Scalars vs vectors — definition, examples: Question bank

Figure — Scalars vs vectors — definition, examples

Formula The one formula every item leans on — built from a picture

Place $A$ and $B$ tail-to-tail and complete the parallelogram; the diagonal is $R$ . Its length is $∣ R ∣ = A^{2} + B^{2} + 2 A B cos θ .$ Why this and not $A + B$ ? Slide $B$ so its tail sits at the tip of $A$ (that is the triangle law — the figure below shows the slide). Now $A$ , the slid $B$ , and the closing arrow $R$ form a triangle. The law of cosines — the exact tool for "find the third side from two sides and the angle wedged between them" — needs the interior angle of that triangle at the corner where $A$ ends and $B$ begins.

Why that interior angle is $(18 0^{\circ} - θ)$ , not $θ$ : the original angle $θ$ is measured tail-to-tail. But in the triangle, $A$ arrives at that corner head-first and $B$ leaves it tail-first — the two arrows are laid nose-to-tail, so the arrow directions are flipped relative to the tail-to-tail picture. The corner opening is therefore the supplement of $θ$ , i.e. $18 0^{\circ} - θ$ . The figure marks both $θ$ (blue, tail-to-tail) and its supplement (plum, inside the triangle) so you can see they add to a straight line. Plugging in: $∣ R ∣^{2} = A^{2} + B^{2} - 2 A B cos (18 0^{\circ} - θ) = A^{2} + B^{2} + 2 A B cos θ,$ since $cos (18 0^{\circ} - θ) = - cos θ$ . The $+ 2 A B cos θ$ term is exactly "how much of $B$ points along $A$ ".

True or false — justify

Temperature is a scalar because it is fully given by one number and a unit.

True. "27 °C" needs no direction; you never place two temperatures tail-to-tail and take a diagonal, so no parallelogram law applies — it is a pure scalar.

Electric current in a circuit wire is a vector because it flows in a definite direction.

False (as a circuit quantity). At a junction, currents add arithmetically — you literally count charges per second flowing in and out,

I_{1} + I_{2} = I_{3}

(Kirchhoff), with no diagonal and no

cos θ

. So the circuit current

I

is a scalar. (Careful: the related current density

J

, "current per unit area at a point", is a genuine vector field in electromagnetism — but that is a different quantity from the single number

I

carried by a wire. The trap conflates the two.)

The magnitude of a vector can be negative.

False. A magnitude is a length (found via

), so it is

\geq 0

. A minus sign on a vector flips its direction, it does not make its length negative.

If two vectors have equal magnitude, their sum must be larger than either one.

False. Put them tail-to-tail at

θ = 18 0^{\circ}

(pointing opposite): the diagonal collapses,

A^{2} + A^{2} + 2 A^{2} cos 18 0^{\circ} = 2 A^{2} - 2 A^{2} = 0

. The sum is zero, smaller than either.

Distance travelled can be smaller than the magnitude of displacement.

False. Distance is total path length and displacement is the straight-line shortcut, so distance

\geq ∣ displacement ∣

always — equality only when the path is a straight line with no reversal.

On a complete lap of a track, average speed is zero.

False. Speed uses distance (the full lap length, nonzero), so it is nonzero. It is the average velocity that is zero, because displacement is zero.

Adding a scalar and a vector, like "5 kg + a force", is a meaningful operation.

False. Different categories (and often different dimensions, see Units and Dimensions) cannot be added; only same-type quantities combine.

Two vectors always produce a resultant whose magnitude lies between

∣ A - B ∣

and

A + B

True. As

θ

runs

0^{\circ} \to 18 0^{\circ}

cos θ

runs

+ 1 \to - 1

, so

∣ R ∣

sweeps continuously from the maximum

A + B

(arrows aligned, diagonal fully stretched) down to the minimum

∣ A - B ∣

(arrows opposed, diagonal shrunk).

Spot the error

"Displacement of 3 km East plus 4 km North gives 7 km."

The error is adding magnitudes of non-collinear vectors arithmetically. Draw them tail-to-tail at

θ = 9 0^{\circ}

: the diagonal is

∣ R ∣ = 3^{2} + 4^{2} + 2 \cdot 3 \cdot 4 cos 9 0^{\circ} = 9 + 16 = 5

km (the

cos 9 0^{\circ} = 0

kills the cross term), not

7

"Speed is just velocity without the arrow, so they always have the same number."

They agree only over a straight, non-reversing path. On any curved or back-and-forth motion the distance exceeds the shortcut, so average speed

>

the magnitude of the average velocity (you must compare the scalar speed to

∣ average velocity ∣

, not to a signed vector). See Speed vs Velocity.

"Since force is a vector,

6 N + 8 N

can never equal

14

N."

Vectors can add like scalars when aligned. Set

θ = 0^{\circ}

cos 0^{\circ} = 1

; then the diagonal lies straight along both arrows:

36 + 64 + 2 \cdot 6 \cdot 8 \cdot 1 = 196 = 14

N. The parallelogram rule reduces to plain addition in that special case.

"Work has direction because force has direction, so work is a vector."

Work is the dot product of force and displacement — you multiply the two lengths and the cosine of the angle between them, and a length-times-length-times-number is just a plain number (Dot and Cross Products). Work is a scalar — it can be positive or negative, but that sign means energy in vs out, not a direction.

"A negative charge is a vector because the sign points somewhere."

The sign on charge is a type/polarity label (below-zero on a number line), not a spatial direction. Two charges add as

q_{1} + q_{2}

with no angle involved, so charge is a scalar.

"Because velocity is a vector, its magnitude (speed) is also a vector."

Taking the magnitude of a vector strips the direction and yields a scalar. Speed is the scalar

∣ v ∣

Why questions

Why isn't "having a direction" enough to call something a vector?

Because counted quantities like circuit current also have a direction yet add arithmetically (

I_{1} + I_{2}

, no diagonal). The decisive test is whether they obey the triangle/parallelogram law when placed tail-to-tail.

Why does the resultant formula contain

cos θ

rather than just

A + B

Because the resultant is the triangle's third side, and the law of cosines needs the angle between the other two sides. The

cos θ

term encodes how much of $B$ lines up with $A$ : when they align it adds fully (

cos 0^{\circ} = 1

), when perpendicular it contributes nothing (

cos 9 0^{\circ} = 0

Why can a vector's average value be zero over a trip even when "something clearly happened"?

Directed quantities can cancel: equal-and-opposite contributions (tail-to-tail at

18 0^{\circ}

) sum to zero. A round trip's forward and return displacements annihilate, giving net zero.

Why do we split vectors into components (scalar pieces) at all?

Picture dropping a straight shadow of the arrow onto the horizontal and vertical axes — each shadow is a signed number, and numbers along one axis just add. That turns hard parallelogram addition into easy per-axis sums, the trick expanded in Components of a Vector.

Why does distance being a scalar guarantee it can never decrease as you keep moving?

Distance accumulates path length, which is always a positive amount, so it only ever grows. Displacement can shrink because a directed quantity can reverse.

Edge cases

What is the resultant of two equal vectors

A

θ = 18 0^{\circ}

Zero.

A^{2} + A^{2} + 2 A^{2} cos 18 0^{\circ} = 2 A^{2} - 2 A^{2} = 0

— the minimum case, full cancellation (arrows opposed, diagonal vanishes).

What is

∣ R ∣

when the two vectors are perpendicular (

θ = 9 0^{\circ}

The cross term dies because

cos 9 0^{\circ} = 0

, leaving the clean Pythagorean diagonal

∣ R ∣ = A^{2} + B^{2}

— e.g.

A = 3

B = 4

gives

5

Is a vector of zero magnitude still a vector?

Yes — the zero vector has magnitude

0

and an undefined/arbitrary direction, but it still lives in the vector category and is the additive identity (

A + 0 = A

Can a scalar be negative and still be a scalar?

Yes. Temperature

- 5

°C or potential

- 3

V are scalars whose minus sign means "below the reference", not a spatial direction.

If two vectors are anti-parallel (

θ = 18 0^{\circ}

) with

A = 8

B = 5

, what is

∣ R ∣

∣ R ∣ = 8^{2} + 5^{2} + 2 \cdot 8 \cdot 5 cos 18 0^{\circ} = 64 + 25 - 80 = 9 = 3

, directed along the larger vector

A

. This is the lower bound

∣ A - B ∣

of the resultant range.

What happens to

∣ R ∣ = A^{2} + B^{2} + 2 A B cos θ

if one vector has magnitude zero?

It collapses to

A^{2} = A

: adding the zero vector leaves the other unchanged, as expected.

Does average speed ever equal the magnitude of average velocity?

Yes — only when motion is along a straight line without reversing, so distance equals

∣ displacement ∣

and the two ratios coincide.

Recall One-line summary of every trap

The pattern behind all of these is: direction is a symptom, the addition rule is the diagnosis. Ask "do two of these, placed tail-to-tail, combine by the parallelogram law?" — if yes, vector; if they just add up as counted numbers, scalar.

Connections

Scalars vs vectors — definition, examples — parent note these traps drill.
Vector Addition — Triangle & Parallelogram Law — source of the $A^{2} + B^{2} + 2 A B cos θ$ rule.
Distance vs Displacement — trap set on path-length vs shortcut.
Speed vs Velocity — the round-trip cancellation traps.
Components of a Vector — why scalar components make addition easy.
Dot and Cross Products — why work (a dot product) is a scalar.
Units and Dimensions — you still cannot add unlike quantities.