Find an equation for the Plane that contains both the Point

Find an equation for the Plane that contains both the Point P(-3, 6, -1) and the line with equation x - 4/2 = x - 4/1, z = 2

Solution

The directional vector v, of the line can be taken from the coefficients of the parameter t. v = <4, 4, 0> we have direction cosines as Q(2,1,2) A second point P, in the desired plane can be calculated by setting the line parameter t equal to zero. P(-3, 6, -1) Now we can calculate a second directional vector u = PQ, of the desired plane. u = QP =

= <-3-2, 6-1, -1-2> = <-5, 5, -3> The normal vector n, of the desired plane is orthogonal to both directional vectors u and v. Take the cross product. n = u X v = <-5, 5, -3> X <4, 4, 0> = <12, -12, 40> With the normal vector n and a point in the plane we can write the equation of the plane. Remember, the normal vector of the plane is orthogonal to any vector that lies in the plane. And the dot product of orthogonal vectors is zero. Let\'s choose an arbitrary point R(x,y,z), in the plane and the point P to define the vector in the plane. n • PR = 0 n • = 0 <12, -12, 40> • = 0 12(x - 4) - 12(y - 4) + 40(z - 0) = 0 12x - 48 - 12y + 48 + 40z = 0 12x - 12y + 40z= 0........eq of plane