Question

Q = (0,6, -4) R= (5,-4, -5) Consider the triangle with vertices: P= (-2,0, -1) (a) Find the vectors PO, PŘ, and QŘ (b) What is the measure of the angle at P (ZQPR)? (c) What is the perimeter of the triangle APQR ? (d) What is the area of the triangle APQR? (e) Find a vector that is perpendicular to the plane containing P, Q, and R Verify that the vector you have found is perpendicular to PO (f) Find the equation of the line in 3-space that passes through the points P and Q (g) Find the equation of the plane containing the points P, Q, and R

Answer 1

- î P = (-2,0,-1); ² 9 = (0,6,-4) o = 69 -40 (5,-4,-5); 5 î -47-5 R = (a) P9 11 zî t6 9 -3 PR Ř - = rî -4Î , 4 Q R = R - = 5î - 10 § - 1 p8 l = 52²+6² +63)* √49 2 √(7)2 +-4) 2+ (4) 49+16 +16 = 81 9 (10) + (-1)2 126 2 11.22 o Po PR = 1861 18k I lose 2) (2^ t6] - 3ů). (77-45-4Q) = 7.9 losa -) Losa 63

= cos23) =178 ) +18RIT PR 7+11-22 +9 e say the vectors Š perpendicular to the 63 ZOPR = 88,18' (e) Perimeter of triangle - 27.22 Area of triangle, A = // 129 X PR = 1 1 1 19 1 | FR lx sina = 7.9 31.5 (d) Cosa = 2²3 sind=fi-Cosa = 0.999 1 blane P, Q, R. So, Š = P8XP û unr - 4 - 4 S š = -36ê - 13] - 50 m po iš - = (2î +6] - 3). (-360-137-50M ) =-72-78 + 150 = 0

pg. So, the rector's is perpendicular to P=(-2,0-1) 9 = (0,6,-4) R=(5,-4,-5) line passes through P fg in 30 space y-YI x-24 Z-ZI 22-24 Y2-Y, a Zz-Z, Zt - 4+1 a+2 y-o o't2 x+2 2 -3 ..R Z+1 6 This is the eas of line in 3 space passes through P& q. (g) To find can of plane the general form of ear in $ A(x-2) + B(y-Y,) + C(2-2) = 0 For point P(-2, 0, -1) kan of plane ic A(x+2) + B(4-0) + @ (2+1)=0-0 This ears of plane satisfying for point Q&R. For 9, A (0+2) & B(6-0) + C (-4+1) = 0 = 2A + 6B - 300 For R15, -4,-5) А 11 A ( 5+2) + B (-4-0) +(( -5+1) = 0 = 7A-4B-4C =0 (in A, B, C are coust where

To solve these can we take the canations in determinant form x+2 у Z+1 2 -3 7 -4 -4 7 - 36(x+2) - 13y - 50 (2+1) = o. 736(x+2) +134 + 50 (Z+1)