Question

Displacement 1 is in the yz plane 65.4 ° from the positive direction of the y axis, has a positive z component, and has a magnitude of 5.31 m. Displacement d2 is in the xz plane 38.3 ° from the positive direction of the x axis, has a positive z component, and has magnitude 2.23 m. What are (a) did2 . (b) the x component of dixd2 .(e) they component of dixd2 , (d) the z component of dix d2 , and (e) the angle between di and d2 7 Units C (a) Number (b) Number (c) Number (d) Number (e) Number Units Units Units Units

Answer 1

here,

d1 = 5.31 * ( cos(65.4) j + sin(65.4) k)

d1 = 2.21 j m + 4.83 k m

d2 = 2.23 * ( cos(38.3) i + sin(38.3) k)

d2 = 1.75 i + 1.38 K m

a)

d1 . d2 = ( 2.21 j m + 4.83 k m ) . (1.75 i m + 1.38 k m ) = 6.67 m^2

b)

d1 X d2 = ( 2.21 j m + 4.83 k m ) X (1.75 i m + 1.38 k m )

d1 X d2 = ( - 2.21 * 1.75 k + 2.21 * 1.38 i + 4.83 * 1.75 j)

d1 X d2 = 3.05 i m + 8.45 j m - 3.87 k m

the x- component is 3.05 m

c)

the y- component is 8.45 m

d)

the z- component is - 3.87 m

e)

the angle between d1 and d2 , theta = arccos(d1.d2/(|d1|*|d2|))

theta = arccos(6.67 /( sqrt(2.21^2 + 4.83^2)) * sqrt(1.75^2 + 1.38^2))

theta = 55.7 degree