Multivariate Calculus - Bruce E. Shapiro

LECTURE 2. VECTORS IN 3D 17
and the direction angles are
( v · i
α = arccos
‖v‖
)
, β = arccos
( ) v · j
, γ = arccos
‖v‖
( ) v · k
‖v‖
(2.29)
Example 2.6 Find the direction cosines and angles of the vector v = (4, −2, −4)
Solution. First, we calculate the magnitude of v,
Thus the three direction cosines are
‖v‖ = √ (4) 2 + (−2) 2 + (−4) 2 = 6
cos α = v · i
‖v‖ = 4 6 = 2 3
cos β = v · j
‖v‖ = −2 6 = −1 3
cos γ = v · k
‖v‖ = −4 6 = −2 3
and the direction cosines are
( 2
α = arccos ≈ 48.19 deg
3)
(
β = arccos − 1 )
≈ 109.47 deg
3
( 2
γ = arccos ≈ 131.81 deg
3)
Projection of one vector on another vector.
Consider any two vectors u and v, as illustrated in figure 2.6. We can always express
v as the sum of two vectors:
v = m + n (2.30)
where m is parallel to u and n is perpendicular to u. Let θ bet the angle between
u and v. Then
‖m‖ = ‖v‖ cos θ (2.31)
Since m and u are parallel, then they must have the same unit vectors, so that
Therefore
and since v = m + n,
ˆm = û
m = ‖m‖ ˆm = (‖v‖ cos θ) ˆm = (‖v‖ cos θ) û = (v · û) û (2.32)
n = v − m (2.33)
Math 250, Fall 2006 Revised December 6, 2006.