Answer to Question #211904 in Analytic Geometry for Faith

Question

Let L be the line given by <3,-1,2>+t<1,1,-1>,for tER.

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

2.find an equation for the plane through the point p=(3,-2,4) that is perpendicular to the line <-8,2,0>+t<-3,2,-7>.

Answer 1

1.

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

$$-2x+3y-4z+1$$

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

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

$$=5t-16$$

Consider $t=0$

$$5t-16=5(0)-16=-16\not=0$$

Therefore the line L does not lie on the plane $-2x+3y-4z+1=0.$

2.

$$\vec n=\langle-3, 2, -7\rangle$$

Point $P=(3,-2,4).$

The equation for the plane through the point $P=(3,-2,4)$ that is perpendicular to the line $\langle-8, 2, 0\rangle+t\langle-3, 2, 7\rangle$ is

$$-3(x-3)+2(y-(-2))-7(z-4)=0$$

or

$$3x-2y+7z-41=0$$