equqtion of plane

Question

Find equation of the plane through (2,4,5) perpendicular to the line x=5+t , y=1+3t , z=4t

Answer 1

We can find two points on the given line by setting parameter t=0 and 1:

P=(5,1,0) and Q=(6,4,4). The plane is perpendicular to segment PQ, and, as a vector PQ, represents the normal to the plane. The vector PQ=<6-5,4-1,4-0>=<1,3,4>. Let R(2,4,5) be the given point on the plane.

This gives us the coefficients of the variable expressions in the plane equation: 1(x-2)+3(y-4)+4(z-5)=0.

x-2+3y-12+4z-20=0, x+3y+4z=34. If we plug in point Q we get 6+12+16=34, therefore Q also lies on the plane and PQ should be perpendicular to QR (which lies on the plane).

QR=<2-4,4-4,5-4>=<-4,0,1>. QR.PQ=<-4,0,1>.<1,3,4>=-4+0+4=0, which proves that a line on the plane is perpendicular to the given line which in turn shows that the plane x+3y+4z=34 is perpendicular to the given line.