Dot Product & Angle Between Vectors Lesson

In this lesson, we will focus on the unit vector, the dot product, and the angle between two vectors. A unit vector is a vector that has a magnitude of 1. Two extremely useful unit vectors are defined using i and j: $$\hat{\text{i}}=\langle 1, 0 \rangle$$ $$\hat{\text{j}}=\langle 0, 1 \rangle$$ Note: The i is a unit vector in the x-direction. Similarly, j is the unit vector in the y-direction. A unit vector is typically denoted using boldface font or a hat "^" above the letter.

i, j Form for Vectors

We can write a vector v with a horizontal component of a and a vertical component of b, using the i, j form for vectors. This is also known as a linear combination of the vectors i and j. Any vector in the plane can be written as a linear combination of the standard unit vectors i and j. $$\overrightarrow{v}=\langle a, b \rangle$$ $$\hat{\text{i}}=\langle 1, 0 \rangle$$ $$\hat{\text{j}}=\langle 0, 1 \rangle$$ Let's write our vector v as a linear combination of unit vectors: $$\overrightarrow{v}=\langle a, b \rangle$$ $$\overrightarrow{v}=a \langle 1, 0 \rangle + b \langle 0, 1 \rangle$$ $$\overrightarrow{v}=a \hat{\text{i}}+ b \hat{\text{j}}$$ Let's look at an example.
Example #1: Write the vector p as a linear combination of unit vectors. $$\overrightarrow{p}=\langle -5, 4 \rangle$$ To write this vector as a linear combination of unit vectors, we simply take the horizontal component and multiply by the unit vector i and add this to the vertical component multiplied by the unit vector j: $$\overrightarrow{p}=-5\hat{\text{i}}+ 4\hat{\text{j}}$$

Finding the Unit Vector

In some applications, it is useful to find a unit vector u that has the same direction as a given nonzero vector v. To perform this action, we will divide our vector v by its magnitude to obtain: $$\hat{\text{u}}=\frac{\overrightarrow{v}}{\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}=\overrightarrow{v}\cdot \frac{1}{\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}$$ The vector u is called a unit vector in the direction of vector v. Let's look at an example.
Example #2: Find a unit vector u in the direction of v. $$\overrightarrow{v}=\langle -5, -12 \rangle$$ Let's reference our formula from above: $$\hat{\text{u}}=\frac{\overrightarrow{v}}{\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}=\overrightarrow{v}\cdot \frac{1}{\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}$$ Let's find the magnitude of our vector v: $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{(-5)^2 + (-12)^2}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{25 + 144}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{169}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=13$$ Now, let's plug into the formula: $$\hat{\text{u}}=\overrightarrow{v}\cdot \frac{1}{\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}$$ $$\hat{\text{u}}=\langle -5, -12 \rangle \cdot \frac{1}{13}$$ $$\hat{\text{u}}=\left\langle -\frac{5}{13}, -\frac{12}{13}\right\rangle$$ We can verify that the magnitude of our unit vector u is 1: $$\left|\hat{\text{u}}\right|=\sqrt{\left(-\frac{5}{13}\right)^2 + \left(-\frac{12}{13}\right)^2}$$ $$\left|\hat{\text{u}}\right|=\sqrt{\left(\frac{25}{169}\right) + \left(\frac{144}{169}\right)}$$ $$\left|\hat{\text{u}}\right|=\sqrt{\left(\frac{169}{169}\right)}$$ $$\left|\hat{\text{u}}\right|=\sqrt{1}$$ $$\left|\hat{\text{u}}\right|=1$$

How to Find the Dot Product & Angle Between Two Vectors

The dot product of two vectors is a real number, which is also known as the inner product. Dot products are used to determine the angle between two vectors.

Dot Product

$$\overrightarrow{u}=\langle a, b \rangle$$ $$\overrightarrow{v}=\langle c, d \rangle$$ The dot product of two vectors u and v is denoted as: $$\overrightarrow{u}\cdot \overrightarrow{v}=ac + bd$$ To find the dot product of two vectors, we find the sum of the product of the first components and the product of the second components. Let's look at a few examples.
Example #3: Find the dot product of u and v. $$\overrightarrow{u}=7\hat{\text{i}}+ 6\hat{\text{j}}$$ $$\overrightarrow{v}=3\hat{\text{i}}- 4\hat{\text{j}}$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=7 \cdot 3 + 6 \cdot (-4)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=21 + (-24)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=-3$$ Example #4: Find the dot product of u and v. $$\overrightarrow{u}=-2\hat{\text{i}}+ 8\hat{\text{j}}$$ $$\overrightarrow{v}=4\hat{\text{i}}- 6\hat{\text{j}}$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=-2 \cdot 4 + 8 \cdot (-6)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=-8 + (-48)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=-56$$

Finding the Angle Between Two Vectors

If θ is the angle between two nonzero vectors u and v, where: $$0° ≤ θ ≤ 180°$$ We can use the following formula to find cos θ: $$\text{cos}\hspace{.1em}θ=\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right| \left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}$$ If we want to find θ, we can use our inverse cosine function: $$θ=\text{cos}^{-1}\left(\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right| \left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}\right)$$ Note: We will derive this formula in the practice test section. Let's look at a few examples.
Example #5: Find the angle between the two vectors. $$\overrightarrow{u}=-6\hat{\text{i}}- 5 \hat{\text{j}}$$ $$\overrightarrow{v}=4\hat{\text{i}}+ 8 \hat{\text{j}}$$ Let's grab our formula from above for reference: $$θ=\text{cos}^{-1}\left(\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right| \left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}\right)$$ Let's find the dot product of the two vectors, and the magnitude of each vector: $$\overrightarrow{u}\cdot \overrightarrow{v}=(-6)(4) + (-5)(8)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=-24 + (-40)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=-64$$ $$\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right|=\sqrt{(-6)^2 + (-5)^2}$$ $$\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right|=\sqrt{36 + 25}$$ $$\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right|=\sqrt{61}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{(4)^2 + (8)^2}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{16 + 64}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{80}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=4\sqrt{5}$$ Now, let's plug into our formula: $$θ=\text{cos}^{-1}\left(\frac{-64}{\sqrt{61}\cdot 4\sqrt{5}}\right)$$ $$θ=\text{cos}^{-1}\left(\frac{-64}{4\sqrt{305}}\right)$$ $$θ ≈ 156.37°$$ Example #6: Find the angle between the two vectors. $$\overrightarrow{u}=2\hat{\text{i}}- 8 \hat{\text{j}}$$ $$\overrightarrow{v}=3\hat{\text{i}}- 5 \hat{\text{j}}$$ Let's grab our formula from above for reference: $$θ=\text{cos}^{-1}\left(\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right| \left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|}\right)$$ Let's find the dot product of the two vectors, and the magnitude of each vector: $$\overrightarrow{u}\cdot \overrightarrow{v}=(2)(3) + (-8)(-5)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=6 + 40$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=46$$ $$\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right|=\sqrt{(2)^2 + (-8)^2}$$ $$\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right|=\sqrt{4 + 64}$$ $$\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right|=\sqrt{68}$$ $$\left| \hspace{.25em}\overrightarrow{u}\hspace{.25em}\right|=2\sqrt{17}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{(3)^2 + (-5)^2}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{9 + 25}$$ $$\left| \hspace{.25em}\overrightarrow{v}\hspace{.25em}\right|=\sqrt{34}$$ Now, let's plug into our formula: $$θ=\text{cos}^{-1}\left(\frac{46}{2\sqrt{17}\cdot \sqrt{34}}\right)$$ $$θ=\text{cos}^{-1}\left(\frac{46}{34\sqrt{2}}\right)$$ $$θ ≈ 16.93°$$

Orthogonal Vectors

When the dot product for two nonzero vectors u and v is zero, then cos θ is 0 and θ is 90°. This tells us that our vectors u and v are perpendicular vectors, which is also known as orthogonal vectors. Let's look at an example.
Example #7: Determine if the two vectors are orthogonal. $$\overrightarrow{u}=4 \hat{\text{i}}+ 8 \hat{\text{j}}$$ $$\overrightarrow{v}=6 \hat{\text{i}}- 3\hat{\text{j}}$$ To determine if the two vectors are orthogonal, we will find the dot product of the two vectors. $$\overrightarrow{u}\cdot \overrightarrow{v}=(4)(6) + (8)(-3)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=24 + (-24)$$ $$\overrightarrow{u}\cdot \overrightarrow{v}=0$$ Since the dot product is 0, the two vectors are orthogonal.

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