Answer to Question #216650 in Analytic Geometry for Niceman

Question

Let L be the line given by (3,-1,2) + t (1,1,-1).Show that the above line L lies on the plane -2x + 3y -4z + 1 = 0

Answer 1

$$<3,−1,2>+t<1,1,−1>$$

$$x=3+t,y=−1+t, z=2−t$$

Show that the Line lies on the plane 

$$-2(3+t) + 3(-1+t) -4(2-t)+ 1 = 0$$

$$-6-2t-3+3t-8+4t+1=0$$

$$5t=16$$

$$t=3.2$$

$$x=6.2,y=2.2,z=-1.2$$

Let $t=0$

$$x=3,y=-1,z=2$$

Check whether the point $(3, -1, 2)$ lies on the plane

$$-2(3)+3(-1)+4(2)+1=0, True$$

Therefore the line L lies on the plane -2x + 3y -4z + 1 = 0.