Equations of planes We have touched on equations of planes previously. Here we will ﬁll in some of the details. Planes in point-normal form The basic data which determines a plane is a point P0 in the plane and a vector N orthogonal to the plane. We call N a normal to the plane and we will sometimes say N is normal to the plane, instead of orthogonal. Now, suppose we want the equation of a plane and we have a point P0 = (x0 , y0 , z0 ) in the − → plane and a vector N = �a, b, c� normal to the plane. −−→ Let P = (x, y, z) be an arbitrary point in the plane. Then the vector P0 P is in the plane and therefore orthogonal to N. This means −−→ N · P0 P = 0 ⇔ �a, b, c� · �x − x0 , y − y0 , z − z0 � = 0 ⇔ a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 We call this last equation the point-normal form for the plane. P = (x, y, z) N P0 = (x0 , y0 , z0 ) Example 1: Find the plane through the point (1,4,9) with normal �2, 3, 4�. Answer: Point-normal form of the plane is 2(x − 1) + 3(y − 4) + 4(z − 9) = 0. We can also write this as 2x + 3y + 4z = 50. Example 2: Find the plane containing the points P1 = (1, 2, 3), P2 = (0, 0, 3), P3 = (2, 5, 5). Answer: The goal is to ﬁnd the basic data, i.e. a point in the plane and a normal to the plane. The point is easy, we already have three of them. To get the normal we note −−−→ −−−→ (see ﬁgure below) that P1 P2 and P1 P3 are vectors in the plane, so their cross product is orthogonal (normal) to the plane. That is, ⎛ ⎞ i j k −−−→ −−−→ ⎝ −1 −2 0 ⎠ = −4i − j(−2) + k(−1) = �−4, 2, −1�. N = P1 P2 × P1 P3 = 1 3 2 N P1 P2 P = (x, y, z) P3 Using point-normal form (with point P1 ) the equation of the plane is −4(x − 1) + 2(y − 2) − (z − 3) = 0, or equivalently − 4x + 2y − z = −3. z� • � containing the point (0,0,3) Example 3: Find the plane with normal N = k Eq. of plane: �0, 0, 1� · �x, y, z − 3� = 0 ⇔ z = 3. �y x� Example 4: Find the plane with x, y and z intercepts a, b and c. Answer: We could ﬁnd this using the method example 1. Instead, we’ll use a shortcut that works when all the intercepts are known. In this case, the intercepts are (a, 0, 0), (0, b, 0), (0, 0, c) z� c and we simply write the plane as x/a + y/b + z/c = 1. b �y You can easily check that each of the given points is on the plane. For completeness we’ll do this using the more general method of example 1. The 3 points give us 2 vectors in the plane, �−a, b, 0� and �−a, 0, c�. ⇒ N = �−a, b, 0� × �−a, 0, c� = �bc, ac, ab�. Point-normal form: bc(x − a) + ac(y − 0) + ab(z − 0) = 0 ⇔ bc x + ac y + ab z = abc ⇔ x/a + y/b + z/c = 1. a x� Lines in the plane While we’re at it, let’s look at two ways to write the equation of a line in the xy-plane. Slope-intercept form: Given the slope m and the y-intercept b the equation of a line can be written y = mx + b. Point-normal form: We can also use point-normal form to ﬁnd the equation of a line. Given a point (x0 , y0 ) on the line and a vector �a, b� normal to the line the equation of the line can be written a(x − x0 ) + b(y − y0 ) = 0. y� � �a, b� (x0 , y0 )• �x