\), to be determined.

Since the plane passes through the point \(P_0\), then \(P_0\) satisfies the equation of the plane. Substituting \( P_0~(3, 4, -1) \) into this equation, we get

Therefore, the scalar equation of the plane is \( x - 2y + 5z + 10 = 0 \).

Find the scalar equation of the plane containing the line \( (x, y, z) = (-4, 0, 5) + t(3, 1, 2) \) and the point \( (1, 1, 0) \).

Solution

Using the direction vector of the line, one vector in the plane is \( (3, 1, 2) \).

The vector extending from \( (-4, 0, 5) \) to \( (1, 1, 0) \) is another vector in the plane (since both points lie in the plane).

That is, \( (1, 1, 0) - (-4, 0, 5) = (5, 1, -5)\) is a second direction vector for the plane.

By inspection, these two vectors are not collinear.

The cross product of these two vectors gives a vector that is perpendicular to both, and hence perpendicular to the plane (that is, a normal to the plane).

Thus, the scalar equation of the plane is of the form \( -7x + 25y - 2z + D = 0 \).

Substituting the point \( (1, 1, 0) \), we get

The scalar equation of the plane is \( -7x + 25y - 2z - 18 = 0 \) or \( 7x - 25y + 2z + 18 = 0 \).