Question

1.) Find an equation for the plane that passes through the origin (0, 0, 0) and is parallel to the plane - x + 3y - 2z = 6

2.) Find the distance between the point (-1, - 2, 0) and the plane 3x - y + 4z =-2

Answer 1

Q1

For the line to be parallel to the line $-x+3y-2z=6$ .Then,

the would be $-x+3y-2z=k$ . To get k,

Since the line passes through the origin.

$k=0$

The line is therefore, $-x+3y-2z=0$

Q2

$(-1,-2,0)=(x_o,y_o,z_o)\\ The\ distance\ from\ the\\ equation\ Ax+By+Cz+D=0\ is\\ d=\dfrac{|Ax_o+By_o+Cz_o+D|}{\sqrt{A^2+B^2+C^2}}\\ A=3,B=-1,C=4,D=2\\ d=\dfrac{|3(-1)-1(-2)+4(0)+2|}{\sqrt{3^2+(-1)^2+4^2}}\\ d=\dfrac{1}{\sqrt{26}}unit$