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Cartesian components and direction cosines

(Definition)

Cartesian Components and Direction Cosines

Choose a right-handed orthonormal Cartesian basis

ˆ ˆ ˆ
i,j,k.

Every vector has the unique component form

|--------------------|
u--=-uxˆi +-uyˆj +-uzˆk.-                              (1)

PIC

Figure 11, modernized: decomposition into Cartesian components.

For a point P = (x,y,z),

rP = xˆi + yˆj + zˆk.

Hence

−P− →P  = (x  − x )ˆi + (y − y )ˆj + (z − z )ˆk.
  1 2     2    1      2    1       2   1

The magnitude is

    2    2    2    2
∥u ∥  = ux + uy + uz.                               (2)

If α,β,γ are the direction angles,

u  = ∥u ∥cos α,  u  =  ∥u∥ cosβ,   u  = ∥u∥ cosγ,
 x                 y                z

so

   2       2        2
cos α + cos  β + cos γ = 1.

For planar vectors it is better numerically to use

𝜃 = atan2 (uy,ux)

rather than reconstructing the quadrant from tan 𝜃 = uy∕ux.

Source examples

For u = (−2, 1, 2),

∥u ∥ = 3,

with direction cosines (−2∕3, 1∕3, 2∕3).

If u = (5, 2) and v = (−3,−4), then

                                 √ --
w  = u + v =  (2,− 2),    ∥w ∥ =   8,

and 𝜃 = −45∘ (equivalently 315∘).

A line through P1 = (x1,y1,z1) parallel to ℓ = (a,b,c) can be written parametrically as

r = r1 + λ ℓ,

or, when the denominators are nonzero,

x-−-x1-= y-−-y1 =  z-−-z1.
  a         b        c

Source problems

  1. Find u + v + w graphically and analytically for:
    1. ∥u∥ = 6,𝜃u = 60∘; ∥v∥ = 10,𝜃 v = 120∘; ∥w∥ = 8,𝜃 w = 270∘.
    2. Three unit vectors at 0∘, 120∘, 240∘.
    3. u = 2î + 3ĵ, v = −5î + 2ĵ, w = −ĵ.
    4. u = î, v = −ĵ, w = î −ĵ.
  2. Add
             ˆ         ˆ            ˆ
2ˆi + 3 ˆj − k, ˆi + 2 k, ˆi + ˆj − 3 k,

    and find the resultant magnitude and direction cosines.

  3. Given A = (−1, 2,−1), B = (−3, 6, 6), C = (4, 3, 1), D = (0, 0, 2), find the length and direction cosines of −→
AB + − →
AC + −−→
AD.
  4. Using the same points: find the point dividing AB in the ratio 2 : 1; find the centroid of A,B,C; and show that the side midpoints of the skew quadrilateral ABCD form a parallelogram.
  5. Prove
            2       2      2
∥u +  v∥  = ∥u ∥ +  ∥v∥  + 2∥u ∥∥v∥ cos𝜃.
  6. Deduce
    cos 𝜃 = uxvx-+-uyvy-+-uzvz.
             ∥u ∥∥v ∥
  7. Find the Cartesian equations of the line through A = (−2, 0, 2) and B = (2, 1,−3).

Modern notation references

The notation and terminology in this modernized article follow standard present-day mechanics and vector-analysis usage, particularly:

  1. J. R. Taylor, Classical Mechanics, University Science Books, 2005.
  2. D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.
  3. H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison–Wesley, 2002.

Source

This article is a modernized restatement of the corresponding Public Domain article in Louis Brand, Vectorial Mechanics, John Wiley & Sons, New York, 1930, Chapter I, “Vector Algebra.” The original 1930 edition is the source basis.


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See Also: scalar component and vector projection on an Axis, vectors in space, vectors in a plane, vector subtraction and position vectors, negative of a vector, equality of vectors, vector, vector algebra, vector addition, point division and position vectors, centroids and weighted position vectors, vector product, dot product, dot product algebra and geometric applications, cross product, cross product algebra and applications, scalar triple product, summary of vector algebra


Cross-references: mechanics, unit vectors, direction cosines, magnitude, vector

This is version 3 of Cartesian components and direction cosines, born on 2026-08-21, modified 2026-08-21.
Object id is 1076, canonical name is CartesianComponentsAndDirectionCosines.
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Classification:
Physics Classification: 02. (Mathematical methods in physics)

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