Distance from A (where a line and plane intersect) to P (which is a point perpendicular to the plane...
A coordinates = [-2, 1, 3]
P coordinates = [2, 3, -1]
(original equation of plane is 2x + y -2z + 9 = 0)
..thanks!
Distance from A (where a line and plane intersect) to P (which is a point perpendicular to the plane...
A coordinates = [-2, 1, 3]
P coordinates = [2, 3, -1]
(original equation of plane is 2x + y -2z + 9 = 0)
..thanks!
I understand that you simply need the distance from P to the plane.
Use the following formula:
Given a plane $\displaystyle Q: ux+vy+wz+t=0 $, the distance from a point $\displaystyle M(x_0,y_0,z_0)$ to Q is:
$\displaystyle d(M,Q)=\frac{|ux_0+vy_0+wz_0+t|}{\sqrt{u^2+v^2+w^2 }}$
In your case, you have: $\displaystyle d=\frac{|(2)(2)+(1)(3)+(-2)(-1)+9|}{\sqrt{2^2+1^2+2^2}}=\frac{18}{3}=6$
hope this helps
The way you have stated this, you are just asking for the distance from A to B: $\displaystyle \sqrt{(-2-2)^2+ (1- 3)^2+ (3+ 1)^2}$.
That will be the "distance from point to plane" only if the line connecting the two points is perpendicular to the plane. The line from (-2, 1, 3) to (2, 3, -1) is given by x= -2+ 4t, y= 1+ 2t, z= 3- 4t which has "direction vector" <4, 2, -4>. That is, in fact, in the same direction as the normal vector, <2, 1, -2> of the plane so, apparently, you had already done the hard work!
The distance from point (2, 3, -1) to the plane 2x + y -2z + 9 = 0 is $\displaystyle \sqrt{(-2-2)^2+ (1- 3)^2+ (3+ 1)^2}= \sqrt{16+ 4+ 16}= 6$