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

One Arrow, Two Triangles: The Vector Language of Space

IKey Formulas

Formula	Name	What you need
$\vec{A} = \langle a_1, a_2, a_3 \rangle = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$	Components — two notations	How far the arrow goes along $x$ , $y$ , $z$
$\|\vec{A}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}$	Length (magnitude)	Every component
$\overrightarrow{PQ} = \langle q_1 - p_1,\ q_2 - p_2,\ q_3 - p_3 \rangle$	Vector from one point to another	The two endpoints
$\operatorname{dir}\vec{A} = \dfrac{\vec{A}}{\|\vec{A}\|}$	Direction (unit vector)	The vector and its own length

Key Insight: The length formula is the Pythagorean theorem used twice — $|\vec{B}|^2 = a_1^2 + a_2^2$ on the floor, then $|\vec{A}|^2 = |\vec{B}|^2 + a_3^2$ in the vertical slice. Nothing else is going on.

IIVisualization 1 — One Arrow, Two Right Triangles

The length of $\vec{A}$ hides in two right triangles: one lying on the floor, one standing in a vertical slice.

IIIVisualization 2 — A Vector Has No Starting Point

A vector is its components alone — the point the arrow starts from carries no information.

IVVisualization 3 — Square, Add, Take the Root

Each component contributes its square as an area; the total area is $|\vec{A}|^2$ .

Dimension n 3

💡 The same three moves run unchanged for $n = 10{,}000$ components: only the picture stops at three dimensions, never the formula.

VQuiz Questions

Problem 1 · Length from Components

Given: $\vec{A} = \langle 2, 3, 6 \rangle$ — find $|\vec{A}|$ .

✅ Correct!

2^2 + 3^2 + 6^2 = 4 + 9 + 36 = 49

, and

\sqrt{49} = 7

❌ A component was left out.

\sqrt{13} = \sqrt{2^2 + 3^2}

is the length of the shadow

\vec{B}

in the

xy

-plane. The climb

a_3 = 6

still has to be added in the vertical slice:

13 + 36 = 49

❌ That adds the components instead of squaring them.

2 + 3 + 6 = 11

measures a walk along three edges of the box; the arrow cuts straight across, so it must come out shorter.

❌ That is $|\vec{A}|^2$ , not $|\vec{A}|$ . The sum of squares is

49

; the length is its square root.

Show solution

Square each component, add them up, take the square root:

|\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} = \sqrt{2^2 + 3^2 + 6^2}

= \sqrt{4 + 9 + 36} = \sqrt{49} = 7

The two triangles are visible inside that arithmetic: the floor gives $|\vec{B}|^2 = 2^2 + 3^2 = 13$ , and the vertical slice gives $|\vec{A}|^2 = 13 + 6^2 = 49$ .

Problem 2 · A Negative Component and a Zero

Given: $\vec{A} = \langle -3, 0, 4 \rangle$ — find $|\vec{A}|$ .

✅ Correct!

(-3)^2 = 9

, the zero component contributes nothing, and

9 + 0 + 16 = 25

gives

|\vec{A}| = 5

❌ The sign was carried through the squaring.

(-3)^2 = +9

, not

-9

— squaring destroys every sign, which is exactly why a length can never come out negative. You computed

-9 + 0 + 16 = 7

❌ That adds the components.

-3 + 0 + 4 = 1

; the formula squares first, so the

-3

contributes

9

, not

-3

❌ That adds the two leg lengths.

3 + 4 = 7

is the trip along the legs; the hypotenuse of a

3

–

4

right triangle is

5

Show solution

A zero component is allowed and a negative component is allowed — the formula never changes:

|\vec{A}| = \sqrt{(-3)^2 + 0^2 + 4^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5

Because $a_2 = 0$ , this arrow lies flat in the $xz$ -plane: it goes $3$ units backward along $x$ and climbs $4$ units, and its length is the hypotenuse of that single $3$ – $4$ – $5$ right triangle.

Problem 3 · From Two Points to a Length

Given: the points $P = (1, -2, 4)$ and $Q = (3, 1, 8)$ — find the vector $\overrightarrow{PQ}$ and its length $|\overrightarrow{PQ}|$ .

What are the components of $\overrightarrow{PQ}$ ?

What is $|\overrightarrow{PQ}|$ ?

✅ Correct!

\overrightarrow{PQ} = \langle 2, 3, 4 \rangle

and

|\overrightarrow{PQ}| = \sqrt{4 + 9 + 16} = \sqrt{29} \approx 5.385

❌ That is $\overrightarrow{QP}$ . The arrow from

P

Q

is tip minus tail:

Q - P

, not

P - Q

. Reversing the arrow flips every component's sign (though not the length).

❌ A double negative slipped in the second component.

q_2 - p_2 = 1 - (-2) = 1 + 2 = 3

, not

1 - 2 = -1

❌ The coordinates were added, not subtracted.

\overrightarrow{PQ}

measures the change from

P

Q

, so each component is a difference:

3 - 1

1 - (-2)

8 - 4

❌ Only two components were used.

\sqrt{13} = \sqrt{2^2 + 3^2}

is the shadow length; the third component

4

still has to enter:

13 + 16 = 29

❌ That adds the components.

2 + 3 + 4 = 9

; the formula squares them first, giving

4 + 9 + 16 = 29

❌ That is $|\overrightarrow{PQ}|^2$ . The sum of squares is

29

; the length is

\sqrt{29}

, a little under

6

Show solution

Step 1 — the components are differences, tip minus tail:

\overrightarrow{PQ} = \langle 3 - 1,\ 1 - (-2),\ 8 - 4 \rangle = \langle 2,\ 3,\ 4 \rangle

Step 2 — apply the length formula to those components:

|\overrightarrow{PQ}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \approx 5.385

Once the components are in hand, $P$ and $Q$ are no longer needed: the same arrow drawn from the origin, or from anywhere else, is the same vector with the same length.

Problem 4 · Four Dimensions, Same Formula

Given: the four-dimensional vector $\vec{A} = \langle 1, 2, 2, 4 \rangle$ — find $|\vec{A}|$ and then $\operatorname{dir}\vec{A}$ .

What is $|\vec{A}|$ ?

What is $\operatorname{dir}\vec{A}$ ?

✅ Correct!

1 + 4 + 4 + 16 = 25

|\vec{A}| = 5

, and dividing by

5

gives

\left\langle \tfrac{1}{5}, \tfrac{2}{5}, \tfrac{2}{5}, \tfrac{4}{5} \right\rangle

— whose own length is

\sqrt{\tfrac{25}{25}} = 1

❌ One component was dropped.

\sqrt{21} = \sqrt{1 + 4 + 16}

uses only three of the four entries; both

2

's must be squared:

1 + 4 + 4 + 16 = 25

❌ That adds the components.

1 + 2 + 2 + 4 = 9

; squaring first gives

25

, and

\sqrt{25} = 5

❌ That is $|\vec{A}|^2$ .

25

is the sum of the squares; the length is

\sqrt{25} = 5

❌ That is $\vec{A}$ itself, and its length is $5$ , not $1$ . The direction is

\vec{A}

rescaled to unit length, so every component must be divided by

|\vec{A}|

❌ The vector was multiplied by $|\vec{A}|$ instead of divided. That makes it five times longer — length

25

— when the goal is length

1

❌ Divided by $|\vec{A}|^2 = 25$ instead of by $|\vec{A}| = 5$ . Its length is

\tfrac{1}{25} \cdot 5 = \tfrac{1}{5}

, not

1

Show solution

Nothing about the machinery changes in four dimensions — the sum simply runs one term longer:

|\vec{A}| = \sqrt{1^2 + 2^2 + 2^2 + 4^2} = \sqrt{1 + 4 + 4 + 16} = \sqrt{25} = 5

The direction is that vector scaled down to unit length, which means dividing by its own length:

\operatorname{dir}\vec{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{1}{5}\langle 1, 2, 2, 4 \rangle = \left\langle \tfrac{1}{5}, \tfrac{2}{5}, \tfrac{2}{5}, \tfrac{4}{5} \right\rangle

Verify: $\sqrt{\tfrac{1}{25} + \tfrac{4}{25} + \tfrac{4}{25} + \tfrac{16}{25}} = \sqrt{\tfrac{25}{25}} = 1$ ✓ — it points the same way as $\vec{A}$ and has length $1$ .

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