Dot Products and Projections

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Objectives:

1.: Know how to compute a dot product.
2.: Know the cosine formula for the dot product.
3.: Understand what the dot product formula says about the angle between vectors.
4.: Know what the projection looks like geometrically and how to compute it.

Recap Video

You can watch watch the following video which recaps the ideas of the section.

Test your understanding with the following questions.

If $\textbf {v} = \left \langle v_1,v_2,v_3 \right \rangle$ and $\textbf {w} = \left \langle w_1,w_2,w_3 \right \rangle$ , then the dot product $\textbf {v} \cdot \textbf {w}$ is equal to which of the following (select all that apply):

$v_1w_1+v_2w_2+v_3w_3$ $\left \langle v_1w_1,v_2w_2,v_3w_3 \right \rangle$ $\| \textbf {v} \| \| \textbf {w} \| \cos (\theta )$ , where $\theta$ is the angle between $\textbf {v}$ and $\textbf {w}$ . $\| \textbf {v} \| \| \textbf {w} \| \sin (\theta )$ , where $\theta$ is the angle between $\textbf {v}$ and $\textbf {w}$ .

If $\textbf {v}$ and $\textbf {w}$ are two nonzero vectors, then which of the following is true of the projection $\text {proj}_{\textbf {w}} \textbf {v}$ ? Select all that apply.

If it is nonzero, then it is always parallel to $\textbf {w}$ . $\text {proj}_{\textbf {w}} \textbf {v} = \frac {\textbf {v} \cdot \textbf {w}}{\textbf {w} \cdot \textbf {w}} \textbf {w}$ . If it is nonzero, then it is always parallel to $\textbf {v}$ .

Problems

If $\textbf {v} = \left \langle 1,2,1 \right \rangle$ and $\textbf {w} = \left \langle -2,-1,3 \right \rangle$ , then $\textbf {v} \cdot \textbf {w} = \answer {-1}$ .

The angle between the two vectors is AcuteRightObtuse .

What it the angle between the vectors $\textbf {v} = \left \langle 1,2,1 \right \rangle$ and $\textbf {w}=\left \langle 0,1,-3 \right \rangle$ ?

We will use the formula $\textbf {v} \cdot \textbf {w} = \| \textbf {v} \| \| \textbf {w}\| \cos (\theta ).$

The dot product of $\textbf {v}$ and $\textbf {w}$ is $\textbf {v} \cdot \textbf {w} = \answer {-1}$ .
The length of $\textbf {v}$ is $\| \textbf {v} \| = \answer {\sqrt {6}}$ and the length of $\textbf {w}$ is $\|\textbf {w}\| = \answer {\sqrt {10}}$
The dot product formula says that $\cos (\theta ) = \frac {\textbf {v} \cdot \textbf {w}}{\|\textbf {v}\| \|\textbf {w}\|} = -\frac {\answer {1}}{\answer {\sqrt {60}}}$
Therefore, $\theta = \answer {\arccos \left (\frac {-1}{\sqrt {60}} \right )}$

If $\textbf {u}$ and $\textbf {v}$ are unit vectors with an angle of $\pi /3$ between them, then $\textbf {u} \cdot \textbf {v} = \answer {1/2}$ .

Use $\textbf {u} \cdot \textbf {v} = \| \textbf {u} \| \| \textbf {v} \| \cos (\theta )$ .

Problems

Let $\textbf {v} = \left \langle 1,2,0 \right \rangle$ an $\textbf {w} = \left \langle 3,5,1 \right \rangle$ .

The dot product of $\textbf {v}$ and $\textbf {w}$ is $\textbf {v} \cdot \textbf {w} = \answer {13}$ .
The angle between $\textbf {v}$ and $\textbf {w}$ is acuterightobtuse .
The projection $\text {proj}_{\textbf {w}} \textbf {v}$ will be in the direction as $\textbf {w}$ .
The component $\text {comp}_{\textbf {w}} \textbf {v} = \answer {\frac {13}{\sqrt {35}}}$ .
The projection $\text {proj}_{\textbf {w}} \textbf {v} = \left \langle \answer {\frac {39}{35}}, \answer {\frac {65}{35}},\answer {\frac {13}{35}} \right \rangle .$

Suppose $\textbf {v}$ and $\textbf {w}$ are nonzero three-dimensional vectors with $\textbf {v}$ orthogonal to $\textbf {w}$ . Then $\text {proj}_{\textbf {w}} \textbf {v} = \left \langle \answer {0},\answer {0},\answer {0} \right \rangle .$

True/False

If $\textbf {v}$ is a unit vector, then it must be true that $\textbf {v} \cdot \textbf {v} = 1$ . TrueFalse

Assume $\textbf {v}$ and $\textbf {w}$ are nonzero vectors with $\textbf {v} \cdot \textbf {w} = 0$ . Then it must be true that $\| \textbf {v} + \textbf {w} \|^2 = \| \textbf {v} \|^2+ \| \textbf {w} \|^2.$ TrueFalse

Assume $\textbf {v}$ and $\textbf {w}$ are nonzero vectors. Then it must be true that $\text {proj}_{2\textbf {w}} \textbf {v} = 2\text {proj}_{\textbf {w}}\textbf {v}.$ TrueFalse

If $\textbf {v}$ , $\textbf {w}$ , $\textbf {u}$ are nonzero vectors with $\textbf {v} \cdot \textbf {w} = 0$ and $\textbf {w} \cdot \textbf {u} = 0$ , then it must be true that $\textbf {v} \cdot \textbf {u}$ is always zero. TrueFalse

Optional: Dot Product Cosine Formula

For those who are interested, below is a video explaining where the dot product cosine formula comes from.

Optional: Projection Formula Explained

Below is a video explaining how to arrive at the projection formula.