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Vector Dot Product

Vector Dot Product in 1D y-axis

The Dot Product of two vectors is one of two methods of vector multiplication, the other is the cross product. A dot product is denoted by the multiplication sign (⋅) between two vectors. The dot product of two vectors is a scalar quantity of the two vector's components multiplied individually based on their axes and then summed together. It is worth noting that the dot product of two vectors is also known as a scalar product as the resultant of the dot product of vectors is a scalar quantity.

Below is a simulation of the dot product in the y-axis. There is a red vector of variable magnitude and angle based on the movement of your mouse, and a second vector, a blue vector of magnitude (length) 17 and an angle of 90 degrees. Since these vectors are on the y-axis, when calculating the vector normal, we get a value of 0. When figuring out the scalar vector we multiply the vector normal by the magnitude (length) of the red vector, which still comes out to be 0. The dot product of the red vector and the blue vector now becomes the product of the component vectors of the blue vector multiplied by the component vectors of the red vector.

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

\[ \vec{A} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 0 \end{bmatrix} \]

or

\[ \vec{A} = 0\hat{i} + 2\hat{j} + 0\hat{k} \]

or

\[ \vec{A} = \langle 0, 2, 0 \rangle \]

and

\[ \vec{B} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 17 & 0 \end{bmatrix} \]

or

\[ \vec{B} = 0\hat{i} + 17\hat{j} + 0\hat{k} \]

or

\[ \vec{B} = \langle 0, 17, 0 \rangle \]

The dot product can be calculated using the following equations:

\[ \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos(\theta) \]

\[ \vec{A} \cdot \vec{B} = (A_x B_x) + (A_y B_y) + (A_z B_z) \]

The projection and angle of a vector in the dot product can be calculated using the following equations:

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \vec{B} \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|^2} \vec{A} \]

\[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| \cdot |\vec{B}|} \]

Now let's see what happens when we calculate the dot product of the red vector and blue vector, the vector projection, and the angle between the vectors:

\[ \vec{A} = 0\hat{i} + 2\hat{j} + 0\hat{k} \]

\[ \vec{B} = 0\hat{i} + 17\hat{j} + 0\hat{k} \]

\[ \vec{A} \cdot \vec{B} = ((0)(0)) + ((2)(17)) + ((0)(0)) \]

\[ \vec{A} \cdot \vec{B} = (0) + (34) + (0) \]

\[ \vec{A} \cdot \vec{B} = 34 \]

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \frac{34}{|17|^2} \langle 0, 17, 0 \rangle \]

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \frac{34}{289} \langle 0, 17, 0 \rangle \]

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \frac{34}{289} 17 \]

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = 2 \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \frac{34}{|2|^2} \langle 0, 2, 0 \rangle \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \frac{34}{4} \langle 0, 2, 0 \rangle \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \frac{34}{4} 2 \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = 17 \]

\[ \cos(\theta) = \frac{34}{|2| \cdot |17|} \]

\[ \cos(\theta) = \frac{34}{|34|} \]

\[ \cos(\theta) = 1 \]

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

\[ \theta = 0 \]

As we can see, the result of the dot product is 34, just like in the simulation, the projections of the vectors are 2 and 17, which are the component vectors of both vectors, and the angle between the vectors is 0 degrees.

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