Distance From a Point to a PlaneDate: 03/31/98 at 16:57:06 From: Keir Larock Subject: Distance between a point (x0,y0,z0) and the plane ax+by+cz+d Hi, In my calculus class the other day, we went over the proof that the distance from the point (x0,y0,z0) to the plane ax + by + cz = d is |ax0 + by0 + cz0|/((a^2 + b^2 + c^2)^(1/2)) I was wondering if you could do it out for me, as I have lost my notes. Thank you again for your time and help; it is greatly appreciated! Keir Larock Date: 04/01/98 at 10:24:43 From: Doctor Rob Subject: Re: Distance between a point (x0,y0,z0) and the plane ax+by+cz+d Hi Keir, Thanks for writing to Dr. Math. Consider the plane a*x + b*y + c*z = d. The vector v = (a,b,c) is normal to the plane. Pick any point on the plane, P1 = (x1,y1,z1), and you are given the point P0 = (x0,y0,z0) in space. Then a*x1 + b*y1 + c*z1 = d. The vector from P1 to P0 is: P0 - P1 = (x0 - x1, y0 - y1, z0 - z1) This vector, projected onto v, will have length the distance from P0 to the plane. That length is: d = |P0 - P1|*cos(theta) = |v.(P0 - P1)|/|v| = |a*(x0 - x1) + b*(y0 - y1) + c*(z0 - z1)|/sqrt(a^2 + b^2 + c^2) = |a*x0 + b*y0 + c*z0 - d|/sqrt(a^2 + b^2 + c^2) -Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 04/01/98 at 13:16:05 From: Keir Larock Subject: Re: [Re: two problems (x^y = y^x and distance between a point (x0,y0,z0) and the plane ax+by+cz+d)] Hi, Thank you very much for your help! Every time I write I am amazed at how quickly and thoroughly my questions are answered! I cannot thank you guys enough... Thank you, Keir Larock |
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