# Using a normal vector and point to find distance and the closest point

• Apr 6th 2011, 05:02 AM
cottontails
Using a normal vector and point to find distance and the closest point
"Find the distance from P(3,0,-1) to the plane P described by the equation
4x + 2y - z = 6.
Find the closest point to P which lies on P."

With this question, I basically do not know how you would answer the question to find the distance (given that point and the equation). I had a look at the answer in hope for some direction although, I do not understand why and how you would be able to figure out to find the length of vector P to Q (dot product) n-hat to get the distance. I know that you are able to find the normal vector from the equation although, how are you able to find the point Q that lies on P (this was given in the answer)? Moreover, I'm unsure with how you find the closest point. I would assume that you would need the distance to find it though.

Unfortunately, this was not something that was covered in my lectures nor it is in my textbook so, any help would be appreciated.
• Apr 6th 2011, 06:44 AM
HallsofIvy
To measure the distance from a point, A, to a plane, p, you measure along the line through A perpendicular to the plane, p. That's true because a leg of a right triangle is always shorter than the hypotenuse. Further, if the vector is written as Ax+ By+ Cz= D as yours is, then the vector <A, B, C> is perpendicular to the plane. So you want to measure the distance from A to p, you want a line that passes through P and is in the direction of the vector <A, B, C>. Okay, a line through point $(x_0, y_0, z_0)$ in the direction of vector <A, B, C> is given by the parametric equations $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$. Here, you are given the equation of the plane so you know A, B, and C. You are given the point so you know $x_0$, $y_0$, and $z_0$. Use that information to write the equation of the line through (3, 0, -1) perpendicular to 4x+ 2y- z= 6. The point at which that line intersects the plane (replace x, y, and z in the equation of the plane by there expressions in terms of t for the line and solve for t) is the point on the plane closest to .