Question

Q4 / Given the points M (0.1, -0.2, -0.1), Ñ (-0.2, 0.1, 0.3), and Ž (0.4, 0, 0.1) in Cartesian coordinate.
Find: 1. The Vector RMN
2. The dot product Rmn. RMP
3. The angle between Rmn. RMP
4. Spherical coordinate for vector M

Answer 1

ip-fm971) Giver M = 10011-0.2 -0.1) N = (-0.2, 0.1,0.3) P=6004, 0,001) ros sereymui RMN RN-RM S fona +0:16+0•36)- Coola -0.26 -0.6). 11 - (-0.29. coola J + Coolb+ Odb) + (0-36Foll) -0.3a + 0.35 to 40 AMN - RMP - - (Ooha tobto.c)-Cola -0.20-0-0) + (0-4a-ola) + (05 tovab) + Coole + Olc) AHP 0.39 +0.25 +0dc. Eine

step & to find dot product blw Rn No RM p ad aloo ongle las (RmnoRMP)= (Rmn• (RMP) 18mml/RMP) -> 1RMALE (2003)² + 10-37% + 10.41² - 0 0.583 65 0.4123 Irrpl = 10.31²+ 10.23² + 10021²= The dot produt blw RMN. PAPA RMN PTP = (-0.3 +0.3 jt (0-3+ox] + (0403) O + (0.5) + (0-6). Ooll To find angle substitute eq ② i ca all in > Cos (RNDOR MP) = To-s83). 10-4123/. tobo = Pala] cool RMN PMP/= 0.487.x0.5. Cao (PMN PMP) = 0.5 x Yao cao (BMN. PMP) = along Cao (60) RM NORM P= 60° [2006- bol. The angle blw RMN - RMP =

do 9 to find the sphetical co-ordinates of H. 200 Ma 10.1, 2012, 0.3. 1200 e y=-0.2 3 = 0) 8) loma JO-9410.234710.0" 0.244 . 8 = 0.244. Com ton Gle ton (-0.2) = -63-43° +607-) : -6 31439 ton ²+ y² 6 - 6 = tem 10-027107 = -65° 90