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Vector Dot Product

Vector Dot Product in 3D

The Dot Product of two vectors is one of two methods of vector multiplication, the other is the cross product. A 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 the same simulation as the one previously of the dot product in 3D. There is a red vector of magnitude (length) 3.742, and a second vector, a blue vector of magnitude (length) 4.243. (To not confuse the student for the 3D example the scalar and normal Vectors have been ommited). 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. As a student you will be able to change this vector's angle, direction, and magnitude to alter the result of the vector's dot product.



Red Vector:


Blue Vector:


Dot Product:

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


\[ \vec{A} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \end{bmatrix} \]

or

\[ \vec{A} = -1\hat{i} + 2\hat{j} + 3\hat{k} \]

or

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


and


\[ \vec{B} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \end{bmatrix} \]

or

\[ \vec{B} = 4\hat{i} - 1\hat{j} + 1\hat{k} \]

or

\[ \vec{B} = \langle 4, -1, 1 \rangle \]


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:


\[ \vec{A} = -1\hat{i} + 2\hat{j} + 3\hat{k} \]

\[ \vec{B} = 4\hat{i} - 1\hat{j} + 1\hat{k} \]


\[ \vec{A} \cdot \vec{B} = ((-1)(4)) + ((2)(-1)) + ((3)(1)) \]

\[ \vec{A} \cdot \vec{B} = (-4) + (-2) + (3) \]

\[ \vec{A} \cdot \vec{B} = -3 \]


\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \frac{-3}{|4.243|^2} \langle 4, -1, 1 \rangle \]

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \frac{-3}{18.003} \langle 4, -1, 1 \rangle \]

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \langle \frac{-3}{18.003} 4, \frac{-3}{18.003} -1, \frac{-3}{18.003} 1 \rangle \]

\[ \text{Projection of $\vec{A}$ on $\vec{B}$} = \langle -\frac{2}{3}, \frac{1}{3}, -\frac{1}{3} \rangle \]


\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \frac{-3}{|3.742|^2} \langle -1, 2, 3 \rangle \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \frac{-3}{14.002} \langle -1, 2, 3 \rangle \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \langle \frac{-3}{14.002} -1, \frac{-3}{14.002} 2, \frac{-3}{14.002} 3 \rangle \]

\[ \text{Projection of $\vec{B}$ on $\vec{A}$} = \langle \frac{3}{14}, -\frac{3}{7}, -\frac{9}{14} \rangle \]


\[ \cos(\theta) = \frac{-3}{|3.742| \cdot |4.243|} \]

\[ \cos(\theta) = \frac{-3}{|15.877|} \]

\[ \cos(\theta) = -0.189 \]

\[ \theta = \cos^{-1}(-0.189) \]

\[ \theta = 100.89 \]


As we can see, the result of the dot product is -3, just like in the simulation, the projections of the vectors are 〈-2/3, 1/3, -1/3〉 and 〈3/14, -3/7, -9/14〉, and the angle between the vectors is 100.89 degrees.

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




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