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

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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

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

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