Question

1. (15 points) (a) (5 points) Find the equation of the plane a that contains points A(1,5,4) B(1,0, 1) and C(4, 0,5) (b) (5 points) Find the distance from the point D(2, 1,7) to this plane (c) (5 points) If plane 3 has equation y -3z+2x = 5, find a unit vector that is parallel to the intersection of a and B.

Answer 1

art+byoz+dFO . Auns han find the equation of plane a, thoough A (15,4). B(1,0,1),¢4,0,5). Solution The equation of plane o di aspyrne Isla relatia in plane passing through (x, 49, 21) 2 tủ gx + b + c + d = 0 . . . > 01x-x1) +614-41) + 6(724) =O firot 9 will check the point are non-collineas 1.541 igen mil ioi Txions) - 5 (5-4) +4109 SI =-570'. The determinent is not eero. => the pointe ase not collinear Now eqn of plane in cortesian (22x) (7-4,) (2-21) 121-79) 18,-99) (77-79) 1-0 1. XX9-13) ($8-13) (3:23) 2-1 -15 2-41 (2-1). [-30) -(8-5) [ +9]+(2-4) (15770.. Le coin 720(x-1) +918-5) - 15 (2-4)=0 ) is the In required equation of plane...

o find the distance from the point D (2,1,7) to this plane. solution . The distance of the pt old, 1,7) to plane 2014-1) + 914-5)-15 (2-4)=0 20%-90 +94-45-152+60=0 20x+3y-152 -5=0 -0 d = 20 X 9 + 9 x 1541-5 62f V 46+81 +225 706 d= 4079-705-5 706 1706 V706 © If plane: op has egn ex+y = 32=5 - from © © 19 found the normal vector ( 20,9,-15). and 109,+, -3). The cross poordwet As 20 9-15 - i ( €50 -27 +15)-11-60+30) + f ( 20-18) -191+30-17 +2k Now check both vectos, ir thogonal to..(-12,30,2) cindnormalzed, -127+2,04 + 129+309tek vi c V1 + + + 4 ^ V10 78 !