Mathematics

27/3848–12FE Reference Handbook 10.6 · Mathematics · p. 60 Length of a vector and direction cosines

Handbook formula

Vector Length and Direction Cosines

∣ A ∣ = a_{x}^{2} + a_{y}^{2} + a_{z}^{2}, cos α = \frac{a _{x}}{∣ A ∣}, cos^{2} α + cos^{2} β + cos^{2} γ = 1

p. 60 prints |A| = √(ax² + ay² + az²) and the direction cosines cos α = ax/|A|, cos β = ay/|A|, cos γ = az/|A|, with cos²α + cos²β + cos²γ = 1. Those cosines are the components of the unit vector. A positive y-component points up, not down. The vector below is an original example.

$a_{x}, a_{y}, a_{z}$: Components of A
$∣ A ∣$: Length of A
$α, β, γ$: Angles with the x, y, and z axes

Step-by-step solved example

A = ⟨2, 3, 6⟩. Find |A| and the three direction cosines.

A = ⟨2, 3, 6⟩ in the first octant: positive y is up. |A| = 7. Direction cosines 2/7, 3/7, 6/7.

1. Length
|A| = √(4 + 9 + 36) = √49 = 7.
$∣ A ∣ = 7$
2. Cosines
cos α = 2/7, cos β = 3/7, cos γ = 6/7. All components are positive, so the arrow is in the first octant (y up).
$2/7, 3/7, 6/7$
3. Check
(4 + 9 + 36)/49 = 1.
$1$

Answer: |A| = 7; cosines 2/7, 3/7, 6/7

8 practice questions

0/8 correct

1.|⟨2, 3, 6⟩| equals

2.cos α for that vector is

3.cos β is

4.cos γ is

5.cos²α + cos²β + cos²γ equals

6.The unit vector along A is

7.A negative ay component is drawn

8.The length formula on p. 60 is