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Vector Subtraction

Vector Subtraction in 3D

Vector Subtraction involves calculating the difference of vectors. This is done by subtracting the vector components of each individual vector and once the result is obtained, those components are now the components of the new vector that is equal to the difference of the vectors.

Below is the same simulation as the one previously of vector subtraction in 3D. There is a red vector of magnitude 5.385, at an angle of 33.85 degrees. This vector is added to a blue vector of magnitude 6.403, at an angle of 8.99 degrees. The x-axis component of the red vector is +4, the y-axis component of the vector is +2, and the z-axis component of the vector is +3 because the red vector is pointing in the positive direction for the x, y, and z axes. The x-axis component of the blue vector is +6, the y-axis component of the vector is -2, and the z-axis component of the vector is +1 because the blue vector is pointing in the positive direction for the x and z axes and in the negative direction for the y-axis. The new vector, which is purple, now has a magnitude of 4.899, and an angle of 24.09 degrees. The x-axis component of this vector is -2, the y-axis component of this vector is +4, and the z-axis component of this vector is +2 because the purple vector is pointing in the positive direction for the y and z axis and in the negative direction for the x-axis. As a student you will change the component vectors of the blue vector and see how those new components result in a new purple vector, which is the difference of the red and blue vectors.

Red Vector:

x, y, z:

Blue Vector:

x, y, z:

Purple Vector:

The vectors in the simulation can be represented in numerous ways. Here are the three most common ways of portraying them:

\[ \vec{A} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 2 & 3 \end{bmatrix} \]

or

\[ \vec{A} = 4\hat{i} + 2\hat{j} + 3\hat{k} \]

or

\[ \vec{A} = \langle 4, 2, 3 \rangle \]

and

\[ \vec{B} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & -2 & 1 \end{bmatrix} \]

or

\[ \vec{B} = 6\hat{i} - 2\hat{j} + 1\hat{k} \]

or

\[ \vec{B} = \langle 6, -2, 1 \rangle \]

Vector Subtraction can be calculated using the following equations:

\[ \vec{A} - \vec{B} = \langle A_x - B_x, A_y - B_y, A_z - B_z \rangle \]

or

\[ \vec{A} - \vec{B} = (A_x - B_x)\hat{i} + (A_y - B_y)\hat{j} + (A_z - B_z)\hat{k} \]

Now let's see what happens when we subtract the red and blue vectors:

\[ \vec{A} = 4\hat{i} + 2\hat{j} + 3\hat{k} \]

\[ \vec{B} = 6\hat{i} - 2\hat{j} + 1\hat{k} \]

\[ \vec{A} - \vec{B} = \langle 4 - 6, 2 - -2, 3 - 1 \rangle \]

\[ \vec{A} - \vec{B} = \langle -2, 4, 2 \rangle \]

\[ \vec{C} = \langle -2, 4, 2 \rangle \]

or

\[ \vec{A} - \vec{B} = (4 - 6)\hat{i} + (2 - -2)\hat{j} + (3 - 1)\hat{k} \]

\[ \vec{A} - \vec{B} = (-2)\hat{i} + (4)\hat{j} + (2)\hat{k} \]

\[ \vec{C} = -2\hat{i} + 4\hat{j} + 2\hat{k} \]

As we can see from the formula, when we subtract the components of the red vector and the blue vector, we get new vector components as a result, which become the vector components for the purple vector, which is the difference of the red and blue vectors.

You have now learned about vector subtraction. Go back now to the vector page to move onto the next lesson.

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