Question

1. Find the scalar component of A in the direction of B and find the angle between the two vectors where: A = (-ỉ+] - 4k)N B = (-i + 10j - 6k)mm 2. Find a unit normal vector to a triangle who's vertices are located at points: Point A: (12, 5, 10) mm Point B: (2,4,-2) mm Point C: (-22, 5, -10) mm

Answer 1

Given À = –? + 7j - 4K N B = - +108 -66 mm scalar component of Å in Al coso Å is direction of - Å. B BI -fi += -48).(-î +1o-C) lî -10 8 6²) - E06) +(+) (0)+(-476) V 240? +6² scalar component of direction of à in = 8.116398 B Al coso - 95 137 code - 97 +1 3+- 4K) 95 95 Vist TEG o-cost (as R & B is 2.4851° ) = 2.4851 : Angle between

2 Given, A = (12, 5, 10) = 12 h +58 Hor B=(2, 4,-2) = 27 +48 - 2² c = (-22, 5, 10) = -227 tsh - lok AB = OBOR -47 - 47 - 2) - (12+ +oº) = -107 - 9 - 12k Here ö is origin BC - oc -a oi +oftor - (22+54 -Jok) - (23 +4ģ-28) - -249 +ġ +(-88) Normal vector - AB x B2 - i Í K H0 - 12 1-24 i -8 = i (8+12) – j (80-288) + (10-24) - 200 + 208-34k Unit normal vector - ABBE TAB BC) 20.1 +208] -34k 1202 + 205* +34

Unit normal vector = l Tagsco (20? +2067 (20i +208} -34R). 144820

I hope this will help you.