Precalculus

The dot product and the angle between vectors

Two arrows leave the same point. Can one number, computed straight from their components, say how closely they point the same way, and recover the angle between them?

You need

Law of cosinesPrecalculus Component form of a vectorPrecalculus

Unlocks

Orthogonal vectorsPrecalculus Vector and scalar projectionPrecalculus

Where you are 3 in The dot product

Trigonometry7

Polar coordinates and vectors10

The dot product3

The dot product and the angle between vectors

Orthogonal vectors Vector and scalar projection

Why this exists

Vectors can be added, subtracted and stretched by a number. None of those operations answers a question that comes up as soon as two vectors share a starting point: how far apart are their directions? You could find each direction angle and subtract, but that means two arctangents, a quadrant check for each, and a correction whenever the difference comes out above 180^\circ.

There is a better tool. Multiply the matching components of the two vectors and add. The result is a single number, and that number turns out to be the two lengths times the cosine of the angle between the vectors. So one line of arithmetic on components reports how well two vectors line up, and the angle comes out with one inverse cosine and no quadrant check at all.

Definition

The lesson runs over two days. Day 1 is the dot product computed from components and the algebra it obeys. Day 2 is what the number means: the cosine formula, the angle between two vectors, and what the sign of the dot product says.

For \mathbf{u} = \langle a_1, b_1 \rangle and \mathbf{v} = \langle a_2, b_2 \rangle in component form, the dot product is

\mathbf{u} \cdot \mathbf{v} = a_1 a_2 + b_1 b_2.

Multiply the horizontal components, multiply the vertical components, add. The answer is a real number (a scalar), not a vector, which is why the dot product is also called the scalar product. In Unit vectors and i, j notation the same rule reads

(a_1\mathbf{i} + b_1\mathbf{j}) \cdot (a_2\mathbf{i} + b_2\mathbf{j}) = a_1 a_2 + b_1 b_2.

The angle between two nonzero vectors \mathbf{u} and \mathbf{v} is the angle \theta between their arrows when both are drawn from the same tail, with

0 \le \theta \le \pi \qquad (0^\circ \le \theta \le 180^\circ).

It is the smaller of the two angles the arrows make, so it never exceeds a straight angle. It is 0 when the vectors point the same way and \pi when they point opposite ways. Since a vector can be drawn from any tail, moving both arrows to a common starting point changes nothing.

Intuition

Figure 1Drawn from one tail, \mathbf{u}, \mathbf{v} and \mathbf{u} - \mathbf{v} form a triangle, and \theta is the angle between the sides \|\mathbf{u}\| and \|\mathbf{v}\|. The law of cosines applies to it.

Drawn from one tail, two vectors \mathbf{u} and \mathbf{v} have a third vector running from the head of \mathbf{v} to the head of \mathbf{u}. By the triangle rule it is \mathbf{u} - \mathbf{v}, because \mathbf{v} + (\mathbf{u} - \mathbf{v}) = \mathbf{u}. That triangle has two sides of lengths \|\mathbf{u}\| and \|\mathbf{v}\| with the angle \theta between them, and the law of cosines already knows the length of its third side. The dot product is exactly the piece of that law that carries the angle.

Figure 2The sign of \mathbf{u} \cdot \mathbf{v} is the sign of \cos\theta. An acute angle gives a positive dot product, a right angle gives zero, an obtuse angle gives a negative one.

Read the formula \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta (proved below) as a measure of agreement. Lengths are positive, so everything about direction sits in \cos\theta. Arrows pointing roughly the same way make a small angle and a large positive dot product. As the angle opens toward 90^\circ the dot product shrinks to 0, and past 90^\circ it turns negative, most negative when the arrows point in opposite directions.

Results

Properties. For vectors \mathbf{u}, \mathbf{v}, \mathbf{w} and any real number k:

Property	Statement
commutative	\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}
distributive	\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}
scalars move out	(k\mathbf{u}) \cdot \mathbf{v} = k(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot (k\mathbf{v})
zero vector	\mathbf{0} \cdot \mathbf{v} = 0
length	\mathbf{v} \cdot \mathbf{v} = \\|\mathbf{v}\\|^2

The last line says a vector dotted with itself is its squared length, so \displaystyle \|\mathbf{v}\| = \sqrt{\mathbf{v} \cdot \mathbf{v}}. Expanding with the first three lines gives the length of a sum or difference without components:

\|\mathbf{u} \pm \mathbf{v}\|^2 = \|\mathbf{u}\|^2 \pm 2\,\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2.

The dot product and the angle. If \mathbf{u} and \mathbf{v} are nonzero and \theta is the angle between them, then

\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta.

The angle between two vectors. Dividing by the two lengths,

\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|}, \qquad \theta = \arccos\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|}.

The inverse cosine returns an angle from 0 to \pi, which is exactly the range of \theta, so the calculator’s answer is always the angle wanted.

The sign of the dot product. For nonzero \mathbf{u} and \mathbf{v}:

\mathbf{u} \cdot \mathbf{v}	\theta
positive	acute, 0 \le \theta < 90^\circ
zero	right, \theta = 90^\circ
negative	obtuse, 90^\circ < \theta \le 180^\circ

(The ends 0^\circ and 180^\circ are the parallel cases, not acute or obtuse angles in the strict sense; they sit in the positive and negative rows.)

Why it is true

The properties. Each one is a property of real numbers applied one component at a time. For commutativity, a_1 a_2 + b_1 b_2 = a_2 a_1 + b_2 b_1 because multiplying real numbers is commutative. For the distributive law, with \mathbf{w} = \langle a_3, b_3 \rangle,

\begin{aligned} \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) &= a_1(a_2 + a_3) + b_1(b_2 + b_3) \\ &= (a_1 a_2 + b_1 b_2) + (a_1 a_3 + b_1 b_3), \end{aligned}

by the distributive property of numbers and then reordering and regrouping the sum. Scalars move out because (ka_1)a_2 + (kb_1)b_2 = k(a_1 a_2 + b_1 b_2). Finally, with \mathbf{v} = \langle a, b \rangle, \mathbf{v} \cdot \mathbf{v} = a^2 + b^2, which is \|\mathbf{v}\|^2 by the magnitude formula.

The cosine formula. Suppose first that \mathbf{u} and \mathbf{v} do not lie along one line, so 0 < \theta < \pi. Drawn from a common tail as in Figure 1, they form a triangle whose third side is \mathbf{u} - \mathbf{v}. The law of cosines, applied to the side across from \theta, gives

\|\mathbf{u} - \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 - 2\|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta.

Now compute the same length with the dot product. By the length property and the distributive law,

\begin{aligned} \|\mathbf{u} - \mathbf{v}\|^2 &= (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) \\ &= \mathbf{u} \cdot \mathbf{u} - \mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} \\ &= \|\mathbf{u}\|^2 - 2\,\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2, \end{aligned}

where the two middle terms combine because the dot product is commutative. The two right-hand sides are equal. Subtract \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 from both and divide by -2:

\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta.

If the vectors do lie along one line there is no triangle, so check directly. Then \mathbf{v} = k\mathbf{u} for some k \ne 0 (parallel vectors are scalar multiples), and \mathbf{u} \cdot \mathbf{v} = k\|\mathbf{u}\|^2. If k > 0 the angle is 0, \cos\theta = 1 and \|\mathbf{v}\| = k\|\mathbf{u}\|, so the right side is also k\|\mathbf{u}\|^2. If k < 0 the angle is \pi, \cos\theta = -1 and \|\mathbf{v}\| = -k\|\mathbf{u}\|, so the right side is \|\mathbf{u}\|(-k\|\mathbf{u}\|)(-1) = k\|\mathbf{u}\|^2 again. The formula holds for every pair of nonzero vectors.

The sign. In \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta the two lengths are positive, so the dot product has the sign of \cos\theta. On [0, \pi] cosine is positive before \displaystyle \frac{\pi}{2}, zero at \displaystyle \frac{\pi}{2} and negative after it.

Worked examples

Examples 1 to 3 go with Day 1 (components and properties), Examples 4 to 7 with Day 2 (the angle).

Example 1

Three dot products from components

Let \mathbf{u} = \langle 6, -5 \rangle, \mathbf{v} = \langle 2, 3 \rangle and \mathbf{w} = \langle 5, 6 \rangle. Find \mathbf{u} \cdot \mathbf{v}, \mathbf{u} \cdot \mathbf{w} and \mathbf{u} \cdot \mathbf{u}, and use the last to find \|\mathbf{u}\|.

Solution

Multiply matching components and add:

\mathbf{u} \cdot \mathbf{v} = (6)(2) + (-5)(3) = 12 - 15 = -3,

\mathbf{u} \cdot \mathbf{w} = (6)(5) + (-5)(6) = 30 - 30 = 0,

\mathbf{u} \cdot \mathbf{u} = 6^2 + (-5)^2 = 36 + 25 = 61.

Each answer is a number. Since \mathbf{u} \cdot \mathbf{u} = \|\mathbf{u}\|^2, the length is \displaystyle \|\mathbf{u}\| = \sqrt{61} \approx 7.81, the same value the magnitude formula gives. The zero in \mathbf{u} \cdot \mathbf{w} is no accident: Day 2 shows it means the two arrows are perpendicular, and Orthogonal vectors is built on that fact.

Example 2

Expanding a dot product with the properties

All that is known about \mathbf{u} and \mathbf{v} is \|\mathbf{u}\| = 3, \|\mathbf{v}\| = 5 and \mathbf{u} \cdot \mathbf{v} = 4. Find (2\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + 3\mathbf{v}).

Solution

There are no components, so the definition cannot be used. The properties can. Distribute each term of the first vector over the second, pulling the scalars out:

\begin{aligned} &(2\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + 3\mathbf{v}) \\ &= 2\,\mathbf{u} \cdot \mathbf{u} + 6\,\mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} - 3\,\mathbf{v} \cdot \mathbf{v}. \end{aligned}

By commutativity \mathbf{v} \cdot \mathbf{u} = \mathbf{u} \cdot \mathbf{v}, so the middle terms combine to 5\,\mathbf{u} \cdot \mathbf{v}. By the length property \mathbf{u} \cdot \mathbf{u} = 3^2 = 9 and \mathbf{v} \cdot \mathbf{v} = 5^2 = 25. So

2(9) + 5(4) - 3(25) = 18 + 20 - 75 = -37.

The expansion looks like multiplying out (2x - y)(x + 3y), and it is allowed for the same reason: the dot product distributes and is commutative.

Example 3

The length of a sum without components

Vectors \mathbf{u} and \mathbf{v} have \|\mathbf{u}\| = 4, \|\mathbf{v}\| = 6 and \mathbf{u} \cdot \mathbf{v} = -3. Find \|\mathbf{u} + \mathbf{v}\|.

Solution

A length is the square root of a vector dotted with itself, so expand (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}):

\begin{aligned} \|\mathbf{u} + \mathbf{v}\|^2 &= \|\mathbf{u}\|^2 + 2\,\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2 \\ &= 16 + 2(-3) + 36 = 46. \end{aligned}

So \displaystyle \|\mathbf{u} + \mathbf{v}\| = \sqrt{46} \approx 6.78. Notice it is less than 4 + 6 = 10: the negative dot product pulls the sum shorter than it would be if the arrows pointed the same way.

Example 4

An angle that comes out exactly

Find the angle between \mathbf{u} = \langle 3, 1 \rangle and \mathbf{v} = \langle 1, 2 \rangle.

Solution

The formula needs three numbers. The dot product is \mathbf{u} \cdot \mathbf{v} = 3 + 2 = 5. The lengths are \displaystyle \|\mathbf{u}\| = \sqrt{10} and \displaystyle \|\mathbf{v}\| = \sqrt{5}. So

\cos\theta = \frac{5}{\sqrt{10}\,\sqrt{5}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}.

The only angle in [0^\circ, 180^\circ] with cosine \displaystyle \frac{1}{\sqrt{2}} is \theta = 45^\circ, or \displaystyle \frac{\pi}{4}. Check with direction angles: \mathbf{u} points at \displaystyle \arctan\frac{1}{3} and \mathbf{v} at \arctan 2, and by the tangent difference formula \displaystyle \tan(\arctan 2 - \arctan\frac{1}{3}) is \displaystyle \frac{2 - 1/3}{1 + 2/3} = 1, so the difference is 45^\circ as well. The dot product needed no tangents and no quadrant reasoning.

Example 5

An obtuse angle

Find the angle between \mathbf{u} = \langle 4, -3 \rangle and \mathbf{v} = \langle -1, 6 \rangle to the nearest tenth of a degree.

Solution

\mathbf{u} \cdot \mathbf{v} = (4)(-1) + (-3)(6) = -4 - 18 = -22. The sign already says the angle is obtuse. The lengths are \displaystyle \|\mathbf{u}\| = \sqrt{16 + 9} = 5 and \displaystyle \|\mathbf{v}\| = \sqrt{1 + 36} = \sqrt{37}. So

\cos\theta = \frac{-22}{5\sqrt{37}} \approx -0.7234, \qquad \theta \approx 136.3^\circ.

The inverse cosine of a negative number lies between 90^\circ and 180^\circ, in agreement with the sign. The inverse cosine is the right inverse here: an arctangent would need a quadrant fix, and an arcsine could never return an obtuse angle.

Example 6

From lengths and an angle back to a length

Vectors \mathbf{u} and \mathbf{v} have lengths 6 and 4 and make an angle of 120^\circ. Find \mathbf{u} \cdot \mathbf{v} and \|\mathbf{u} - \mathbf{v}\|.

Solution

By the cosine formula,

\mathbf{u} \cdot \mathbf{v} = (6)(4)\cos 120^\circ = 24\left(-\tfrac{1}{2}\right) = -12.

Then the length property and the distributive law give

\begin{aligned} \|\mathbf{u} - \mathbf{v}\|^2 &= \|\mathbf{u}\|^2 - 2\,\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2 \\ &= 36 + 24 + 16 = 76, \end{aligned}

so \displaystyle \|\mathbf{u} - \mathbf{v}\| = \sqrt{76} = 2\sqrt{19} \approx 8.72. This is the law of cosines run in reverse: the dot product carried the angle into the length of the third side.

Example 7

The angles of a triangle from its corners

Nkiruka plots the triangle A(-2, 1), B(4, 3), C(0, 6) and wants all three angles without a protractor. Find them to the nearest tenth of a degree.

Solution

Figure 3Example 7. Each angle of the triangle is the angle between the two vectors that leave that corner along its sides.

The angle at a corner is the angle between the two vectors that leave that corner along the sides. Both must start at the corner: using \overrightarrow{BA} at A instead of \overrightarrow{AB} would give the supplement.

At A: \overrightarrow{AB} = \langle 6, 2 \rangle and \overrightarrow{AC} = \langle 2, 5 \rangle, with dot product 12 + 10 = 22 and lengths \displaystyle \sqrt{40} and \displaystyle \sqrt{29}:

\cos A = \frac{22}{\sqrt{40}\,\sqrt{29}} \approx 0.6459, \qquad A \approx 49.8^\circ.

At B: \overrightarrow{BA} = \langle -6, -2 \rangle and \overrightarrow{BC} = \langle -4, 3 \rangle, with dot product 24 - 6 = 18 and lengths \displaystyle \sqrt{40} and 5:

\cos B = \frac{18}{5\sqrt{40}} \approx 0.5692, \qquad B \approx 55.3^\circ.

At C: \overrightarrow{CA} = \langle -2, -5 \rangle and \overrightarrow{CB} = \langle 4, -3 \rangle, with dot product -8 + 15 = 7 and lengths \displaystyle \sqrt{29} and 5:

\cos C = \frac{7}{5\sqrt{29}} \approx 0.2600, \qquad C \approx 74.9^\circ.

Check: 49.8^\circ + 55.3^\circ + 74.9^\circ = 180.0^\circ. All three dot products are positive, so the triangle is acute, which the figure confirms.

Boundaries

The angle needs two nonzero vectors. The zero vector has no direction, so “the angle between \mathbf{0} and \mathbf{v}” means nothing, and the angle formula would divide by \|\mathbf{0}\| = 0. The dot product itself is still defined: \mathbf{0} \cdot \mathbf{v} = 0.
The result is a number, not a vector. So (\mathbf{u} \cdot \mathbf{v}) \cdot \mathbf{w} is not a dot product at all, since its first factor is a number. The expressions (\mathbf{u} \cdot \mathbf{v})\mathbf{w} and \mathbf{u}(\mathbf{v} \cdot \mathbf{w}) are scalar multiples of different vectors and are usually different.
No cancelling. \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{w} with \mathbf{u} \ne \mathbf{0} does not force \mathbf{v} = \mathbf{w}, because there is no division by a vector.
Both vectors need the same number of components. The definition pairs components one to one. In three dimensions the same rule a_1 a_2 + b_1 b_2 + c_1 c_2 applies and the cosine formula holds with the same proof, since three points still span a flat triangle.
The angle is between 0^\circ and 180^\circ and has no direction of turning. The dot product cannot tell whether \mathbf{v} lies counterclockwise or clockwise from \mathbf{u}: swapping the vectors leaves it unchanged.

Where this goes next

The zero row of the sign table becomes its own result in Orthogonal vectors: two nonzero vectors are perpendicular exactly when their dot product is 0, a test that needs no angles and no slopes. Dividing \mathbf{u} \cdot \mathbf{v} by one length instead of two gives \|\mathbf{u}\|\cos\theta, the signed length of the shadow \mathbf{u} casts on \mathbf{v}; that shadow is the Vector and scalar projection, and it is also how physics measures the work a force does along a path. The parent idea, The dot product, gathers these together.

Exercises

Exercises 1 to 9 go with Day 1 (Examples 1 to 3), Exercises 10 to 20 with Day 2 (Examples 4 to 7), and the reasoning group draws on both.

Day 1: computing dot products

Find \langle -6, 1 \rangle \cdot \langle 2, 7 \rangle.
Find \langle -3, 4 \rangle \cdot \langle -5, -2 \rangle.
Find \left\langle \tfrac{1}{2}, 3 \right\rangle \cdot \left\langle 8, -\tfrac{2}{3} \right\rangle.
For \mathbf{u} = \langle 5, -12 \rangle, find \mathbf{u} \cdot \mathbf{u} and use it to find \|\mathbf{u}\|.
Find (2\mathbf{i} - 5\mathbf{j}) \cdot (4\mathbf{i} + \mathbf{j}).
Find k if \langle k, 3 \rangle \cdot \langle 2, -5 \rangle = 11.

Day 1: using the properties

Given \|\mathbf{u}\| = 2, \|\mathbf{v}\| = 7 and \mathbf{u} \cdot \mathbf{v} = -3, find (\mathbf{u} + 2\mathbf{v}) \cdot (3\mathbf{u} - \mathbf{v}).
Given \|\mathbf{u}\| = 5, \|\mathbf{v}\| = 3 and \mathbf{u} \cdot \mathbf{v} = 6, find \|\mathbf{u} - \mathbf{v}\|.
Given \|\mathbf{u} + \mathbf{v}\| = 10 and \|\mathbf{u} - \mathbf{v}\| = 6, find \mathbf{u} \cdot \mathbf{v}.

Day 2: the angle between two vectors

Find \mathbf{u} \cdot \mathbf{v} if \|\mathbf{u}\| = 8, \|\mathbf{v}\| = 5 and the angle between them is 60^\circ.
Find \mathbf{u} \cdot \mathbf{v} if \|\mathbf{u}\| = 5, \displaystyle \|\mathbf{v}\| = 3\sqrt{2} and the angle between them is 135^\circ.
Find the exact angle between \displaystyle \langle \sqrt{3}, 1 \rangle and \displaystyle \langle -\sqrt{3}, 1 \rangle.
Find the angle between \langle -1, 4 \rangle and \langle 5, 2 \rangle to the nearest tenth of a degree.
Find the angle between \langle 7, -1 \rangle and \langle 3, 4 \rangle to the nearest tenth of a degree.
Find the exact angle, in radians, between \displaystyle \langle -1, \sqrt{3} \rangle and \displaystyle \langle \sqrt{3}, -1 \rangle.
Without finding any angle, say whether the angle between each pair is acute, right or obtuse: (a) \langle 5, -2 \rangle and \langle 3, 7 \rangle; (b) \langle -4, 6 \rangle and \langle 9, 6 \rangle; (c) \langle 2, -3 \rangle and \langle -1, 1 \rangle.

Day 2: angles at work

Find the three angles of the triangle P(1, -1), Q(7, -1), R(3, 3) to the nearest tenth of a degree.
Ifunanya’s vector \mathbf{a} has length 10 and direction angle 25^\circ; her vector \mathbf{b} has length 6 and direction angle 145^\circ. Find \mathbf{a} \cdot \mathbf{b}.
For which values of k is the angle between \langle k, 2 \rangle and \langle 3, -4 \rangle obtuse?
Vectors have \|\mathbf{u}\| = 3, \|\mathbf{v}\| = 4 and \|\mathbf{u} + \mathbf{v}\| = 6. Find the angle between them to the nearest tenth of a degree.

Reasoning

Prove that (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = \|\mathbf{u}\|^2 - \|\mathbf{v}\|^2, and say what it gives when \mathbf{u} and \mathbf{v} have the same length.
Explain why |\mathbf{u} \cdot \mathbf{v}| \le \|\mathbf{u}\|\,\|\mathbf{v}\| for nonzero vectors, and say when the two sides are equal.
Nkiruka says that if \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{w} and \mathbf{u} \ne \mathbf{0}, then \mathbf{v} = \mathbf{w}. Show with an example that she is wrong, and explain what the equation does say.
Obiora says that if the angle between \mathbf{u} and \mathbf{v} is \theta, the angle between \mathbf{u} and -\mathbf{v} is 180^\circ - \theta. Prove it with the dot product.
Explain why \mathbf{v} \cdot \mathbf{v} is never negative, and why it is 0 only for \mathbf{v} = \mathbf{0}.

Answers

-5. (-6)(2) + (1)(7) = -12 + 7 = -5.
7. (-3)(-5) + (4)(-2) = 15 - 8 = 7.
2. \tfrac{1}{2} \cdot 8 + 3\left(-\tfrac{2}{3}\right) = 4 - 2 = 2.
\mathbf{u} \cdot \mathbf{u} = 169 and \|\mathbf{u}\| = 13. 25 + 144 = 169, and \displaystyle \|\mathbf{u}\| = \sqrt{169}.
3. The components are \langle 2, -5 \rangle and \langle 4, 1 \rangle, so 8 - 5 = 3.
k = 13. 2k - 15 = 11, so 2k = 26.
-101. Expanding gives 3\,\mathbf{u} \cdot \mathbf{u} + 5\,\mathbf{u} \cdot \mathbf{v} - 2\,\mathbf{v} \cdot \mathbf{v}, which is 3(4) + 5(-3) - 2(49), that is 12 - 15 - 98.
\displaystyle \sqrt{22} \approx 4.69. \|\mathbf{u} - \mathbf{v}\|^2 = 25 - 2(6) + 9 = 22.
16. Subtracting \|\mathbf{u} - \mathbf{v}\|^2 = \|\mathbf{u}\|^2 - 2\,\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2 from \|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + 2\,\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2 leaves 4\,\mathbf{u} \cdot \mathbf{v} = 100 - 36 = 64.
20. (8)(5)\cos 60^\circ = 40 \cdot \tfrac{1}{2}.
-15. \displaystyle (5)(3\sqrt{2})\cos 135^\circ = 15\sqrt{2}\left(-\tfrac{\sqrt{2}}{2}\right) = -15.
120^\circ (\tfrac{2\pi}{3}). The dot product is -3 + 1 = -2, both lengths are 2, so \cos\theta = -\tfrac{2}{4} = -\tfrac{1}{2}.
82.2^\circ. The dot product is -5 + 8 = 3, the lengths are \displaystyle \sqrt{17} and \displaystyle \sqrt{29}, so \displaystyle \cos\theta = \tfrac{3}{\sqrt{493}} \approx 0.1351.
61.3^\circ. The dot product is 21 - 4 = 17, the lengths are \displaystyle 5\sqrt{2} and 5, so \displaystyle \cos\theta = \tfrac{17}{25\sqrt{2}} \approx 0.4808.
\tfrac{5\pi}{6}. The dot product is \displaystyle -\sqrt{3} - \sqrt{3} = -2\sqrt{3}, both lengths are 2, so \displaystyle \cos\theta = -\tfrac{\sqrt{3}}{2}.
(a) acute, since 15 - 14 = 1 > 0; (b) right, since -36 + 36 = 0; (c) obtuse, since -2 - 3 = -5 < 0.
P \approx 63.4^\circ, Q = 45^\circ, R \approx 71.6^\circ. At P: \langle 6, 0 \rangle \cdot \langle 2, 4 \rangle = 12 and \displaystyle \cos P = \tfrac{12}{6 \cdot 2\sqrt{5}} = \tfrac{1}{\sqrt{5}}. At Q: \langle -6, 0 \rangle \cdot \langle -4, 4 \rangle = 24 and \displaystyle \cos Q = \tfrac{24}{6 \cdot 4\sqrt{2}} = \tfrac{1}{\sqrt{2}}. At R: \langle -2, -4 \rangle \cdot \langle 4, -4 \rangle = 8 and \displaystyle \cos R = \tfrac{8}{\sqrt{20}\,\sqrt{32}} = \tfrac{1}{\sqrt{10}}. The three add to 180^\circ.
-30. The angle between the arrows is 145^\circ - 25^\circ = 120^\circ, so \mathbf{a} \cdot \mathbf{b} = (10)(6)\cos 120^\circ = 60\left(-\tfrac{1}{2}\right).
k < \tfrac{8}{3} with k \ne -\tfrac{3}{2}. The angle is more than 90^\circ when the dot product 3k - 8 is negative. At k = -\tfrac{3}{2} the vector is \langle -\tfrac{3}{2}, 2 \rangle = -\tfrac{1}{2}\langle 3, -4 \rangle, which points the opposite way, so the angle is 180^\circ, a straight angle rather than an obtuse one.
62.7^\circ. 36 = 9 + 2\,\mathbf{u} \cdot \mathbf{v} + 16 gives \mathbf{u} \cdot \mathbf{v} = \tfrac{11}{2}, so \cos\theta = \tfrac{11/2}{12} = \tfrac{11}{24} \approx 0.4583.
Distributing gives four terms, \mathbf{u} \cdot \mathbf{u}, -\mathbf{u} \cdot \mathbf{v}, \mathbf{v} \cdot \mathbf{u} and -\mathbf{v} \cdot \mathbf{v}, and the middle two cancel by commutativity, leaving \|\mathbf{u}\|^2 - \|\mathbf{v}\|^2. When the lengths are equal the product is 0: the vectors \mathbf{u} + \mathbf{v} and \mathbf{u} - \mathbf{v} (the two diagonals of the parallelogram on \mathbf{u} and \mathbf{v}, a rhombus in this case) have dot product zero, which Orthogonal vectors reads as perpendicular.
|\mathbf{u} \cdot \mathbf{v}| = \|\mathbf{u}\|\,\|\mathbf{v}\|\,|\cos\theta| and |\cos\theta| \le 1. Equality needs |\cos\theta| = 1, so \theta = 0 or \pi: the vectors are parallel, one a scalar multiple of the other. This inequality is also why the number inside the arccosine always lies in [-1, 1].
Take \mathbf{u} = \langle 1, 0 \rangle, \mathbf{v} = \langle 2, 5 \rangle, \mathbf{w} = \langle 2, -1 \rangle: both dot products are 2, yet \mathbf{v} \ne \mathbf{w}. The equation says \mathbf{u} \cdot (\mathbf{v} - \mathbf{w}) = 0, a statement about the angle between \mathbf{u} and \mathbf{v} - \mathbf{w}, not that \mathbf{v} - \mathbf{w} is zero.
Let \varphi be the angle between \mathbf{u} and -\mathbf{v}. Since \|-\mathbf{v}\| = \|\mathbf{v}\| and \mathbf{u} \cdot (-\mathbf{v}) = -\,\mathbf{u} \cdot \mathbf{v}, \cos\varphi = -\cos\theta = \cos(180^\circ - \theta). Both \varphi and 180^\circ - \theta lie in [0^\circ, 180^\circ], where cosine takes each value once, so \varphi = 180^\circ - \theta.
With \mathbf{v} = \langle a, b \rangle, \mathbf{v} \cdot \mathbf{v} = a^2 + b^2, a sum of two squares, and a square of a real number is never negative. The sum is 0 only if both squares are 0, that is a = b = 0.