Section 12.5: Lines & Planes | When are planes called parallel? If they are not parallel, What is the angle between two non-parallel planes?

When are planes called parallel? If they are not parallel, What is the angle between two non-parallel planes?

Two planes are parallel if their normal vectors are parallel. For instance, the planes \(x + 2y - 3z = 4\) and \(2x + 4y - 6z = 3\) are parallel because their normal vectors are \(\mathbf{n}_1 = \langle 1, 2, -3 \rangle\) and \(\mathbf{n}_2 = \langle 2, 4, -6 \rangle\) and \(\mathbf{n}_2 = 2\mathbf{n}_1\). If two planes are not parallel, then they intersect in a straight line and the angle between the two planes is defined as the acute angle between their normal vectors (see angle \(\theta\) in Figure 9).

EXAMPLE 7 (a) Find the angle between the planes \(x + y + z = 1\) and \(x - 2y + 3z = 1\). (b) Find symmetric equations for the line of intersection L of these two planes.

SOLUTION (a) The normal vectors of these planes are \[ \mathbf{n}_1 = \langle 1, 1, 1 \rangle \quad \mathbf{n}_2 = \langle 1, -2, 3 \rangle \] and so, if \(\theta\) is the angle between the planes, Corollary 12.3.6 gives \[ \cos\theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{1(1) + 1(-2) + 1(3)}{\sqrt{1+1+1}\sqrt{1+4+9}} = \frac{2}{\sqrt{3}\sqrt{14}} = \frac{2}{\sqrt{42}} \] \[ \theta = \cos^{-1}\left(\frac{2}{\sqrt{42}}\right) \approx 72^\circ \] (b) We first need to find a point on L. For instance, we can find the point where the line intersects the xy-plane by setting \(z = 0\) in the equations of both planes. This gives the equations \(x + y = 1\) and \(x - 2y = 1\), whose solution is \(x = 1, y = 0\). So the point \((1, 0, 0)\) lies on L.

Now we observe that, since L lies in both planes, it is perpendicular to both of the normal vectors. Thus a vector v parallel to L is given by the cross product \[ \mathbf{v} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = 5\mathbf{i} - 2\mathbf{j} - 3\mathbf{k} \] and so the symmetric equations of L can be written as \[ \frac{x - 1}{5} = \frac{y}{-2} = \frac{z}{-3} \] NOTE Since a linear equation in \(x, y,\) and \(z\) represents a plane and two nonparallel planes intersect in a line, it follows that two linear equations can represent a line. The points \((x, y, z)\) that satisfy both \(a_1x + b_1y + c_1z + d_1 = 0\) and \(a_2x + b_2y + c_2z + d_2 = 0\) lie on both of these planes, and so the pair of linear equations represents the line of intersection of the planes (if they are not parallel). For instance, in Example 7 the line L was given as the line of intersection of the planes \(x + y + z = 1\) and \(x - 2y + 3z = 1\). The symmetric equations that we found for L could be written as \[ \frac{x - 1}{5} = \frac{y}{-2} \quad \text{and} \quad \frac{y}{-2} = \frac{z}{-3} \] which is again a pair of linear equations. They exhibit L as the line of intersection of the planes \((x - 1)/5 = y/(-2)\) and \(y/(-2) = z/(-3)\).