Wize University Linear Algebra Textbook > Products of Vectors

Angle Between Vectors

Angle Between Vectors

The dot product can be used to measure the angle between vectors $\vec{u}$ and $\vec{v}$ in $\reals^n$ . We call this angle $\theta$ .

Another formula for the dot product involves $\theta$ :

\boxed{\quad \vec{u} \cdot \vec{v} = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \cos(\theta) \quad}

Notes

Since the norms of the vectors on the RHS are non-negative, the sign of $\cos\theta$ determines the sign of the dot product.

$\vec{u} \cdot \vec{v} = \underbrace{ \lVert \vec{u} \rVert \lVert \vec{v} \rVert }_{\text{positive}} \cos(\theta)$

$\begin{array}{r l | c} & \bm{\theta} & \bm{\vec u \cdot \vec v}\\ \hline\\ 0<\theta<\frac{\pi}{2},&\text{acute}&\quad \vec u \cdot \vec v>0\\[1.5em] \frac{\pi}{2}<\theta<\pi, &\text{obtuse}&\quad \vec u \cdot \vec v<0\\[1.5em] \theta=\frac{\pi}{2}, &\text{orthogonal}&\quad \vec u \cdot \vec v=0\\[1.5em] \theta=0, &\text{parallel (same direction)}&\quad \vec u \cdot \vec v=\lVert \vec u \rVert \lVert \vec v \rVert\\[1.5em] \theta=\pi, &\text{parallel (opposite direction)}&\quad\ \ \ \vec u \cdot \vec v=-\lVert \vec u \rVert \lVert \vec v \rVert\\ \end{array}$

Finding the Angle Between Vectors

Combining the two equations for $\vec{u} \cdot \vec{v}$ gives us the following:

\boxed{\quad u_1v_1+u_2v_2+\dots+u_nv_n=\|\vec{u}\|\|\vec{v}\|\cos(\theta) \quad}

Steps

LHS: Calculate the dot product $\vec{u} \cdot \vec{v}\ = \ u_1v_1\ +\ u_2v_2 \ +\ \dots \ +\ u_nv_n$
RHS: Calculate $\lVert \vec u \rVert$ and $\lVert \vec v \rVert$
Solve for $\cos (\theta)$
Solve for $\theta$ using special triangles or the inverse cosine, $\cos^{-1}$

Example: Angle Between Vectors

Find the angle between the vectors $\vec u = \begin{bmatrix} -2\\ 1\\ \end{bmatrix}$ and $\vec v = \begin{bmatrix} -3\\ -1\\ \end{bmatrix}$ .

Image illustrating Example

The formula is $\vec{u} \cdot \vec{v} = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \cos(\theta)$

LHS: $\vec u \cdot \vec v \ =\ \begin{bmatrix} -2\\ 1\\ \end{bmatrix} \cdot \begin{bmatrix} -3\\ -1\\ \end{bmatrix} \ =\ (-2)(-3) + (1)(-1) \ =\ 5$

RHS: $\begin{aligned} &\lVert \vec u \rVert = \sqrt{(-2)^2+1^2} = \sqrt{5}\\ &\lVert \vec v \rVert = \sqrt{(-3)^2+(-1)^2} = \sqrt{10}\\ \end{aligned}$

Subsititute these values into the formula, then solve for $\cos \theta$ and simplify:

$\begin{aligned} 5 \ =\ \sqrt{5} \sqrt{10} &\cos \theta\\ \dfrac{5}{\sqrt{5}\sqrt{10}} \ =\ &\cos\theta\\ \dfrac{5}{\sqrt{50}} \ =\ &\cos\theta\\ \dfrac{5}{\sqrt{25}\sqrt2} \ =\ &\cos\theta\\ \dfrac{\cancel{5}}{\cancel{5}\sqrt2} \ =\ &\cos\theta\\ \dfrac{1}{\sqrt{2}} \ =\ &\cos\theta\\ \end{aligned}$

Finally, we can recall a special triangle to find $\theta$ :

$\therefore \ \ \boxed{\theta = \dfrac{\pi}{4}}$

Example: Angle Between Vectors

Given $\vec{u} = \lang 1,4,-2\rang$ and $\vec{v}=\lang 0,-2,5\rang$ , find the angle between $\vec{u}$ and $\vec{v}$ .

$\begin{aligned} \vec{u} \cdot \vec{v} \ =\ (1)(0) + (4)(-2) + (-2)(5) \ =\ -18 \end{aligned}$

$\lVert \vec u \rVert =\sqrt{\left(1\right)^2+\left(4\right)^2+\left(-2\right)^2}=\sqrt{21}$

$\lVert \vec v \rVert =\sqrt{\left(0\right)^2+\left(-2\right)^2+\left(5\right)^2}=\sqrt{29}$

Substituting into the formula:

$-18 = \sqrt{21}\sqrt{29}\cos\theta$

So we can solve for $\cos\theta$ , then use the inverse cosine to find $\theta$ :

$\begin{aligned} &\cos\theta = \dfrac{-18}{\sqrt{21}\sqrt{29}} \approx -0.7294\\[1.5em] &\theta = \cos^{-1} \left( \dfrac{-18}{\sqrt{609}} \right) \approx \boxed{2.388 \text{ rad} \ \ (136.8\degree)}\\ \end{aligned}$

Practice: Angle Between Vectors

Given $\vec{u}=\lang1,7\rang$ , $\vec{u}\cdot\vec{v}=6$ , and $\cos\theta=\frac{1}{2}$ , find $\lVert \vec v \rVert$ .

A) $\sqrt{50}$

B) $\frac{3}{\sqrt{50}}$

C) $\frac{12}{\sqrt{50}}$

D) $\frac{6}{\sqrt{50}}$

E) None of the above

I don't know

Practice: Types of Angles Between Vectors

Match the correct angle to each expression.

$\theta = \pi$

$\theta = \frac{\pi}{6}$

$\theta = \frac{\pi}{2}$

$\theta = \frac{3\pi}{4}$

$\theta = 0$

$\vec u \cdot \vec v = \lVert \vec u \rVert \lVert \vec v \rVert$

$\vec u \cdot \vec v = -\frac{1}{2}$

$\vec u \cdot \vec v = -\lVert \vec u \rVert \lVert \vec v \rVert$

$\vec u \cdot \vec v = 7$

$\vec u \cdot \vec v = 0$

I don't know

Practice: Angle Between Vectors

Suppose $\vec{u},\ \vec v \in \reals^n$ are non-zero vectors that are not parallel, and let $\vec{w}=\lVert\vec{v}\rVert\vec{u} -\lVert\vec{u}\rVert\vec{v}$ .

Find a simplified expression for the cosine of the angle between $\vec{v}$ and $\vec{w}$ .

$\dfrac{\|\vec{u}\|\|\vec{v}\|}{\|\vec{w}\|}$

$\dfrac{\|\vec{u}\|-\|\vec{v}\|-\vec{u}\cdot\vec{v}}{\|\vec{w}\|}$

$\dfrac{\|\vec{v}\|(\vec{u}\cdot\vec{v})}{\|\vec{w}\|}$

$\dfrac{\vec{u}\cdot\vec{v}-\|\vec{u}\|\|\vec{v}\|}{\|\vec{w}\|}$

$\dfrac{\vec{u}\cdot\vec{v}-\|\vec{v}\|^2}{\|\vec{w}\|}$

I don't know

Extra Practice

Additional practice problems--vector products

Additional Practice Problems--Vector Products

1. If $\vec{u}=\left(-2,3,-1\right),\ \ \vec{v}=\left(-1,-2,3\right)$ , find

a) $\vec{u}\ \bullet\ \vec{v}$

Find the value of cos 𝜃 where 𝜃 is the acute angle betwee $𝑢⃗=(1,−2)$ an $\vec𝑣 = (3,1).$

Consider three vectors $\vec{A}=2\hat{i}+\hat{j},\,\,\vec{B}=\hat{i}-2\hat{j}$ , and $\vec{C}=-2\hat{i}+\hat{j}$ . The angle between $\vec{C}$ and $\vec{A}\times\vec{B}$ is?

133 - FML 3 - 18.1W e.g. 47

Given $\bcb{\vec{u} = \left< 1, 0, 1\right>}$ and $\bcb{\vec{v} = \left< 1, 0, 0\right>}$ , compute the acute angle between $\bcb{\vec{u}}$ and $\bcb{\vec{v}}$ .

Angle Between Vectors

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