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Vector Components

Vector Components in 3D

Vector Components are smaller vectors that help split an angled vector towards the coordinate axes in a two and three dimensional coordinate system.

Below is the same simulation as the one previously of a vector in the x, y, and z axis. It has a magnitude (length) of 4.690, and angles of 33.69 and 50.23 and its direction is northeast. Vectors portrayed in three dimensions have three vector components. As a student you will be able to change this vector's angle, direction, and magnitude.

Vector:

Magnitude: Theta: Phi:

Vector Components:

A vector and its vector components can be represented in numerous ways. Here are the three most common ways of portraying them:

\[ \vec{D} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \end{bmatrix} \]

or

\[ \vec{D} = 3\hat{i} + 2\hat{j} + 3\hat{k} \]

or

\[ \vec{D} = \langle 3, 2, 3 \rangle \]

The angles between the vector components in the x, y and z axis can be calculated using the following equations:

\[ \tan(\theta) = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{v_y}{v_x} \]

or

\[ \theta = \tan^{-1}\left( \frac{v_y}{v_x} \right) \]

and

\[ \cos(\phi) = \frac{\text{z-axis component}}{\text{Magnitude}} = \frac{v_z}{v} \]

or

\[ \phi = \cos^{-1} \left( \frac{v_z}{v} \right) \]

Now let's see what happens when we use the numbers chosen:

\[ \theta = \tan^{-1} \left( \frac{2}{3} \right) \]

\[ \theta = \tan^{-1} (0.66666666666) \]

\[ \theta = 33.69 \]

and

\[ \phi = \cos^{-1} \left( \frac{3}{4.690} \right) \]

\[ \phi = \cos^{-1} (0.63965884861) \]

\[ \phi = 50.23 \]

Vector Components in the x and y axis can be calculated using the following equations:

\[ v_x = v \cos(\theta)\sin(\phi) \]

\[ v_y = v \sin(\theta)\sin(\phi) \]

\[ v_z = v \cos(\phi) \]

Now let's see what happens when we use the numbers chosen:

\[ v_x = 4.690 \cos(33.69)\sin(50.23) \]

\[ v_x = 4.690 \cdot (0.83205094808) \cdot (0.76861857891) \]

\[ v_x = 3 \]

\[ v_y = 4.690 \sin(33.69)\sin(50.23) \]

\[ v_y = 4.690 \cdot (0.55469921561) \cdot (0.76861857891) \]

\[ v_y = 2 \]

\[ v_z = 4.690 \cos(50.23) \]

\[ v_z = 4.690 \cdot (0.63970733944) \]

\[ v_z = 3 \]

In the first case, we take the x-axis component of the vector to have a magnitude of 4.690, with angles 33.69 and 50.23 degrees Following the formula, we get that the x-axis component of the vector is 3. Similarly, in the second case, we take the y-axis component of the vector to have a magnitude of 4.690, with the same angles of 33.69 and 50.23 degrees. Following the formula, we get that the y-axis component of the vector is 2. Similarly, in the third case, we take the z-axis component of the vector to have a magnitude of 4.690, with the angle of 50.23 degrees. Following the formula, we get that the z-axis component of the vector is 3. These three results make sense when we look back at the simulation.

The magnitude (length) of the vector once you know the vector components of the vector can be calculated by the following equation:

\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]

Let's calculate the magnitude of the vector in the simulation:

\[ |\vec{v}| = \sqrt{3^2 + 2^2 + 3^2} \]

\[ |\vec{v}| = \sqrt{9 + 4 + 9} \]

\[ |\vec{v}| = \sqrt{22} \]

\[ |\vec{v}| = 4.690 \]

As we can see from the formula, regardless if a vector component is positive or negative, the component number value gets squared, making it a positive number. Solving the rest of the equation, we find the magnitude, is a positive scalar, which is 4.690 above, which is correct.

You have now learned the fundamentals of vector components. Go back now to the vector page to move onto the next lesson.

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