Vector Review

Vector Gap Review

A fast, targeted review for vector moves that go beyond simply being comfortable with the worksheet.

Transfer

Use worksheet skills in new problem formats.

Prevent

Catch common wrong answers before they happen.

Rehearse

Practice the exact decisions needed for vector problems.

Core skill → Common twist

Core skill	Common twist	Risk
Find component form from a graph	The vector may not start at the origin.	Using endpoint as the vector.
Find magnitude and direction	Reverse the process: build components from magnitude and angle.	Forgetting cosine/sine roles.
Add and scale vectors	Use a fractional scalar such as ½.	Halving only one component.
Sketch resultant vectors	Sketch must match the signs of the components.	Drawing a vector in the wrong quadrant.
Find direction angle	Quadrant IV requires adjustment to a positive standard angle.	Leaving a negative angle or using the reference angle.

Use the same 4-part pattern

Question

What information is given?

Method

Which formula or decision rule applies?

Answer

Compute carefully, then round only if instructed.

Check

Do the signs and graph direction make sense?

The vector problems is not mainly harder algebra. It is about choosing the correct interpretation.

Gap 1 Read movement, not location

Component form from two points

Question

Vector from A(−4, 3) to B(1, −2).

Method

Terminal − initial.

Answer

v = ⟨1−(−4), −2−3⟩
v = ⟨5, −5⟩

Check

Right 5, down 5.

⟨horizontal change, vertical change⟩ = ⟨x₂ − x₁, y₂ − y₁⟩

Gap 1 Visual check

Do not use the endpoint as the vector

Common wrong habit:
“The arrow ends at (1, −2), so the vector is ⟨1, −2⟩.”

Correct habit:
“The vector is the movement from tail to head: right 5 and down 5.”

Gap 1 One-minute drill

Component form from points

Find the component form of the vector from A(−2, 5) to B(4, 1).

v = ⟨x₂ − x₁, y₂ − y₁⟩

v = ⟨4 − (−2), 1 − 5⟩ = ⟨6, −4⟩. Check: right 6, down 4.

Gap 2 Unit vectors

Scale the vector to length 1

Question

Find a unit vector in the direction of v = ⟨a,b⟩.

Method

Divide both components by the magnitude.

Answer

unit = ⟨a/‖v‖, b/‖v‖⟩

Check

The new vector should have magnitude 1.

unit vector = vector ÷ magnitude

Gap 2 Worked example

Unit vector example

Find a unit vector in the direction of w = ⟨8, −6⟩.

Question

w = ⟨8,−6⟩

Method

‖w‖ = √(8² + (−6)²)

Answer

‖w‖ = 10
unit = ⟨8/10,−6/10⟩ = ⟨0.8,−0.6⟩

Check

√(.8² + (−.6)²)=1

Gap 2 Unit-vector model

Targeted unit-vector model

Find a unit vector in the direction of w = ⟨6, −3⟩.

‖w‖ = √(6² + (−3)²)
‖w‖ = √45 = 3√5

unit = ⟨6/(3√5), −3/(3√5)⟩

unit = ⟨2/√5, −1/√5⟩

decimal form:
unit ≈ ⟨0.894, −0.447⟩
to one decimal: ⟨0.9, −0.4⟩

Exact radical form is often safest unless the problem specifically asks for a decimal.

Gap 2 Wrong-answer diagnosis

Unit vector mistakes

Wrong answer	What happened?	Fix
⟨6, −3⟩	Gave the original vector.	Divide by the magnitude.
3√5	Gave only the magnitude.	The answer must be a vector.
⟨2/√5, 1/√5⟩	Lost the negative sign.	Keep direction: right and down.
⟨6/√45, −3⟩	Divided only one component.	Divide both components.

Gap 2 One-minute drill

Unit vector

Find a unit vector in the direction of ⟨3, 4⟩.

Magnitude = 5, so the unit vector is ⟨3/5, 4/5⟩ = ⟨0.6, 0.8⟩.

Gap 3 Magnitude + angle to components

Use cosine for x and sine for y

Question

Given length r and direction θ measured from the positive x-axis.

Method

Break the vector into horizontal and vertical pieces.

Answer

v = ⟨r cos θ, r sin θ⟩

Check

The signs should match the quadrant.

x-component = r cos θ | y-component = r sin θ

Gap 3 Worked example

Components from magnitude and direction

Write the vector in component form if ‖v‖ = 9 and θ = 40°.

Question

r = 9, θ = 40°

Method

v = ⟨r cosθ, r sinθ⟩

Answer

v = ⟨9cos40°, 9sin40°⟩
v ≈ ⟨6.9, 5.8⟩

Check

40° is Quadrant I, so both components are positive.

Gap 3 Magnitude-angle model

Targeted magnitude-angle conversion

Write v in component form if ‖v‖ = 8 and θᵥ = 75°. Round to one decimal place.

v = ⟨8cos75°, 8sin75°⟩

v ≈ ⟨8(0.2588), 8(0.9659)⟩

v ≈ ⟨2.0704, 7.7272⟩

v ≈ ⟨2.1, 7.7⟩

Check: 75° is Quadrant I, so both components should be positive.

Gap 3 Wrong-answer diagnosis

Magnitude/angle mistakes

Wrong answer	What happened?	Fix
⟨8sin75°, 8cos75°⟩	Sine and cosine were switched.	Cosine is x; sine is y.
⟨75cos8°, 75sin8°⟩	Magnitude and angle were switched.	Use length × trig(angle).
⟨2.0704, 7.7272⟩	Not rounded as requested.	Round to one decimal place.

Gap 3 One-minute drill

Magnitude and angle

Write the vector in component form if ‖v‖ = 10 and θ = 30°. Round to one decimal place.

v = ⟨10cos30°, 10sin30°⟩ ≈ ⟨8.7, 5.0⟩.

Gap 4 Direction angle conventions

Report the standard direction angle

Unless told otherwise, report the direction angle as a positive angle measured counterclockwise from the positive x-axis, between 0° and 360°.

reference angle = tan⁻¹(|y|/|x|), then adjust by quadrant

QI

⟨+, +⟩
use reference

QII

⟨−, +⟩
180° − ref

QIII

⟨−, −⟩
180° + ref

QIV

⟨+, −⟩
360° − ref

Gap 4 Calculator mode

Before using sin, cos, or tan⁻¹

Calculator checklist

Use degree mode, not radians.
Use parentheses around negative components.
Round only at the end.
After tan⁻¹, check the quadrant.

If the calculator is in radians, the numbers may look precise but be completely wrong for degree-based vector problems.

Gap 4 Quadrant IV model

Targeted Quadrant IV angle

Find ‖v‖ and θᵥ if v = ⟨12, −9⟩.

‖v‖ = √(12² + (−9)²)
‖v‖ = √225 = 15

reference = tan⁻¹(9/12)
reference ≈ 36.9°

⟨12, −9⟩ is Quadrant IV.

θ = 360° − 36.9°
θ ≈ 323.1°

Gap 4 Wrong-answer diagnosis

Direction angle mistakes

Answer for ⟨12, −9⟩	Problem
36.9°	Ignores that the vector points down.
−36.9°	Correct location, but not the standard positive direction angle.
143.1°	Uses Quadrant II instead of Quadrant IV.
323.1°	Correct standard direction angle.

Gap 4 One-minute drill

Magnitude and direction

Find ‖v‖ and θ if v = ⟨−5, −12⟩.

‖v‖ = 13. Reference angle = tan⁻¹(12/5) ≈ 67.4°. Quadrant III, so θ = 180° + 67.4° = 247.4°.

Gap 5 Fractional scalar multiplication

Multiply every component first

Question

Find ½w + u.

Method

Distribute ½ to both components of w.

Answer

½⟨a,b⟩ = ⟨a/2,b/2⟩

Check

The final vector should match the sketch.

Gap 5 Worked example

Fractional scalar example

If w = ⟨10, 14⟩ and u = ⟨−3, 4⟩, find ½w + u.

½w + u = ½⟨10, 14⟩ + ⟨−3, 4⟩
= ⟨5, 7⟩ + ⟨−3, 4⟩
= ⟨2, 11⟩

Sketch check: ⟨2,11⟩ should point right and steeply upward.

Gap 5 Wrong-answer diagnosis

Fraction mistakes

Wrong move	Problem	Correct move
½⟨10,14⟩ = ⟨5,14⟩	Only x was halved.	½⟨10,14⟩ = ⟨5,7⟩
½w + u = ½(w+u)	Changed the expression.	Scale w first, then add u.
⟨5−3, 7−4⟩	Subtracted the second component incorrectly.	⟨5 + (−3), 7 + 4⟩

Gap 5 One-minute drill

Fractional scalar

Find ½⟨8, −6⟩ + ⟨1, 5⟩.

½⟨8, −6⟩ = ⟨4, −3⟩. Then ⟨4, −3⟩ + ⟨1, 5⟩ = ⟨5, 2⟩.

Graphing the resultant

Once you have component form, draw the resultant from the origin.

⟨a,b⟩ means a right/left and b up/down

The sketch is a check on the signs, not a separate mystery step.

Examples

⟨5,2⟩ right and up

⟨−5,2⟩ left and up

⟨−5,−2⟩ left and down

⟨5,−2⟩ right and down

Visual plan for graphical vector addition

What vector problems is testing visually

Each arrow has its own component form. The location of the arrow on the grid does not change its component form.

Count each arrow as ⟨right/left, up/down⟩.
Write a component vector for each arrow.
Add all x-components and all y-components.
Draw one resultant vector using the final components.

Key visual idea:

You are allowed to “move” vectors without rotating, stretching, or flipping them. This lets you place them tail-to-head and draw the resultant from the first tail to the final head.

Common-tail visual: count, add, draw

Same example as the guided notes

Count each arrow

Arrow	Component form	Meaning
A	⟨−4, 1⟩	4 left, 1 up
B	⟨3, 2⟩	3 right, 2 up
C	⟨2, 4⟩	2 right, 4 up
Resultant	⟨1, 7⟩	1 right, 7 up

r = ⟨−4+3+2, 1+2+4⟩ = ⟨1,7⟩

On the guided notes, students fill in these components from the graph before drawing r.

Scattered-vector visual: location does not matter

Same example as the guided notes

Count each tail-to-head movement

Arrow	Component form	Meaning
A	⟨2, −3⟩	2 right, 3 down
B	⟨−4, −1⟩	4 left, 1 down
C	⟨5, 3⟩	5 right, 3 up
Resultant	⟨3, −1⟩	3 right, 1 down

r = ⟨2−4+5, −3−1+3⟩ = ⟨3,−1⟩

The vector can be moved without changing its component form, as long as it is not rotated, stretched, or flipped.

Graphical addition one-minute drill

Count the arrows

Suppose three arrows have components:

A = ⟨−4,2⟩, B = ⟨7,−1⟩, C = ⟨−2,5⟩

Find the resultant r.

Sketch check

After you calculate r, decide whether it should point:

right/left and up/down

r = ⟨−4+7−2, 2−1+5⟩ = ⟨1,6⟩. Sketch: 1 right and 6 up.

Practice simulation: 5 quick items

Find the component form and magnitude for the vector from A(−3, 4) to B(2, 1).
Add vectors ⟨−2, 5⟩ + ⟨6, −1⟩ and describe the resultant direction.
Find a unit vector in the direction of ⟨−9, 12⟩.
If w = ⟨16, −10⟩ and u = ⟨−5, 4⟩, find ½w + u.
Write v in component form if ‖v‖ = 6 and θ = 210°.

1) ⟨5,−3⟩, ‖v‖=√34 ≈ 5.8. 2) ⟨4,4⟩, right and up. 3) ⟨−3/5,4/5⟩ = ⟨−0.6,0.8⟩. 4) ⟨3,−1⟩. 5) ⟨−5.2,−3.0⟩.

Final mistake-proofing checklist

Before turning it in

Did I subtract terminal − initial?
Did I divide both components for a unit vector?
Did I use cos for x and sin for y?
Did I adjust the angle quadrant?

Sanity checks

Unit vector length should be 1.
Positive x means right.
Negative y means down.
The sketch should match the signs.

Most lost points come from interpretation errors, not difficult computation.

Printable one-page summary

Component form from points

v = ⟨x₂ − x₁, y₂ − y₁⟩

Terminal point minus initial point.

Graphical vector addition

r = ⟨sum of x’s, sum of y’s⟩

Count each arrow’s movement, then add components.

Magnitude

‖v‖ = √(a²+b²)

For v = ⟨a,b⟩.

Unit vector

unit = ⟨a/‖v‖, b/‖v‖⟩

Divide both components by the magnitude.

Magnitude/angle to components

v = ⟨r cosθ, r sinθ⟩

Cosine is x; sine is y.

Direction angle

ref = tan⁻¹(|b|/|a|)

Then adjust by quadrant.

Quadrant adjustment

QI ref QII 180° − ref

QIII 180° + ref QIV 360° − ref

Final checks

Unit vector length = 1. Positive x = right. Negative y = down. Sketch must match signs.