Question

1. Consider the points A(4,0,0), B(0,3,0),C(0,0,5), and P(1,2,8). a. Write the equation of the plane passing through the points A, B, and C. b. Write parametric equations of a line passing through point P and orthogonal to the plane described in part a. c. Find the exact distance between point and the plane described in part a.

Answer 1

Solving the problem we get
The given points A (4,0,0), B(930) CL005) 4. The equation f the plane passing through three points (1) Y, Z,), m 3, 2) and (ny, z) is n-ny m-4 yoy, z-z, 9,-), Zz-Z1 = 0 yz-y, y-ty Since the plane passing through the three points A (9,00) B (030) and (2005) i. The equation of the plane is n-4 yo 2-0 0-0 3-0 0-0 S-O 0-4 0-4 n-4 y 2 0 کرا و -4 5 → () [305)-10] -y [9(5)-(-400)] +7 (24) Co)-(403)]=0 >(1-4) [15-0] --Y[-20-0] + Z [0+12] = 0 7 15(x-4) + 20y + 122=0

→ 150–60 + 20y + 122 =0 >154 +20y + 12 Z = 60 b) The parametric equation of a line passing through a point p ca) and parallel to a lector r is r = attb. Here the orthogonal line passing through the point P (1,2,8) and parauel to direction ratio of the normal of the plane ♡ : The direction ratio's of the normal of the plane 15, 2012 are :. The equation of the line is ř = (1+29+8R) + t (15i +209 + 12k) This is the equation of vector parametric form

and let the point M = nityj+ Zk. (ni+y5+Zk) = (1+ 15H) +(2+20+) 9 +18 + 124R) 2 comparing both side we get the parametric equation of the line is N = 1+15+ y = 2 +201, Z=8+126, Where t is a parcameter

© The distance of a point pcrgy, z) to a plane an tbytez=d is arytby, tezind a2+ b²te2 :. The distance of the point P (128) from the plane 15n +206 + 12z=60 is

15(1)+ 2012) + 1218) - 60 = (15)2 + (20)2+(120% 15+40+96-60 5225 + 400 7144 91 √769 unit. 91 769 769 -