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Vector Components

Vector Components in 2D

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

In the simulation below, you can see a vector of variable magnitude based on the movement of your mouse over the simulation. Notice how the x and y axis component lengths of the vectors can be positive or negative and the angles made between the vector component lengths and the vectors themselves can be positive or negative. The numerical value for the magnitude (length) regardless whether or not it is pointing in the positive or negative direction, is always a positive scalar. Note that when drawing a vector with its components, the magnitude will always be positive, but its components can be either positive or negative, both positive, or both negative.

A vector and its vector components can be represented in numerous ways. (Note that in this simulation we have positive and negative directions and these values were chosen at random). Here are the three most common ways of portraying them:

\[ \vec{C} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ -18.4 & 9.8 & 0 \end{bmatrix} \]

or

\[ \vec{C} = -18.4\hat{i} + 9.8\hat{j} + 0\hat{k} \]

or

\[ \vec{C} = \langle -18.4, 9.8, 0 \rangle \]

vector components in the x and y axis can be calculated using the following equations:

\[ \cos(\theta) = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{v_x}{v} \]

or

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

\[ \sin(\theta) = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{v_y}{v} \]

or

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

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

\[ v_x = 20.8 \cos(151.98) \]

\[ v_x = 20.8 \cdot (-0.88278366258) \]

\[ v_x = -18.4 \]

\[ v_y = 20.8 \sin(151.98) \]

\[ v_y = 20.8 \cdot (0.46977974103) \]

\[ v_y = 9.8 \]

In the first case, we take the x-axis component of the vector to have a magnitude of 20.8, with an angle of 151.98 degrees. Following the formula, we get that the x-axis component of the vector is -18.4. Similarly, in the second case, we take the y-axis component of the vector to have a magnitude of 20.8, with the same angle of 151.98 degrees. Following the formula, we get that the y-axis component of the vector is 9.8. These two 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:

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

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

\[ |\vec{v}| = \sqrt{-18.4^2 + 9.8^2} \]

\[ |\vec{v}| = \sqrt{338.56 + 96.04} \]

\[ |\vec{v}| = \sqrt{434.6} \]

\[ |\vec{v}| = 20.8 \]

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 20.8 above, which is correct.

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