Question

Vectors a and b lie in an xy plane. a has magnitude 5.7 and angle 130 degrees relative to +x direction; b has components Bx = -5.34 and By = -4.37. What are the angles between the negative direction of the y axis and (a) the direction of a , (b) the direction of the product a x b , and (c) the direction of a x (b + 3.00k) ?

Answer 1

a)

here

angle between -y and A = 90 + 50 = 140 deg

b)

angle - y , ( A X B) = C -----> angle -j , k because C perpendicular plane (A ,B) = xy ----> 90deg

c)

direction A X ( B + 3k) = D

E = B + 3k = -5.34 i - 4.37 j + 3k

D = A X E

A = 5.7 * cos130 i + 5.7 * sin130 j

A = -3.6 i + 4.36 j

then

D = A X E = i (4.36·3 - 0·(-4.37)) - j ((-3.6)·3 - 0·(-5.34)) + k ((-3.6)·(-4.37) - 4.36·(-5.34))

= i (13.08 + 0) - j (-10.8 + 0) + k (15.732 + 23.2824)

= 13.08 i ; 10.8 j ; 39.01 k

then magnitude of D

D = sqrt( 13.08^2 + 10.8^2 + 39.01^2) = 42.53

- j . D = - j . ( 13.08 i + 10.8 j + 39.01 k ) = - 10.8

then

theta = cos^-1( -10.8 / 42.53) = 104.7 deg