Direction Perpendicular To The Plane Take a pencil, stand it on table top. Perpendicular. Tilt the table and keep the pencil as it was. Still perpendicular. This allows us to find a different vector form of the equation of a plane. You need to know the position vector of a point on the plane. You need to know the direction, n=n1i+n2j+n3k, perpendicular to the plane. We want to find an expression for the position vector r of a point R on the plane. n A AR is on the plane. Therefore AR is perpendicular to direction n. R Position vector r Position vector a Vector AR is r-a Since AR = r – a and AR.n=0 Then (r – a ).n=0 r.n-a.n=0 x n1 y . n2 − a.n = 0 z n 3 n1 x + n2 y + n3 z + d = 0 where d = -a.n - a.n will always be a number, the constant scalar. Example Write down the equation of the plane through the point(5,2,1) given the vector 35 is perpendicular to −1 the plane. 5 3 a = 2 and n = 5 1 − 1 n1x+n2y+n3z - a.n=0 3x+5y-z-(5x3+2x5+(-1)x1)=0 3x+5y-z-24=0 Intersection of a plane and a line Find the point of intersection of the plane 1 2 3x+2y-z=3 and the line r = 4 + λ 1 x 1 2 r = y = 4 + λ 1 z 3 − 1 3 − 1 x = 1 + 2λ y= z= Substitute into the equation of the plane. Distance of a point from a plane The shortest distance of a point, A, from a plane is the distance AP, where P is the point where the line through A perpendicular to the plane intersects the plane. A P Example a. b. c. A is the point (5,2,1) and the plane has equation 4x+2y-z=8. Find: The equation of the line through A perpendicular to the plane. P, the point of intersection with the plane. The distance AP a. The direction, n, is taken from 4x+2y-z=8 and the line through (5,2,1) gives us: 5 4 r = 2 + λ 2 1 − 1 b. x= 5+4λ y= z= Subin to the equation of the plane to find λ. Then use this value to get the point P. c. The vector AP is given by: AP = p − a = | AP |= − = Page 322 Ex 11f 6 onwards