Multivariable-Calculus · Unit 1 · Video 4 · Interactive Practice

How the Dot Product Measures Angles in Space

IKey Formulas

Formula	Name	What you need
$\mathbf{A}\cdot\mathbf{B} = \|\mathbf{A}\|\,\|\mathbf{B}\|\cos\theta$	Geometric form	Two vectors and the angle between them
$\cos\theta = \dfrac{\mathbf{A}\cdot\mathbf{B}}{\|\mathbf{A}\|\,\|\mathbf{B}\|}$	Angle formula	One dot product, two lengths
$\mathbf{A}\cdot\mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3$	Dot product from components	Coordinates only
$\|\mathbf{A}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}$	Length from components	Coordinates only

Key Insight: Lengths are never negative, so dividing by $|\mathbf{A}|\,|\mathbf{B}|$ can never flip a sign — $\mathbf{A}\cdot\mathbf{B}$ and $\cos\theta$ always carry the same sign.

IIVisualization 1 — From Coordinates to the Angle

Two lengths and one dot product are enough to pin the angle down exactly.

💡 Nothing in this arithmetic mentions the plane: a third component simply adds one more product to the sum and one more square under each root.

IIIVisualization 2 — The Angle at a Corner in Space

The angle at $P$ in triangle $PQR$ , from coordinates alone: vectors, lengths, dot product, cosine.

Step 1

Step 1 — The two vectors out of $P$

\overrightarrow{PQ} = Q - P = \langle 0-1,\; 1-0,\; 0-0 \rangle = \langle -1,\, 1,\, 0 \rangle

\overrightarrow{PR} = R - P = \langle 0-1,\; 0-0,\; 2-0 \rangle = \langle -1,\, 0,\, 2 \rangle

Step 2 — Their lengths

|\overrightarrow{PQ}| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2}, \qquad |\overrightarrow{PR}| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{5}

Step 3 — The dot product

\overrightarrow{PQ}\cdot\overrightarrow{PR} = (-1)(-1) + (1)(0) + (0)(2) = 1 + 0 + 0 = 1

Step 4 — The cosine

\cos\theta = \frac{\overrightarrow{PQ}\cdot\overrightarrow{PR}}{|\overrightarrow{PQ}|\,|\overrightarrow{PR}|} = \frac{1}{\sqrt{2}\,\sqrt{5}} = \frac{1}{\sqrt{10}} \approx 0.3162

Step 5 — The angle

\theta = \arccos\!\left(\frac{1}{\sqrt{10}}\right) \approx 71.57^\circ

The dot product was positive, so the angle had to be acute — that was settled before the arccosine was ever taken.

IVVisualization 3 — The Sign Decides the Type

Acute, right, or obtuse is fixed by the sign of $\mathbf{A}\cdot\mathbf{B}$ , before any division happens.

💡 The vanishing case $\mathbf{A}\cdot\mathbf{B} = 0$ is perpendicularity — the exact boundary between the two shaded zones, and where the next video picks up.

VQuiz Questions

Problem 1 · Cosine Straight from Components

Given: $\mathbf{A} = \langle 1, 2, 2 \rangle$ and $\mathbf{B} = \langle 2, 2, 1 \rangle$ — find $\cos\theta$ .

✅ Correct!

\mathbf{A}\cdot\mathbf{B} = 8

and

|\mathbf{A}| = |\mathbf{B}| = 3

, so

\cos\theta = \tfrac{8}{9} \approx 0.889

and

\theta \approx 27.27^\circ

❌ Close, but the denominator is short one factor. The formula divides by the product of both lengths:

|\mathbf{A}|\,|\mathbf{B}| = 3 \cdot 3 = 9

, not by

|\mathbf{A}|

alone.

❌ Not quite.

81 = |\mathbf{A}|^2|\mathbf{B}|^2

— that is the value under the square roots. Take the roots first:

\sqrt{9}\cdot\sqrt{9} = 9

❌ Not quite. The two lengths multiply, they do not add:

3 \cdot 3 = 9

, not

3 + 3 = 6

. And

\tfrac{8}{6} \gt 1

, which no cosine can be.

Show solution

Step 1 — the dot product (multiply matching components, then add):

\mathbf{A}\cdot\mathbf{B} = (1)(2) + (2)(2) + (2)(1) = 2 + 4 + 2 = 8

Step 2 — the two lengths (square, add, take the root):

|\mathbf{A}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3, \qquad |\mathbf{B}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3

Step 3 — divide:

\cos\theta = \frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}|\,|\mathbf{B}|} = \frac{8}{3 \cdot 3} = \frac{8}{9} \approx 0.8889

So $\theta = \arccos\!\left(\tfrac{8}{9}\right) \approx 27.27^\circ$ — a positive dot product, and indeed an acute angle.

Problem 2 · Classify Without Dividing

Given: $\mathbf{A} = \langle 3, -1, 2 \rangle$ and $\mathbf{B} = \langle -2, 4, 5 \rangle$ — classify the angle $\theta$ between them.

✅ Correct!

\mathbf{A}\cdot\mathbf{B} = -6 - 4 + 10 = 0

, so

\cos\theta = 0

and

\theta = 90^\circ

exactly — no lengths required.

❌ Not quite — check the total.

(3)(-2) + (-1)(4) + (2)(5) = -6 - 4 + 10 = 0

. Acute needs a strictly positive dot product; this one lands exactly on the boundary.

❌ Not quite. The two negative products total

-10

, but the third contributes

+10

: the sum is exactly

0

, not negative.

❌ The lengths cannot help here.

|\mathbf{A}|

and

|\mathbf{B}|

are positive, so dividing by them rescales

\cos\theta

without touching its sign. The sign of

\mathbf{A}\cdot\mathbf{B}

alone settles the classification.

Show solution

Only the numerator of the angle formula carries sign information, so compute it alone:

\mathbf{A}\cdot\mathbf{B} = (3)(-2) + (-1)(4) + (2)(5) = -6 - 4 + 10 = 0

Since $\mathbf{A}\cdot\mathbf{B} = |\mathbf{A}|\,|\mathbf{B}|\cos\theta$ with $|\mathbf{A}| \gt 0$ and $|\mathbf{B}| \gt 0$ , a zero dot product forces $\cos\theta = 0$ , hence

\theta = 90^\circ

The two vectors are perpendicular. (For the record, $|\mathbf{A}| = \sqrt{14}$ and $|\mathbf{B}| = \sqrt{45}$ — both irrational, and both irrelevant to the answer.)

Problem 3 · The Angle at a Vertex

Given: the triangle with vertices $P(1, 0, 0)$ , $Q(0, 2, 0)$ and $R(0, 0, 2)$ — find the angle $\theta$ at vertex $P$ .

What is $\overrightarrow{PQ}\cdot\overrightarrow{PR}$ ?

What is $\cos\theta$ ?

✅ Correct!

\cos\theta = \tfrac{1}{5} = 0.2

, so

\theta = \arccos(0.2) \approx 78.46^\circ

— acute, exactly as the positive dot product promised.

❌ Check the products.

\overrightarrow{PQ} = \langle -1, 2, 0 \rangle

and

\overrightarrow{PR} = \langle -1, 0, 2 \rangle

, so the terms are

(-1)(-1)

(2)(0)

and

(0)(2)

❌ Watch the double negative. The first components are both

-1

, and

(-1)(-1) = +1

. The other two products vanish, so the sum is

+1

❌ Check the denominator.

|\overrightarrow{PQ}| = \sqrt{1 + 4 + 0} = \sqrt{5}

and

|\overrightarrow{PR}| = \sqrt{1 + 0 + 4} = \sqrt{5}

, so the product of the lengths is

\sqrt{5}\cdot\sqrt{5} = 5

❌ Those are the video's numbers, not these. Here

Q = (0, 2, 0)

, so

|\overrightarrow{PQ}| = \sqrt{5}

rather than

\sqrt{2}

, and the denominator is

\sqrt{5}\cdot\sqrt{5} = 5

, not

\sqrt{10}

❌ The square roots were skipped.

25 = |\overrightarrow{PQ}|^2\,|\overrightarrow{PR}|^2

— that is the value under the two roots. Take the roots first:

\sqrt{5}\cdot\sqrt{5} = 5

, so

\cos\theta = \tfrac{1}{5}

❌ One length short. Dividing by

|\overrightarrow{PQ}|

alone leaves

1/\sqrt{5}

; the formula divides by both lengths, giving

1/(\sqrt{5}\cdot\sqrt{5}) = 1/5

Show solution

Step 1 — the two edges leaving $P$ (head minus tail):

\overrightarrow{PQ} = Q - P = \langle -1,\, 2,\, 0 \rangle, \qquad \overrightarrow{PR} = R - P = \langle -1,\, 0,\, 2 \rangle

Step 2 — the dot product:

\overrightarrow{PQ}\cdot\overrightarrow{PR} = (-1)(-1) + (2)(0) + (0)(2) = 1 + 0 + 0 = 1

Step 3 — the lengths:

|\overrightarrow{PQ}| = \sqrt{(-1)^2 + 2^2 + 0^2} = \sqrt{5}, \qquad |\overrightarrow{PR}| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{5}

Step 4 — assemble:

\cos\theta = \frac{1}{\sqrt{5}\,\sqrt{5}} = \frac{1}{5} = 0.2 \quad\Longrightarrow\quad \theta = \arccos(0.2) \approx 78.46^\circ

Moving $Q$ from $(0,1,0)$ to $(0,2,0)$ opened the corner from about $71.57^\circ$ to about $78.46^\circ$ — the same procedure, new coordinates.

Problem 4 · Tune One Component to a Right Angle

Given: $\mathbf{A} = \langle 2, -3, 1 \rangle$ and $\mathbf{B} = \langle 4, k, 2 \rangle$ — find the value of $k$ that makes $\theta = 90^\circ$ .

✅ Correct!

\mathbf{A}\cdot\mathbf{B} = 8 - 3k + 2 = 10 - 3k

, which vanishes exactly at

k = \tfrac{10}{3}

— and a vanishing dot product is a right angle.

❌ Sign slip.

10 - 3k = 0

rearranges to

3k = +10

, so

k = +\tfrac{10}{3}

. At

k = -\tfrac{10}{3}

the dot product is

10 + 10 = 20

, firmly acute.

❌ That kills only the middle term. With

k = 0

the sum is

8 + 0 + 2 = 10 \ne 0

; the middle product has to cancel the other two, not merely disappear.

❌ The fraction is inverted. From

3k = 10

, divide by

3

k = \tfrac{10}{3}

. (Check:

\tfrac{3}{10}

gives

10 - 0.9 = 9.1 \ne 0

Show solution

A right angle means $\cos\theta = 0$ . Since $\mathbf{A}\cdot\mathbf{B} = |\mathbf{A}|\,|\mathbf{B}|\cos\theta$ and neither length is zero, this is the same as requiring the dot product to vanish:

\mathbf{A}\cdot\mathbf{B} = (2)(4) + (-3)(k) + (1)(2) = 8 - 3k + 2 = 10 - 3k

10 - 3k = 0 \quad\Longrightarrow\quad k = \frac{10}{3}

Verify: with $\mathbf{B} = \left\langle 4, \tfrac{10}{3}, 2 \right\rangle$ ,

\mathbf{A}\cdot\mathbf{B} = 8 - 3\!\left(\tfrac{10}{3}\right) + 2 = 8 - 10 + 2 = 0 \ \checkmark

Note what was not needed: neither $|\mathbf{A}| = \sqrt{14}$ nor $|\mathbf{B}|$ appears anywhere. Perpendicularity is a statement about the numerator alone.

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