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???
