# How do you find the direction cosines and direction angles of the vector?

Jul 1, 2016

#### Answer:

If the vector is $\left(x , y , z\right) \mathmr{and} r = | \left(x , y , z\right) |$, the direction cosines are$\left(\frac{x}{r} , \frac{y}{r} . \frac{z}{r}\right)$ and the angles are $\left({\cos}^{- 1} \left(\frac{x}{r}\right) , {\cos}^{- 1} \left(\frac{y}{r}\right) , {\cos}^{- 1} \left(\frac{z}{r}\right)\right)$.

#### Explanation:

If the vector is $x i + y j + z k = \left(x , y , z\right)$ and

r = length of ( x, y, z ) =$\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$,

the direction cosines are$\left(\frac{x}{r} , \frac{y}{r} . \frac{z}{r}\right)$ and the angles are

$\left({\cos}^{- 1} \left(\frac{x}{r}\right) , {\cos}^{- 1} \left(\frac{y}{r}\right) , {\cos}^{- 1} \left(\frac{z}{r}\right)\right)$

Example: Vector is (1, 1, 1)..

$r = \sqrt{1 + 1 + 1} = \sqrt{3}$.

Direction cosines are

$\left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}}\right)$.

Angles are

$\left({\cos}^{- 1} \left(\frac{1}{\sqrt{3}}\right) , {\cos}^{- 1} \left(\frac{1}{\sqrt{3}}\right) , {\cos}^{- 1} \left(\frac{1}{\sqrt{3}}\right)\right)$