Multivariable-Calculus · Unit 1 · Video 2 · Interactive Practice

The Double Life of Vectors: Scaling, Adding, and the Parallelogram Rule

IKey Formulas

Formula	Name	What it does
$cA = \langle ca_1,\, ca_2,\, ca_3 \rangle$	Scalar multiplication	Length scales by $\|c\|$ ; a positive $c$ keeps the direction, a negative $c$ flips it
$A = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$	Unit-vector form	Each term is a unit vector stretched by one component
$A + B = \langle a_1 + b_1,\, a_2 + b_2,\, a_3 + b_3 \rangle$	Addition	Head-to-tail; the diagonal out of the shared corner
$A - B = A + (-B) = \langle a_1 - b_1,\, a_2 - b_2,\, a_3 - b_3 \rangle$	Subtraction	The diagonal from the head of $B$ to the head of $A$

Key Insight: One parallelogram carries both operations — the diagonal out of the shared corner is $A + B$ , and the diagonal from the head of $B$ to the head of $A$ is $A - B$ .

IIVisualization 1 — What a Scalar Can and Cannot Change

A scalar changes length, and its sign changes direction — every multiple $cA$ stays on one line.

Scalar c 2

IIIVisualization 2 — Two Routes, One Corner

Walking along $A$ then $B$ , or along $B$ then $A$ , lands on the same corner.

Route Walk 0%

IVVisualization 3 — The Second Diagonal

Subtracting $B$ means adding $-B$ , and the result is the parallelogram's other diagonal.

Step 1

💡 The plane is only a convenience for drawing: every component identity here — $cA$ , $A + B$ , $A - B$ — reads the same with a third slot, or a hundredth.

VQuiz Questions

Problem 1 · Scaling a Vector

Given: $A = \langle 3, 2, 1 \rangle$ — find $-2A$ .

✅ Correct! Every component is multiplied by

-2

: the arrow is twice as long as

A

and points the opposite way.

❌ The length is right, the direction is not. A negative scalar reverses the arrow, so all three components must change sign.

❌ That subtracts $2$ from each component. A scalar multiplies each component;

-2 \cdot 3 = -6

, not

3 - 2 = 1

❌ Only the first component was scaled. The scalar reaches every slot:

-2 \cdot 2 = -4

and

-2 \cdot 1 = -2

as well.

❌ Not quite. Apply the factor

-2

to each of the three components separately.

Show solution

Scalar multiplication acts slot by slot:

-2A = -2\langle 3, 2, 1 \rangle = \langle -2(3),\; -2(2),\; -2(1) \rangle = \langle -6, -4, -2 \rangle

Geometrically the two faces agree: $|-2| = 2$ stretches the arrow to twice its length, and the negative sign reverses it. In unit-vector form,

-2A = -6\hat{i} - 4\hat{j} - 2\hat{k}

Nothing else about the arrow changes — a scalar can stretch, shrink, or reverse, and that is all.

Problem 2 · Componentwise Subtraction

Given: $A = \langle 2, -1, 4 \rangle$ and $B = \langle -3, 5, -2 \rangle$ — find $A - B$ .

✅ Correct! Each slot is handled on its own:

2 - (-3) = 5

-1 - 5 = -6

, and

4 - (-2) = 6

❌ That is $A + B$ . Subtraction flips every sign in

B

first: the leading slot is

2 - (-3) = 5

, not

2 + (-3) = -1

❌ That is $B - A$ — the same length, opposite direction. Order matters:

A - B

runs from the head of

B

to the head of

A

, and

B - A = -(A - B)

runs back the other way.

❌ The last slot lost a double negative.

4 - (-2) = 4 + 2 = 6

, not

4 - 2 = 2

❌ Not quite. Subtract matching slots one at a time, and watch for double negatives.

Show solution

Subtraction is componentwise, first slot to first slot:

A - B = \langle 2 - (-3),\; -1 - 5,\; 4 - (-2) \rangle = \langle 2 + 3,\; -6,\; 4 + 2 \rangle = \langle 5, -6, 6 \rangle

The same answer through the "add the opposite" route, with $-B = \langle 3, -5, 2 \rangle$ :

A + (-B) = \langle 2 + 3,\; -1 + (-5),\; 4 + 2 \rangle = \langle 5, -6, 6 \rangle

Subtraction is not a new operation — it is addition of the reversed vector. ✓

Problem 3 · Two Notations, One Vector

Given: $A = 3\hat{i} + 2\hat{j} + \hat{k}$ and $B = \langle 1, -4, 2 \rangle$ — find $2A - B$ , written in unit-vector form.

✅ Correct!

2A = \langle 6, 4, 2 \rangle

, so

2A - B = \langle 5, 8, 0 \rangle

— and a zero component contributes nothing, so the

\hat{k}

term drops out.

❌ The $\hat{j}$ slot lost a double negative.

4 - (-4) = 8

, not

4 - 4 = 0

; only the

\hat{k}

slot is genuinely zero.

❌ That is $2A + B$ . Subtracting

B

means adding

-B = \langle -1, 4, -2 \rangle

, so the first slot is

6 - 1 = 5

❌ Only the $\hat{i}$ term got doubled. The scalar

2

multiplies all three components of

A

2A = \langle 6, 4, 2 \rangle

, not

\langle 6, 2, 1 \rangle

❌ Not quite. Convert

A

to components, scale it by

2

, then subtract

B

slot by slot.

Show solution

Read $A$ through the lens of scaling: $3\hat{i}$ , $2\hat{j}$ and $\hat{k}$ are the unit vectors stretched by $3$ , $2$ and $1$ , so

A = \langle 3, 2, 1 \rangle, \qquad 2A = \langle 6, 4, 2 \rangle

Now subtract componentwise:

2A - B = \langle 6 - 1,\; 4 - (-4),\; 2 - 2 \rangle = \langle 5, 8, 0 \rangle

Translate back to unit-vector form:

2A - B = 5\hat{i} + 8\hat{j} + 0\hat{k} = 5\hat{i} + 8\hat{j}

The two notations are the same object, so you may compute in whichever one is shorter and report the answer in the other.

Problem 4 · Both Diagonals of One Parallelogram

Given: $A = \langle 5, 2 \rangle$ and $B = \langle 1, 4 \rangle$ drawn from a common corner $O$ , spanning a parallelogram — identify each of its two diagonals.

Which vector runs along the diagonal out of $O$ ?

Which vector runs from the head of $B$ to the head of $A$ ?

✅ Correct! One parallelogram, two operations: the diagonal out of the shared corner is

A + B = \langle 6, 6 \rangle

, and the diagonal read from the head of

B

to the head of

A

A - B = \langle 4, -2 \rangle

❌ Check the diagonal out of $O$ . It is the head-to-tail sum: travel along

A

, then along the slid copy of

B

, and add slot by slot —

\langle 5 + 1,\; 2 + 4 \rangle

❌ That is not the arrow from the head of $B$ to the head of $A$ . Head-to-tail along that diagonal gives

B + D = A

, which forces

D = A - B = \langle 5 - 1,\; 2 - 4 \rangle

; reading the same arrow backwards instead gives

B - A

, the same segment reversed.

Show solution

The diagonal out of $O$ . Slide $B$ until its tail sits at the head of $A$ . The far corner of the parallelogram is where you land:

A + B = \langle 5 + 1,\; 2 + 4 \rangle = \langle 6, 6 \rangle

The other diagonal. Its tail is the head of $B$ , at $(1, 4)$ , and its head is the head of $A$ , at $(5, 2)$ . Call it $D$ and use head-to-tail reasoning: going along $B$ and then along $D$ must land at the head of $A$ , so

B + D = A \quad \Longrightarrow \quad D = A - B = \langle 5 - 1,\; 2 - 4 \rangle = \langle 4, -2 \rangle

The two rejected forms are worth naming. Reading that diagonal backwards — head of $A$ to head of $B$ — gives $B - A = \langle -4, 2 \rangle$ : same length, opposite direction. And $\langle 5, 8 \rangle$ multiplies matching components, which is not one of the vector operations at all.

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