Vector Analysis Problem No.1

Given the two vectors r_A= -ax-3ay-4az and r_B= 2ax+2ay+2az, and point C(1,3,4), find: (a) R_AB; (b)|r_A|; (c)a_A; (d)a_AB; (e) a unit vector directed from C toward A.

Solutions:

(a) find R_AB :

R_AB = (2+1)ax +(2+3)ay + (2+4)az

R_AB = 3ax +5ay+6az

(b) find |r_A| :

rA = -ax-3ay-4az

|r_A| = √(1²+3²+4² ) = √26 = 5.10

(c) find a_A:

a_A = rA∕|rA| = (-ax-3ay-4az)/5.10 = -0.196ax-0.588ay-0.784az

(d) find a_AB:

a_AB = R_AB/|R_AB| = (3ax+5ay+6az)/√(3²+5²+6²) = 0.359ax+0.598ay+0.717az

(e) find a unit vector directed from C toward A:

R_CA = (-1-1)ax+(-3-3)ay+(-4-4)az = -2ax-6ay-8az

|R_CA| = √(2²+6²+8²) = 10.2

a_CA = R_CA/|R_CA| = (-2ax-6ay-8az)/10.2 = -0.1961ax-0.588ay-0.784az