I'm having trouble with following problem:
"Find equation of normal vector from point $A(2,3,-1)$ that is normal to the plane $\pi:2x+y-4z+5=0$"
It doesn't seem that tough but I can't seem to work it out. Like, I know I can get a normal vector out of a plane by using plane's equation, but how can I like move that normal vector until point A is right above (or below it) and then turn that into that equation?
Thanks!
Best Answer
The norm $(2,1,-4)$ is given by the plane equation $2x+y-4z+5=0$. To compute the length of the vector, you need to find the point of perpendicualrity, and this can be solved by parameterizing the line passing through $A$ and normal to the plane: $$ (x,y,z)=(2,3,-1)+\lambda(2,1,-4) $$ You will solve $\lambda$ by plug the point back in the plane equation: $$ 2(2+2\lambda)+(3+\lambda)-4(-1-4\lambda)+5=0 $$ So $\lambda=-16/21$ and the point of perpendicularity is given by $(10/21,47/21,43/21)$.