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Dot product, length and orthogonality in n dimensions

With no picture to measure, how long is a list of six numbers, and how alike are two such lists?

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Dot product, length and orthogonality in n dimensions

Why this exists

Dritan, Valbona and Ilir run a perfumer’s workshop. They describe each scent by the intensities of six notes, citrus, rose, jasmine, cedar, vanilla and musk, so a scent profile is a vector in \mathbb{R}^6 (Adding and scaling vectors in n dimensions). How strong is a profile overall, and how alike are two profiles?

In the plane and in space, geometry answered both first. Length came from the Pythagorean theorem, the angle between two arrows could be measured, and \mathbf{u} \cdot \mathbf{v} = \lVert \mathbf{u} \rVert \lVert \mathbf{v} \rVert \cos\theta was proved from a drawn triangle (The dot product and the angle between vectors). In \mathbb{R}^6 there is nothing to draw. So the order turns around: define the dot product by arithmetic, define length from it, and prove the Cauchy-Schwarz inequality, which is what makes “the angle between two vectors” meaningful in any number of dimensions.

Definition

The dot product of two vectors in the same \mathbb{R}^n multiplies matching entries and adds:

\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + \dots + u_n v_n.

The result is a number, not a vector. Vectors with different numbers of entries have no dot product.

The length, or norm, of \mathbf{v} is

\lVert \mathbf{v} \rVert = \sqrt{\mathbf{v} \cdot \mathbf{v}} = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2},

a square root of a sum of squares, so always defined. The distance between \mathbf{u} and \mathbf{v} is \operatorname{dist}(\mathbf{u}, \mathbf{v}) = \lVert \mathbf{u} - \mathbf{v} \rVert.

A unit vector has length 1. For \mathbf{v} \neq \mathbf{0}, the vector \displaystyle \frac{1}{\lVert \mathbf{v} \rVert}\mathbf{v} is the unit vector in the direction of \mathbf{v}; producing it is normalizing \mathbf{v}.

Two vectors are orthogonal when \mathbf{u} \cdot \mathbf{v} = 0. In the plane this is the perpendicularity test of Orthogonal vectors; in \mathbb{R}^n it is the definition. The zero vector is orthogonal to every vector.

Intuition

Length in \mathbb{R}^n continues the Pythagorean theorem, which \mathbb{R}^3 uses twice: each new entry adds one more perpendicular direction and one more square.

The dot product rewards agreement. A term u_i v_i is positive when the entries share a sign and large when both are large. So \mathbf{u} \cdot \mathbf{v} is large and positive when two lists rise and fall together, near zero when agreements and disagreements cancel, and negative when one is high where the other is low. Dividing by both lengths removes overall size and leaves a measure of direction alone.

Results

Theorem 1 (properties of the dot product). For \mathbf{u}, \mathbf{v}, \mathbf{w} in \mathbb{R}^n and every scalar c:

  1. \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} (symmetry);
  2. (\mathbf{u} + \mathbf{v}) \cdot \mathbf{w} = \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w} (distributive property);
  3. (c\mathbf{u}) \cdot \mathbf{v} = c(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot (c\mathbf{v});
  4. \mathbf{u} \cdot \mathbf{u} \geq 0, with \mathbf{u} \cdot \mathbf{u} = 0 only when \mathbf{u} = \mathbf{0}.

Two consequences are used constantly: \lVert c\mathbf{v} \rVert = |c| \, \lVert \mathbf{v} \rVert, and

\lVert \mathbf{u} + \mathbf{v} \rVert^2 = \lVert \mathbf{u} \rVert^2 + 2\,\mathbf{u} \cdot \mathbf{v} + \lVert \mathbf{v} \rVert^2.

Theorem 2 (Cauchy-Schwarz inequality). For all \mathbf{u}, \mathbf{v} in \mathbb{R}^n,

|\mathbf{u} \cdot \mathbf{v}| \leq \lVert \mathbf{u} \rVert \, \lVert \mathbf{v} \rVert,

with equality exactly when one vector is a scalar multiple of the other.

The angle between two vectors. For nonzero \mathbf{u} and \mathbf{v}, Theorem 2 says

-1 \leq \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{u} \rVert \, \lVert \mathbf{v} \rVert} \leq 1,

so exactly one \theta with 0 \leq \theta \leq \pi has this ratio as its cosine. That \theta is defined to be the angle between \mathbf{u} and \mathbf{v}. In \mathbb{R}^2 and \mathbb{R}^3 it is the angle you would measure. Orthogonal means \displaystyle \theta = \frac{\pi}{2}, and \theta = 0 means one vector is a positive multiple of the other.

Theorem 3 (triangle inequality). For all \mathbf{u}, \mathbf{v} in \mathbb{R}^n, \lVert \mathbf{u} + \mathbf{v} \rVert \leq \lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert. In terms of distance,

\operatorname{dist}(\mathbf{u}, \mathbf{w}) \leq \operatorname{dist}(\mathbf{u}, \mathbf{v}) + \operatorname{dist}(\mathbf{v}, \mathbf{w}).
\mathbf{u}
\mathbf{v}
\mathbf{u} + \mathbf{v}
\mathbf{0}
Figure 1The direct route \displaystyle \lVert \mathbf{u} + \mathbf{v} \rVert = \sqrt{32} is shorter than the detour \displaystyle \lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert = \sqrt{26} + \sqrt{10}.

Theorem 4 (Pythagorean theorem in \mathbb{R}^n). \mathbf{u} and \mathbf{v} are orthogonal if and only if

\lVert \mathbf{u} + \mathbf{v} \rVert^2 = \lVert \mathbf{u} \rVert^2 + \lVert \mathbf{v} \rVert^2.
\mathbf{u}
\mathbf{v}
\mathbf{u} + \mathbf{v}
\mathbf{0}
Figure 2In \mathbb{R}^2, \mathbf{u} = (4, 2) and \mathbf{v} = (-1, 2) have \mathbf{u} \cdot \mathbf{v} = 0, and \lVert \mathbf{u} + \mathbf{v} \rVert^2 = 25 = 20 + 5.
Why it is true

Theorem 1. Each property holds entry by entry. For property 2, the i-th term is (u_i + v_i)w_i = u_i w_i + v_i w_i; add over i. Properties 1 and 3 come from u_i v_i = v_i u_i and (cu_i)v_i = c(u_i v_i) = u_i(cv_i). For property 4, u_1^2 + \dots + u_n^2 is a sum of squares: at least 0, and 0 only if every entry is 0.

The consequences. Property 3 gives \displaystyle \lVert c\mathbf{v} \rVert = \sqrt{c^2 (\mathbf{v} \cdot \mathbf{v})} = |c| \, \lVert \mathbf{v} \rVert. Expanding with property 2 on each side, then symmetry,

\begin{aligned} \lVert \mathbf{u} + \mathbf{v} \rVert^2 &= \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} \\ &\quad + \mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} \\ &= \lVert \mathbf{u} \rVert^2 + 2\,\mathbf{u} \cdot \mathbf{v} + \lVert \mathbf{v} \rVert^2. \end{aligned}

Theorem 4 follows at once: the expansion equals \lVert \mathbf{u} \rVert^2 + \lVert \mathbf{v} \rVert^2 exactly when \mathbf{u} \cdot \mathbf{v} = 0.

Theorem 2. If \mathbf{u} = \mathbf{0}, both sides are 0 and \mathbf{u} = 0\mathbf{v}. Otherwise, for every real t, property 4 and the expansion give

\begin{aligned} 0 &\leq \lVert t\mathbf{u} + \mathbf{v} \rVert^2 \\ &= \lVert \mathbf{u} \rVert^2 t^2 + 2(\mathbf{u} \cdot \mathbf{v})\,t + \lVert \mathbf{v} \rVert^2. \end{aligned}

A quadratic in t with positive leading coefficient that is never negative has at most one real root, so its discriminant is not positive:

4(\mathbf{u} \cdot \mathbf{v})^2 - 4\lVert \mathbf{u} \rVert^2 \lVert \mathbf{v} \rVert^2 \leq 0.

Taking square roots of the nonnegative numbers (\mathbf{u} \cdot \mathbf{v})^2 \leq \lVert \mathbf{u} \rVert^2 \lVert \mathbf{v} \rVert^2 gives the inequality. Equality means the discriminant is 0, so some t_0 makes \lVert t_0\mathbf{u} + \mathbf{v} \rVert = 0, and by property 4, \mathbf{v} = -t_0\mathbf{u}. Conversely, if \mathbf{v} = c\mathbf{u} then |\mathbf{u} \cdot c\mathbf{u}| = |c| \, \lVert \mathbf{u} \rVert^2 = \lVert \mathbf{u} \rVert \, \lVert c\mathbf{u} \rVert.

Theorem 3. Since \mathbf{u} \cdot \mathbf{v} \leq |\mathbf{u} \cdot \mathbf{v}| \leq \lVert \mathbf{u} \rVert \lVert \mathbf{v} \rVert by Theorem 2,

\begin{aligned} \lVert \mathbf{u} + \mathbf{v} \rVert^2 &\leq \lVert \mathbf{u} \rVert^2 + 2\lVert \mathbf{u} \rVert \lVert \mathbf{v} \rVert + \lVert \mathbf{v} \rVert^2 \\ &= \big(\lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert\big)^2. \end{aligned}

Both sides of the theorem are nonnegative, so taking square roots keeps the inequality. For the distance form, apply it to \mathbf{u} - \mathbf{w} = (\mathbf{u} - \mathbf{v}) + (\mathbf{v} - \mathbf{w}).

Worked examples

Example 1

Dot product, length, distance and a unit vector in \mathbb{R}^4

Let \mathbf{u} = (1, -2, 2, 4) and \mathbf{v} = (3, 1, 0, 2). Find \mathbf{u} \cdot \mathbf{v}, \lVert \mathbf{u} \rVert, \lVert \mathbf{v} \rVert, \operatorname{dist}(\mathbf{u}, \mathbf{v}) and the unit vector in the direction of \mathbf{u}.

Solution

Multiply matching entries and add: \mathbf{u} \cdot \mathbf{v} = 3 - 2 + 0 + 8 = 9.

Each length is the square root of the vector dotted with itself: \displaystyle \lVert \mathbf{u} \rVert = \sqrt{1 + 4 + 4 + 16} = \sqrt{25} = 5 and \displaystyle \lVert \mathbf{v} \rVert = \sqrt{9 + 1 + 0 + 4} = \sqrt{14}.

The distance is the length of the difference. Since \mathbf{u} - \mathbf{v} = (-2, -3, 2, 2), the distance is \displaystyle \sqrt{4 + 9 + 4 + 4} = \sqrt{21}.

Normalizing \mathbf{u} means scaling by \displaystyle \frac{1}{5}, which gives \displaystyle \left(\frac{1}{5}, -\frac{2}{5}, \frac{2}{5}, \frac{4}{5}\right). Check: \displaystyle \frac{1 + 4 + 4 + 16}{25} = 1.

Example 2

The angle between two vectors in \mathbb{R}^4

For \mathbf{u} and \mathbf{v} as in Example 1, confirm the Cauchy-Schwarz and triangle inequalities, and find the angle between the vectors.

Solution

Cauchy-Schwarz asks whether \displaystyle 9 \leq 5\sqrt{14} \approx 18.71. It holds, and not with equality, so neither vector is a multiple of the other. For the triangle inequality, \mathbf{u} + \mathbf{v} = (4, -1, 2, 6) has length \displaystyle \sqrt{57} \approx 7.55, less than \displaystyle 5 + \sqrt{14} \approx 8.74.

Because the ratio lies in [-1, 1], the angle is defined:

\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{u} \rVert \, \lVert \mathbf{v} \rVert} = \frac{9}{5\sqrt{14}} \approx 0.4811,

so \theta \approx 61.2^\circ. It is acute because the dot product is positive.

Example 3

Choosing a value that makes two vectors orthogonal

Find k so that \mathbf{x} = (2, k, -1, 3, 0) and \mathbf{y} = (k, 4, 6, -2, 5) are orthogonal in \mathbb{R}^5. Then confirm the Pythagorean theorem for that pair.

Solution

Orthogonal means a zero dot product: \mathbf{x} \cdot \mathbf{y} = 2k + 4k - 6 - 6 + 0 = 6k - 12, which is 0 exactly when k = 2.

With k = 2, \mathbf{x} = (2, 2, -1, 3, 0) and \mathbf{y} = (2, 4, 6, -2, 5), so \lVert \mathbf{x} \rVert^2 = 18 and \lVert \mathbf{y} \rVert^2 = 4 + 16 + 36 + 4 + 25 = 85. The sum is \mathbf{x} + \mathbf{y} = (4, 6, 5, 1, 5), and

\begin{aligned} \lVert \mathbf{x} + \mathbf{y} \rVert^2 &= 16 + 36 + 25 + 1 + 25 \\ &= 103 = 18 + 85, \end{aligned}

as Theorem 4 promises.

Example 4

Working from the properties alone

Vectors \mathbf{u} and \mathbf{v} in some \mathbb{R}^n have \lVert \mathbf{u} \rVert = 3, \lVert \mathbf{v} \rVert = 5 and \mathbf{u} \cdot \mathbf{v} = -4. Find \lVert 2\mathbf{u} - \mathbf{v} \rVert and (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - 3\mathbf{v}).

Solution

No entries are given, so everything comes from Theorem 1. The data are possible, since |-4| \leq 3 \cdot 5. Work with the square of the length, which is a dot product:

\begin{aligned} \lVert 2\mathbf{u} - \mathbf{v} \rVert^2 &= (2\mathbf{u} - \mathbf{v}) \cdot (2\mathbf{u} - \mathbf{v}) \\ &= 4\,\mathbf{u} \cdot \mathbf{u} - 4\,\mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v} \\ &= 36 + 16 + 25 = 77. \end{aligned}

The middle line uses the distributive property, property 3 to bring out the scalars, and symmetry to combine -2\,\mathbf{u} \cdot \mathbf{v} with -2\,\mathbf{v} \cdot \mathbf{u}. So \displaystyle \lVert 2\mathbf{u} - \mathbf{v} \rVert = \sqrt{77}. The same moves give

\begin{aligned} &(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - 3\mathbf{v}) \\ &= \mathbf{u} \cdot \mathbf{u} - 3\,\mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{u} - 3\,\mathbf{v} \cdot \mathbf{v} \\ &= 9 + 12 - 4 - 75 = -58. \end{aligned}
Example 5

How alike are two scent profiles?

In the note order citrus, rose, jasmine, cedar, vanilla, musk, Dritan’s new scent is \mathbf{d} = (4, 2, 0, 2, 1, 0), Valbona’s is \mathbf{s} = (2, 4, 1, 0, 2, 0), and Ilir has mixed \mathbf{r} = (8, 4, 0, 4, 2, 0). Compare \mathbf{s} and \mathbf{r} with \mathbf{d}, first by distance and then by angle.

Solution

By distance: \mathbf{d} - \mathbf{s} = (2, -2, -1, 2, -1, 0), so \displaystyle \operatorname{dist}(\mathbf{d}, \mathbf{s}) = \sqrt{14} \approx 3.74, while \mathbf{d} - \mathbf{r} = -\mathbf{d}, so \operatorname{dist}(\mathbf{d}, \mathbf{r}) = \lVert \mathbf{d} \rVert = 5. Valbona’s profile is nearer.

By angle: \displaystyle \lVert \mathbf{d} \rVert = \sqrt{16 + 4 + 0 + 4 + 1} = 5, \displaystyle \lVert \mathbf{s} \rVert = \sqrt{4 + 16 + 1 + 0 + 4} = 5 and \mathbf{d} \cdot \mathbf{s} = 8 + 8 + 2 = 18. So \displaystyle \cos\theta = \frac{18}{25} = 0.72 and \theta \approx 43.9^\circ. Ilir’s profile is \mathbf{r} = 2\mathbf{d}, the equality case of Cauchy-Schwarz, so its angle with \mathbf{d} is 0.

Distance counts every unit of intensity, so a double-strength scent is far away; the angle compares proportions only. For “do these smell alike”, the angle is the right measure, and its cosine is often called the cosine similarity.

Example 6

A use of Cauchy-Schwarz with no vectors in the question

Show that for all real numbers a_1, \dots, a_n,

(a_1 + \dots + a_n)^2 \leq n\,(a_1^2 + \dots + a_n^2),

with equality only when all the a_i are equal. Then find the least possible value of a_1^2 + \dots + a_5^2 when a_1 + \dots + a_5 = 10.

Solution

The left side is a dot product in disguise. Take \mathbf{a} = (a_1, \dots, a_n) and \mathbf{1} = (1, \dots, 1) in \mathbb{R}^n. Then \mathbf{a} \cdot \mathbf{1} = a_1 + \dots + a_n, \lVert \mathbf{a} \rVert^2 = a_1^2 + \dots + a_n^2 and \lVert \mathbf{1} \rVert^2 = n, so squaring Cauchy-Schwarz gives the claim. Equality needs one vector to be a multiple of the other; since \mathbf{1} \neq \mathbf{0}, that means \mathbf{a} = c\,\mathbf{1}, all entries equal.

With n = 5 and a sum of 10, 100 \leq 5(a_1^2 + \dots + a_5^2), so the sum of squares is at least 20. It equals 20 when every a_i = 2, and only then, so the least value is 20.

Boundaries

  • The dot product needs two vectors with the same number of entries, and its value is a scalar, so (\mathbf{u} \cdot \mathbf{v}) \cdot \mathbf{w} is not a dot product and there is no associative property.
  • There is no cancellation: \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{w} with \mathbf{u} \neq \mathbf{0} says only that \mathbf{u} is orthogonal to \mathbf{v} - \mathbf{w}.
  • The angle and the unit vector need nonzero vectors. The zero vector has no direction, though it is orthogonal to everything.
  • The triangle inequality is an equality only when one vector is a nonnegative multiple of the other; the Pythagorean equation holds only for orthogonal pairs.
  • All of this is for real entries. With complex entries, (1, i) \cdot (1, i) = 1 + i^2 = 0, so property 4 fails; complex vectors need a product with conjugates.

Where this goes next

With orthogonality defined in every \mathbb{R}^n, the point of a subspace closest to a vector is found by dropping a perpendicular, which is Orthogonal bases and projection onto a subspace, and a system with no exact solution gets a best approximate one in Orthogonality and least squares. Later the four properties of Theorem 1 become the definition of an inner product, and since the proofs above used nothing else, Theorems 2 to 4 hold for every such inner product.

Exercises

In Exercises 1 to 7, \mathbf{a} = (2, -1, 0, 2, 4) and \mathbf{b} = (1, 3, -2, 0, 1).

Computing in \mathbb{R}^n

  1. Find \mathbf{a} \cdot \mathbf{b}.
  2. Find \lVert \mathbf{a} \rVert and \lVert \mathbf{b} \rVert.
  3. Find the unit vector in the direction of \mathbf{b}.
  4. Find \operatorname{dist}(\mathbf{a}, \mathbf{b}).
  5. Find the angle between \mathbf{a} and \mathbf{b}, to the nearest tenth of a degree.
  6. Find (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) from the entries, and again from the lengths alone.
  7. Find \lVert -3\mathbf{b} \rVert without writing out -3\mathbf{b}.

Orthogonality and right angles

  1. Which pairs of \mathbf{w}_1 = (1, 1, 1, 1), \mathbf{w}_2 = (1, -1, 1, -1), \mathbf{w}_3 = (1, 1, -1, -1) and \mathbf{w}_4 = (2, 0, 1, 3) are orthogonal?
  2. Find k so that (k, 2, -1, k) and (3, k, 4, 1) are orthogonal.
  3. Find every c for which (c, 2, 1, c) and (c, -3, 4, 1) are orthogonal.
  4. Show that \mathbf{p} = (3, 0, 1, -2, 2) and \mathbf{q} = (1, 4, -1, 1, 0) are orthogonal, and find \lVert \mathbf{p} + \mathbf{q} \rVert both from Theorem 4 and from the entries of \mathbf{p} + \mathbf{q}.
  5. The points A = (2, 1, 0, 3), B = (3, 3, 2, 3) and C = (2, -1, 2, 4) in \mathbb{R}^4 form a triangle. Find its side lengths, show it has a right angle and say where, and find its area.

Working from the properties

In Exercises 13 to 15, \lVert \mathbf{u} \rVert = 2, \lVert \mathbf{v} \rVert = 4 and \mathbf{u} \cdot \mathbf{v} = 3.

  1. Find \lVert \mathbf{u} + \mathbf{v} \rVert.
  2. Find (2\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}).
  3. Find the angle between \mathbf{u} and \mathbf{v}, to the nearest tenth of a degree.
  4. Can two vectors have \lVert \mathbf{u} \rVert = 2, \lVert \mathbf{v} \rVert = 3 and \mathbf{u} \cdot \mathbf{v} = 7? Explain.

Scent profiles

In the note order of Example 5, Valbona records \mathbf{f} = (1, 3, 3, 0, 1, 0), \mathbf{g} = (2, 4, 2, 1, 0, 0) and \mathbf{h} = (0, 1, 0, 5, 2, 4).

  1. Find the cosine of the angle between \mathbf{f} and \mathbf{g}, and the angle to the nearest tenth of a degree.
  2. Which of \mathbf{g} and \mathbf{h} is more alike \mathbf{f} in proportions? Give the angle between \mathbf{f} and \mathbf{h}.

Reasoning

  1. Prove \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} using only properties 1 and 2 of Theorem 1, naming each step.
  2. Prove the polarization identity: \displaystyle \mathbf{u} \cdot \mathbf{v} = \frac{1}{4}\big(\lVert \mathbf{u} + \mathbf{v} \rVert^2 - \lVert \mathbf{u} - \mathbf{v} \rVert^2\big).
  3. Show that if \mathbf{u} is orthogonal to both \mathbf{v} and \mathbf{w}, then \mathbf{u} is orthogonal to c\mathbf{v} + d\mathbf{w} for all scalars c and d.
  4. Show that if \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{w} for every \mathbf{u} in \mathbb{R}^n, then \mathbf{v} = \mathbf{w}.
  5. Give a counterexample in \mathbb{R}^4 to the claim “if \mathbf{u} \neq \mathbf{0} and \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{w}, then \mathbf{v} = \mathbf{w}”.
  6. Use the triangle inequality to prove \big|\lVert \mathbf{u} \rVert - \lVert \mathbf{v} \rVert\big| \leq \lVert \mathbf{u} - \mathbf{v} \rVert.
Answers
  1. \mathbf{a} \cdot \mathbf{b} = 3, from 2 - 3 + 0 + 0 + 4.
  2. \displaystyle \lVert \mathbf{a} \rVert = \sqrt{25} = 5 and \displaystyle \lVert \mathbf{b} \rVert = \sqrt{15}.
  3. \displaystyle \frac{1}{\sqrt{15}}(1, 3, -2, 0, 1), scaling \mathbf{b} by the reciprocal of its length.
  4. \displaystyle \sqrt{34}. Here \mathbf{a} - \mathbf{b} = (1, -4, 2, 2, 3) and 1 + 16 + 4 + 4 + 9 = 34.
  5. About 81.1^\circ, since \displaystyle \cos\theta = \frac{3}{5\sqrt{15}} \approx 0.1549.
  6. 10. From the entries, \mathbf{a} + \mathbf{b} = (3, 2, -2, 2, 5) dotted with \mathbf{a} - \mathbf{b} is 3 - 8 - 4 + 4 + 15. From the lengths, the cross terms cancel by symmetry, leaving 25 - 15 = 10.
  7. \displaystyle 3\sqrt{15}, since \lVert c\mathbf{v} \rVert = |c| \, \lVert \mathbf{v} \rVert.
  8. \mathbf{w}_1 and \mathbf{w}_2, \mathbf{w}_1 and \mathbf{w}_3, \mathbf{w}_2 and \mathbf{w}_3, \mathbf{w}_2 and \mathbf{w}_4. The others give \mathbf{w}_1 \cdot \mathbf{w}_4 = 6 and \mathbf{w}_3 \cdot \mathbf{w}_4 = -2.
  9. \displaystyle k = \frac{2}{3}. The dot product is 3k + 2k - 4 + k = 6k - 4.
  10. c = 1 or c = -2. The dot product is c^2 + c - 2 = (c + 2)(c - 1).
  11. \mathbf{p} \cdot \mathbf{q} = 3 - 1 - 2 = 0. Theorem 4 gives \lVert \mathbf{p} + \mathbf{q} \rVert^2 = 18 + 19 = 37; directly, \mathbf{p} + \mathbf{q} = (4, 4, 0, -1, 2) and 16 + 16 + 1 + 4 = 37. So the length is \displaystyle \sqrt{37}.
  12. AB = 3, AC = 3, \displaystyle BC = 3\sqrt{2}, right angle at A, area \displaystyle \frac{9}{2}. The legs B - A = (1, 2, 2, 0) and C - A = (0, -2, 2, 1) have dot product 0, and C - B = (-1, -4, 0, 1) has squared length 18 = 9 + 9.
  13. \displaystyle \sqrt{26}, from 4 + 2 \cdot 3 + 16 = 26.
  14. -11. Expanding gives 2 \cdot 4 - 2 \cdot 3 + 3 - 16.
  15. About 68.0^\circ, since \displaystyle \cos\theta = \frac{3}{2 \cdot 4} = 0.375.
  16. No. Cauchy-Schwarz requires |\mathbf{u} \cdot \mathbf{v}| \leq 2 \cdot 3 = 6, and 7 > 6.
  17. \displaystyle \cos\theta = \frac{2}{\sqrt{5}} \approx 0.894, so \theta \approx 26.6^\circ. Here \mathbf{f} \cdot \mathbf{g} = 20, \displaystyle \lVert \mathbf{f} \rVert = \sqrt{20}, \lVert \mathbf{g} \rVert = 5.
  18. \mathbf{g}. For \mathbf{f} and \mathbf{h}, the dot product is 5 and \displaystyle \lVert \mathbf{h} \rVert = \sqrt{46}, so \displaystyle \cos\theta = \frac{5}{\sqrt{920}} \approx 0.165 and \theta \approx 80.5^\circ, much wider than 26.6^\circ.
  19. \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = (\mathbf{v} + \mathbf{w}) \cdot \mathbf{u} by symmetry, which is \mathbf{v} \cdot \mathbf{u} + \mathbf{w} \cdot \mathbf{u} by the distributive property, which is \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} by symmetry twice.
  20. As in the proof of Theorem 4, \lVert \mathbf{u} \pm \mathbf{v} \rVert^2 = \lVert \mathbf{u} \rVert^2 \pm 2\,\mathbf{u} \cdot \mathbf{v} + \lVert \mathbf{v} \rVert^2. Subtracting, the squared lengths cancel and leave 4\,\mathbf{u} \cdot \mathbf{v}.
  21. By Exercise 19 and property 3, \mathbf{u} \cdot (c\mathbf{v} + d\mathbf{w}) equals c(\mathbf{u} \cdot \mathbf{v}) + d(\mathbf{u} \cdot \mathbf{w}), which is c \cdot 0 + d \cdot 0 = 0.
  22. Take \mathbf{u} = \mathbf{v} - \mathbf{w}. Then \lVert \mathbf{v} - \mathbf{w} \rVert^2 = \mathbf{u} \cdot \mathbf{v} - \mathbf{u} \cdot \mathbf{w} = 0, so \mathbf{v} - \mathbf{w} = \mathbf{0} by property 4.
  23. \mathbf{u} = (1, 1, 0, 0), \mathbf{v} = (1, 0, 0, 0), \mathbf{w} = (0, 1, 0, 0): both dot products are 1, but \mathbf{v} \neq \mathbf{w}.
  24. Since \mathbf{u} = (\mathbf{u} - \mathbf{v}) + \mathbf{v}, the triangle inequality gives \lVert \mathbf{u} \rVert - \lVert \mathbf{v} \rVert \leq \lVert \mathbf{u} - \mathbf{v} \rVert. Swapping \mathbf{u} and \mathbf{v} bounds \lVert \mathbf{v} \rVert - \lVert \mathbf{u} \rVert by \lVert \mathbf{v} - \mathbf{u} \rVert, the same length. Together these bound the absolute value.
kg1128 · updated 2026-10-22 · CC-BY-SA-4.0