Equation plane
- 1. The Math Center ■ Valle Verde ■ Tutorial Support Services ■ EPCC 1 Finding the Equation of a Plane The equation of a plane in 3-D space can be found by using the Point-Normal Form equation for a plane. a(x-x0)+b(y-y0)+c(z-z0)=0 In order to use this formula, a point in the plane and a vector perpendicular to the plane are needed. A vector perpendicular to the plane is called a normal to the plane. Example: Find an equation for the plane that passes through the given points P1(-2,1,1), P2(0,2,3), and P3(1,0,-1). Step 1: Since the points P1, P2, and P3 lie on the plane, the vectors P1P2 and P1P3 are parallel to the plane. To find the vectors P1P2 and P1P3 the do the following P1(-2,1,1) = P1(x1,y1,z1) P2(0,2,3) = P2(x2,y2,z2) P3(1,0,-1) = P3(x3,y3,z3) P1P2 = u = <x2-x1, y2-y1, z2-z1> = <0-(-2), 2-1, 3-1> = <2, 1, 2> P1P3 = v = <x3-x1, y3-y1, z3-z1> = <1-(-2), 0-1, -1-1> = <3, -1, -2> Therefore, u = <2, 1, 2> and v = <3, -1, -2> are the parallel vectors to the plane
- 2. The Math Center ■ Valle Verde ■ Tutorial Support Services ■ EPCC 2 Step 2: Find the normal vector to the plane using the Cross Product of vectors v and u u x v = 213 212 −− kji = i 21 21 −− - j 23 22 − + k 13 12 − = i(-2-(-2)) - j(-4-6) + k(-2-3) = i(-2+2) - j(-10) + k(-5) = 0i + 10j - 5k Hence the normal vector is 0i + 10j – 5k or <0, 10, -5> Step 3: Use the Point-Normal Form, the normal vector, and the point on the plane to find the equation a(x-x0)+b(y-y0)+c(z-z0)=0 Normal vector = <a, b, c> = <0, 10, -5> Point on the plane = P(x0,y0,z0) = P2(0,2,3) Simplify 0(x-0) + 10(y-2) - 5(z-3) = 0 10y - 20 - 5z + 15 = 0 10y - 5z - 5 = 0 2y - z - 1 = 0 Therefore, the points P1(-2,1,1), P2(0,2,3) and P3(1,0,-1) are all contained by the plane 2y – z – 1 = 0. Note: you may have also used either of the other two points in the Point-Normal Form and still acquired the same equation. You can double check your work by trying the other points.