Question

Find an equation of the plane. The plane passes through the point (1,2,3) and is parallel to the \(y z\) -plane.

Short answer

The equation of the plane is \(x = 1\).

Step-by-step solution

Analyze the Given Information

Understand that a plane parallel to yz-plane will have its normal vector along the x-axis, i.e., normal vector \(n = (a, 0, 0)\). Also, given that the plane passes through the point P(1, 2, 3).

Construct the General Equation of Plane

An equation of a plane in three-dimensional space can be expressed in the form \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\), where a, b, c are the vector components of the normal vector to the plane, and \((x_0, y_0, z_0)\) is a point through which the plane passes. Here, since the plane is parallel to yz-plane, its normal vector is along x-axis. So, the equation becomes \(a(x - x_0) = 0\) as y and z components are equal to 0.

Substitute known parameters

Substitute the normal vector and the given point (1, 2, 3) into the equation. We can say a = 1 without loss of generality because any scalar multiple of a vector is still normal to the plane. So the equation becomes \(1 * (x - 1) = 0\) which simplifies to \(x - 1 = 0\) or \(x = 1\)