Direction Cosines Calculator
Enter the vector components (2D or 3D) to calculate its direction cosines and see the step-by-step solution.
Calculate direction cosines
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Quick Examples
\vec{v} = \langle 2, -3, 6 \rangle
\vec{v} = \langle 1, 2, 1 \rangle
\vec{v} = \langle -4, 0, 3 \rangle
\vec{v} = \langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \rangle
\vec{v} = \langle \sqrt{3}, -\sqrt{2}, \sqrt{5} \rangle
\vec{v} = \langle -5, 12 \rangle
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Solved Exercises
The following are examples of problems solved by the calculator.
Calculate the direction cosines of the vector ⟨2, -3, 6⟩.
Result
The direction cosines of the given vector are:
$$ \displaystyle \cos(\alpha) = \frac{2}{7} \approx 0.29 \\[1em] \displaystyle \cos(\beta) = -\frac{3}{7} \approx -0.43 \\[1em] \displaystyle \cos(\gamma) = \frac{6}{7} \approx 0.86 $$
Step-by-step solution
1. Identify the vector components.
The vector to work with is:
$$ \displaystyle \vec{v} = \langle 2, -3, 6 \rangle $$
Extract the values of its components:
$$ \displaystyle v_x = 2 \\[0.5em] v_y = -3 \\[0.5em] v_z = 6 $$
2. Calculate the magnitude of the vector.
Apply the formula to find the magnitude or length of the vector, which is needed for the direction cosines:
$$ \displaystyle |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \\[1em] |\vec{v}| = \sqrt{(2)^2 + (-3)^2 + (6)^2} = 7 $$
3. Calculate the direction cosines.
Divide each vector component by its magnitude to obtain the respective direction cosine:
$$ \displaystyle \cos(\alpha) = \frac{v_x}{|\vec{v}|} = \frac{2}{7} \approx 0.29 \\[1em] \cos(\beta) = \frac{v_y}{|\vec{v}|} = \frac{-3}{7} = -\frac{3}{7} \approx -0.43 \\[1em] \cos(\gamma) = \frac{v_z}{|\vec{v}|} = \frac{6}{7} \approx 0.86 $$
If, in addition to the direction cosines, we need to find the direction angles, we must apply the arccosine function to each value. You can see the step-by-step process in the vector direction calculator .
Find the direction cosines of the vector with components ⟨1, 2, 1⟩ .
Result
The direction cosines of the given vector are:
$$ \displaystyle \cos(\alpha) = \frac{1}{\sqrt{6}} \approx 0.41 \\[1em] \displaystyle \cos(\beta) = \frac{2}{\sqrt{6}} \approx 0.82 \\[1em] \displaystyle \cos(\gamma) = \frac{1}{\sqrt{6}} \approx 0.41 $$
Step-by-step solution
1. Identify the vector components.
The vector to work with is:
$$ \displaystyle \vec{v} = \langle 1, 2, 1 \rangle $$
Extract the values of its components:
$$ \displaystyle v_x = 1 \\[0.5em] v_y = 2 \\[0.5em] v_z = 1 $$
2. Calculate the magnitude of the vector.
Apply the formula to find the magnitude or length of the vector, which is needed for the direction cosines:
$$ \displaystyle |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \\[1em] |\vec{v}| = \sqrt{(1)^2 + (2)^2 + (1)^2} = \sqrt{6} \approx 2.45 $$
3. Calculate the direction cosines.
Divide each vector component by its magnitude to obtain the respective direction cosine:
$$ \displaystyle \cos(\alpha) = \frac{v_x}{|\vec{v}|} = \frac{1}{\sqrt{6}} \approx 0.41 \\[1em] \cos(\beta) = \frac{v_y}{|\vec{v}|} = \frac{2}{\sqrt{6}} \approx 0.82 \\[1em] \cos(\gamma) = \frac{v_z}{|\vec{v}|} = \frac{1}{\sqrt{6}} \approx 0.41 $$
If, in addition to the direction cosines, we need to find the direction angles, we must apply the arccosine function to each value.
Find the values of the direction cosines for the vector ⟨-5, 12⟩ .
Result
The direction cosines of the given vector are:
$$ \displaystyle \cos(\alpha) = -\frac{5}{13} \approx -0.38 \\[1em] \displaystyle \cos(\beta) = \frac{12}{13} \approx 0.92 $$
Step-by-step solution
1. Identify the vector components.
The vector to work with is:
$$ \displaystyle \vec{v} = \langle -5, 12 \rangle $$
Extract the values of its components:
$$ \displaystyle v_x = -5 \\[0.5em] v_y = 12 $$
2. Calculate the magnitude of the vector.
Apply the formula to find the magnitude or length of the vector, which is needed for the direction cosines:
$$ \displaystyle |\vec{v}| = \sqrt{v_x^2 + v_y^2} \\[1em] |\vec{v}| = \sqrt{(-5)^2 + (12)^2} = 13 $$
3. Calculate the direction cosines.
Divide each vector component by its magnitude to obtain the respective direction cosine:
$$ \displaystyle \cos(\alpha) = \frac{v_x}{|\vec{v}|} = \frac{-5}{13} = -\frac{5}{13} \approx -0.38 \\[1em] \cos(\beta) = \frac{v_y}{|\vec{v}|} = \frac{12}{13} \approx 0.92 $$
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