How do you find the direction cosines of the vector #u=2i -4j+4k# and find the direction angles?

Question

12641 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-02T11:46:24

#alpha = arccos(1/3) = 1.231#
#beta = arccos(-2/3) = 2.3#
#gamma = arccos(2/3) = 0.841#

in radians

Given the unit base vectors #hat i,hat j,hat k# we have

#<< vec u, hat i >> = abs(u) cos(alpha) = 2->cos(alpha) = 2/sqrt(2^2+4^2+4^2) = 1/3#

#<< vec u, hat j >> = abs(u) cos(beta) = -4->cos(beta) = -4/sqrt(2^2+4^2+4^2) = -2/3#

#<< vec u, hat k >> = abs(u) cos(gamma) = 4->cos(gamma) = 4/sqrt(2^2+4^2+4^2) = 2/3#

with

#alpha = arccos(1/3) = 1.231#
#beta = arccos(-2/3) = 2.3#
#gamma = arccos(2/3) = 0.841#