Resolution of vectors — into components (any axes): Question bank

This page assumes only the ideas from Resolution of vectors — into components (any axes): a component is a shadow of a vector along a chosen direction, $A_{x} = A cos θ$ and $A_{y} = A sin θ$ when $θ$ is measured from the x-axis, and the projection onto a unit direction $\overset{n}{^}$ is the dot product $A \cdot \overset{n}{^} = A cos ϕ$ .

Figure — Resolution of vectors — into components (any axes)

Look at the figure: the shadow of the coral vector $A$ onto the lavender ruler $\overset{n}{^}$ is exactly the length $A cos ϕ$ , where $ϕ$ is the angle opening between them.

True or false — justify

Resolving a vector changes the vector itself

False. Resolution only re-describes the same arrow as two (or more) pieces along chosen axes; the physical quantity — its length and direction — is untouched. You swapped the description, not the thing.

A vector always has exactly two components

False. The number of components equals the number of axes you choose. In a plane we usually pick two, but the same vector has different component values on every different pair of axes — and none is "the real one".

A_{x} = A cos θ

is a universal law you can always quote

False. It holds only when

θ

is measured from the axis you are projecting onto. Measure the angle from the other axis and cosine and sine swap. The formula encodes a picture, not a magic string.

The two rectangular components of a vector are always both positive

False. Sign is the direction along the axis. A vector pointing into the second quadrant has

A_{x} < 0, A_{y} > 0

. Dropping signs and keeping only

A cos θ

A sin θ

as positive numbers is a classic error.

The magnitude of a vector is

A_{x} + A_{y}

False. Components lie along perpendicular directions, so they combine by Pythagoras:

A = A_{x}^{2} + A_{y}^{2}

, never by simple addition. Adding them straight would over-count.

If you resolve onto oblique (non-90°) axes, magnitude is still

a^{2} + b^{2}

False. Pythagoras needs a right angle between the axes. For oblique axes with components

a, b

and angle

ψ

between

\overset{u}{^}

and

\overset{v}{^}

, the law of cosines gives

∣ A ∣^{2} = a^{2} + b^{2} - 2 ab cos (18 0^{\circ} - ψ) = a^{2} + b^{2} + 2 ab cos ψ

; only when

ψ = 9 0^{\circ}

does this collapse to Pythagoras.

The component of

A

along

\overset{n}{^}

can be larger than

∣ A ∣

False. Since

A_{\overset{n}{^}} = A cos ϕ

and

∣ cos ϕ ∣ \leq 1

, a shadow can never be longer than the object casting it. Equality (

= A

) happens only when

\overset{n}{^}

points exactly along

A

(i.e.

ϕ = 0

A component can be zero even though the vector is not

True. If

A

is perpendicular to

\overset{n}{^}

then

ϕ = 9 0^{\circ}

, so

A_{\overset{n}{^}} = A cos 9 0^{\circ} = 0

. The vector casts no shadow on a direction at right angles to it.

A component can be negative even though a length is never negative

True. A component is a signed number: it records both how much and which way along the axis.

A cos ϕ < 0

when

ϕ > 9 0^{\circ}

, meaning the shadow falls on the negative side of the axis.

For perpendicular axes, the projection equals the parallelogram component

True. Orthogonality is exactly the condition under which the "shadow" and the "piece that adds back" coincide — there is no leak from one axis into the other.

Spot the error

"On an incline of angle

θ

, the weight along the slope is

W cos θ

Error: it is

W sin θ

. As the figure shows, the angle between the vertical weight

W

and the normal

N

(perpendicular to the plane) is exactly

θ

, so the along-slope part is the opposite side of that triangle → sine. Swapping sin/cos is the most common incline mistake.

"To get the component of

A

along oblique axis

\overset{u}{^}

, just compute

A \cdot \overset{u}{^}

Error: for non-perpendicular axes the projection

A \cdot \overset{u}{^}

picks up a "leak" from the

\overset{v}{^}

-part. The true parallelogram component comes from solving

A = a \overset{u}{^} + b \overset{v}{^}

, not from a single dot product.

θ = tan^{- 1} (A_{y} / A_{x})

gives the direction of any vector directly."

Error:

tan

repeats every

18 0^{\circ}

, so a bare arctan can't tell quadrant II from IV. You must inspect the signs of

A_{x}, A_{y}

and add

18 0^{\circ}

when

A_{x} < 0

, else the calculator returns the arrow's opposite.

"A block held on a smooth incline needs a force

m g

up the slope to stay put."

Error: only the along-slope pull of gravity,

m g sin θ

, must be balanced. The

m g cos θ

part is cancelled by the surface's push. Supplying

m g

up-slope would over-push the block.

"Since

cos 9 0^{\circ} = 0

, a vertical vector has no vertical component."

Error: confusion of reference. A vertical vector makes

0^{\circ}

with the vertical axis, so its vertical component is

A cos 0^{\circ} = A

(full length). The

cos 9 0^{\circ} = 0

applies to its horizontal component.

"Resolving

A

and then adding the components back gives a slightly different vector due to rounding of cos and sin."

Error in principle: exactly,

A_{x} \overset{x}{^} + A_{y} \overset{y}{^} = A cos θ \overset{x}{^} + A sin θ \overset{y}{^}

reconstructs

A

perfectly. Any difference is arithmetic rounding by you, not a flaw in resolution.

Why questions

Why does the adjacent side to the angle use cosine, not sine?

Because

cos θ = adjacent / hypotenuse

by definition on the right triangle. Cosine literally measures the shadow of the hypotenuse onto the direction touching the angle. Mnemonic: "COS hugs the Angle."

Why is the dot product the right tool for a projection, rather than just multiplying magnitudes?

Because a projection asks "how much of

A

points along

\overset{n}{^}

", and

A \cdot \overset{n}{^} = A cos ϕ

builds in the angle

ϕ

between them. Plain

A \times 1

ignores that they may not be aligned.

Why do we tilt the axes on an inclined plane instead of using horizontal–vertical?

Because the motion is along the slope, so tilted axes make one component the pure driving force (

m g sin θ

) and the other a fully balanced force (

m g cos θ

, cancelled by the normal force

N

— the surface's perpendicular push). Horizontal–vertical would smear both effects across both axes.

Why do numbers along a fixed axis add like ordinary numbers, but whole vectors don't?

Once a direction is fixed, "how much along it" is a single signed scalar, and scalars add straight. Whole vectors carry direction too, so they need the parallelogram/triangle rule instead.

Why must you know the angle's reference direction before writing cos or sin?

Because cosine goes with the side touching the angle. If you don't know which axis the angle is measured from, you don't know which axis is "adjacent", and cos/sin can end up on the wrong components.

Why is the projection

A cos ϕ

and not

A tan ϕ

A sin ϕ

Cosine gives the along- $\overset{n}{^}$ shadow (adjacent side of the projection triangle). Sine would give the perpendicular leftover, and tangent isn't a shadow length at all — it compares two sides, not one to the hypotenuse.

Edge cases

What is the component of a vector along a direction perpendicular to it?

Exactly zero:

ϕ = 9 0^{\circ}

gives

A cos 9 0^{\circ} = 0

. Physically the vector throws no shadow onto a screen it is parallel to.

What are the components of the zero vector on any axes?

Both zero, on every choice of axes. The zero vector has no length and no direction, so it casts a zero shadow everywhere — a rare case where the answer is axis-independent.

What happens to

A_{x} = A cos θ

and

A_{y} = A sin θ

θ \to 9 0^{\circ}

A_{x} \to 0

and

A_{y} \to A

: the vector becomes purely vertical, all of it in the y-component. This limiting behaviour is a quick sanity check on any resolution.

For a vector in the third quadrant (

A_{x} < 0, A_{y} < 0

), what does the naive

tan^{- 1} (A_{y} / A_{x})

return?

The ratio of two negatives is positive, so arctan returns a first-quadrant angle — the exact opposite direction. You must add

18 0^{\circ}

because tangent cannot distinguish a direction from its reverse.

\overset{n}{^}

points exactly opposite to

A

, what is

A_{\overset{n}{^}}

ϕ = 18 0^{\circ}

, so

A_{\overset{n}{^}} = A cos 18 0^{\circ} = - A

: the full magnitude but negative, signalling the shadow lands entirely on the negative side of

\overset{n}{^}

On a horizontal surface (

θ = 0

), what are the along-slope and normal components of weight?

Along-slope

= W sin 0^{\circ} = 0

and normal

= W cos 0^{\circ} = W

. No slope means no driving component, and the surface supports the entire weight — matching flat-ground intuition.

On a vertical wall (

θ = 9 0^{\circ}

), what are those components of weight?

Along-slope

= W sin 9 0^{\circ} = W

and normal

= W cos 9 0^{\circ} = 0

. Gravity acts entirely down the "slope" and the wall provides no normal support — the object is in free vertical fall.

Connections

Dot product & scalar projection
Unit vectors and Cartesian coordinates
Inclined plane dynamics
Vectors — addition (parallelogram & triangle law)
Projectile motion
Equilibrium of concurrent forces