How to find the direction cosines and direction angles of a vector

What are the direction cosines and direction angles?

We’ll use the following formulas to find direction cosines and direction angles of a vector.

The direction cosine formulas are

$\cos{\alpha}=\frac{x}{D_a}$

$\cos{\beta}=\frac{y}{D_a}$

$\cos{\Upsilon}=\frac{z}{D_a}$

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and the direction angle formulas are

$\alpha=\arccos{\frac{x}{D_a}}$

$\beta=\arccos{\frac{y}{D_a}}$

$\Upsilon=\arccos{\frac{z}{D_a}}$

where $a=\langle{x},y,z\rangle$ represents the vector and $D_a$ represents the vector length, and where the direction angles are measured in degrees.

Realize that the direction angles are found by taking $\arccos$ of both sides of the direction cosine formulas.

How to find the direction cosines and direction angles of a vector

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Building the direction cosines and direction angles after converting the vector into standard form

Example

Find the direction cosines and direction angles of the vector.

$5x-3y+z=2$

We’ll change the vector function into standard form.

$5x-3y+z=2$

$a=\langle5,-3,1\rangle$

Next we’ll use the distance formula to find the length of the vector $a$. Remember that the initial point of the vector is the origin $(0,0,0)$, and the terminal point is $(5,-3,1)$.

$D_a=\sqrt{(5-0)^2+(-3-0)^2+(1-0)^2}$

$D_a=\sqrt{25+9+1}$

$D_a=\sqrt{35}$

Plugging this value and the vector $a$ into the direction cosine formulas, we get

$\cos{\alpha}=\frac{5}{\sqrt{35}}$

$\cos{\beta}=\frac{-3}{\sqrt{35}}$

$\cos{\Upsilon}=\frac{1}{\sqrt{35}}$

Realize that the direction angles are found by taking arccos of both sides of the direction cosine formulas.

Taking $\arccos$ of both sides of our direction cosines, we’ll be the values for the direction angles, which will be in degrees.

$\alpha=\arccos{\frac{5}{\sqrt{35}}}$

$\alpha=32.3^\circ$

and

$\beta=\arccos{\frac{-3}{\sqrt{35}}}$

$\beta=120.5^\circ$

and

$\Upsilon=\arccos{\frac{1}{\sqrt{35}}}$

$\Upsilon=80.3^\circ$

To summarize our findings, we can say that

the direction cosines are

$\cos{\alpha}=\frac{5}{\sqrt{35}}$

$\cos{\beta}=\frac{-3}{\sqrt{35}}$

$\cos{\Upsilon}=\frac{1}{\sqrt{35}}$

the direction angles are

$\alpha=32.3^\circ$

$\beta=120.5^\circ$

$\Upsilon=80.3^\circ$

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