Question

Displacement D1 is in the yz plane 58.8^o from the positive direction of the y axis, has a positive z component, and has a magnitude of 4.34 m. Displacement D2 is in the xz plane 27.2^o from the positive direction of the x axis, has a positivezcomponent, and has magnitude 2.23 m. What are (a) Dot product of D1 and D2 (b) the x component of D1 x D2 , (c) the y component of , (d) the z component of D1 x D2  , and (e) the angle between vector D1 and D2 ?

Answer 1

The z component of D is, D, sin e -4.34 m 4.34 m 4.34 m - 5.07 m sin θ sin 58.80 The y component of D is, D cos (5.07m)cos 58.8 2.63m Therefore vector D can be expressed as, D 2.63j+4.34k The z component of D,is, D, sin e 2.23 m 2.23m 2.23 m Sin θ sin272 The x component of D,is, D cos (4.88m)cos 27.2 4.34m Therefore vector D, can be expressed as, D2-4.34i+223k

Dot product of the two vectors is, D D (2.63j +4.34k) (4.34i +223)-9.68 Cross product of two vectors is, Dx D (2.63 j+4.34k)x(4.34i+2.23k) --114k+5.86i +18.8j -5.86i +18.8j-11.6k Thus, the x component of the cross product is 5.86i (c) The y component of the cross product is 18.8j (d) The z component of the cross product is -11.6k The angle between two vectors is,

θ = cos 9.68 -1 2.63)' +(4.34)214.34),(2.23)/ cos =67。