🎯⭐ INTERACTIVE LESSON

Vectors in the Plane

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Vectors in the Plane - Complete Interactive Lesson

Part 1: What Is a Vector?

➡️ Vectors in the Plane

Part 1 of 7 — What Is a Vector?

Topics in This Part

Section
Scalars vs. Vectors
Geometric Vectors & Notation
Equal Vectors
Magnitude (Length) of a Vector

🔑 Key Concept: A vector carries two pieces of information at once — a magnitude (how much) and a direction (which way). That single idea models velocity, force, and displacement, which is why vectors show up everywhere in physics and AP Precalculus Unit 4.

Scalars vs. Vectors

A scalar is just a number with units — it has size but no direction.

A vector has both magnitude and direction.

Quantity	Type	Why
A temperature of $72^\circ$	scalar	only a size
A speed of $60$ mph	scalar	size, no direction
A velocity of $60$ mph due north	vector	size and direction
A mass of kg

💡 Speed vs. velocity is the classic distinction. " $60$ mph" is a scalar (speed); " $60$ mph heading east" is a vector (velocity). The direction is what makes it a vector.

Concept Check 🎯

Drawing & Naming Vectors

Geometrically, a vector is a directed line segment — an arrow. It has:

a tail (initial point), where the arrow starts, and
a head (terminal point), where the arrowhead is.

The length of the arrow shows the magnitude; the way it points shows the direction.

Notation

Way to write it	Meaning
$\vec{v}$ or $\mathbf{v}$

Equal Vectors

Two vectors are equal when they have the same magnitude and the same direction — no matter where they are drawn on the page.

A vector has no fixed location. You can slide it anywhere (this is called a translation), and as long as you don't rotate it or stretch it, it stays the same vector.

$\vec{u} = \vec{v} \iff \lVert\vec{u}\rVert = \lVert\vec{v}\rVert \;\text{ and they point the same way}$

Concept Check 🎯

Magnitude (Length)

The magnitude of a vector is the length of its arrow. If you know how far the vector goes horizontally and vertically, the magnitude is just the distance formula — the Pythagorean theorem in disguise.

If a vector moves $a$ units horizontally and $b$ units vertically, then

$\lVert\vec{v}\rVert = \sqrt{a^2 + b^2}$

Find the Magnitude 🧮

Use $\lVert\vec{v}\rVert = \sqrt{a^2 + b^2}$ for a vector that moves right and up.

Part 2: Component Form

➡️ Vectors in the Plane

Part 2 of 7 — Component Form

🔑 The Big Move: Because a vector can slide anywhere, we slide its tail to the origin. Then the vector is described completely by where its head lands: its components.

Component Form

A vector in component form is written with angle brackets:

$\vec{v} = \langle a,\, b \rangle$

Part 3: Adding, Subtracting & Scaling Vectors

➡️ Vectors in the Plane

Part 3 of 7 — Adding, Subtracting & Scaling Vectors

🔑 The Rule: To add, subtract, or scale vectors, just work component by component. The geometry (tip-to-tail arrows) and the algebra (matching components) always agree.

Adding & Subtracting

Add or subtract vectors by combining matching components:

$\langle a_1, b_1\rangle + \langle a_2, b_2\rangle = \langle a_1 + a_2,\; b_1 + b_2\rangle$

Part 4: Unit Vectors & the i, j Form

➡️ Vectors in the Plane

Part 4 of 7 — Unit Vectors & the $\hat{\imath},\,\hat{\jmath}$ Form

🔑 A unit vector has length exactly $1$ . It captures pure direction with no "size baggage," and it lets us write every vector as a clean combination of horizontal and vertical pieces.

Finding a Unit Vector

Part 5: Direction Angle & Magnitude–Direction Form

➡️ Vectors in the Plane

Part 5 of 7 — Direction Angle & Magnitude–Direction Form

🔑 Two languages, one vector. A vector can be described by its components $\langle a, b\rangle$ or by its magnitude and direction angle $(r, \theta)$ . This part is about translating fluently between them.

From Components to Magnitude & Direction

Given $\vec{v} = ⟨ a, b ⟩$ , the is measured counterclockwise from the .

Part 6: Applications: Velocity, Force & Bearings

➡️ Vectors in the Plane

Part 6 of 7 — Applications: Velocity, Force & Bearings

🔑 Why we built all this. Real motion and forces add as vectors, not as plain numbers. A plane in a crosswind, two ropes pulling a crate — you find the resultant by adding components, then convert back to magnitude and direction.

The Resultant

When several vectors act at once, the single vector that has the same total effect is the resultant — just the vector sum.

Worked Example: A Plane in Wind

An airplane heads east at $200$ mph (vector $\langle 200, 0\rangle$ ). A wind blows it north at $40$ mph (vector ). Find the true ground velocity.

Part 7: Mixed Practice & Mastery Check

➡️ Vectors in the Plane

Part 7 of 7 — Mixed Practice & Mastery Check

You can now (1) tell scalars from vectors, (2) find components and magnitude, (3) add, subtract & scale, (4) normalize and use $\hat{\imath},\hat{\jmath}$ form, (5) convert between components and magnitude–direction, and (6) solve velocity/force problems. Let's put it all together.

Quick Reference

Goal	Key formula
Components from points

⚠️ Order matters. $\vec{PQ}$ starts at $P$ and ends at $Q$ . The vector $\vec{QP}$ points the opposite way — same length, reversed direction.

∥v∥ and they point the same way

🔑 Position doesn't matter. Two arrows of equal length pointing in the same direction are equal vectors even if they sit in different parts of the plane. This freedom is what lets us "move" a vector's tail to the origin in Part 2.

A vector that goes $3$ right and $4$ up:

$\lVert\vec{v}\rVert = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

💡 The numbers $3, 4, 5$ form a Pythagorean triple — a right triangle whose legs and hypotenuse are all whole numbers. Memorizing $3\text{-}4\text{-}5$ , $5\text{-}12\text{-}13$ , and $8\text{-}15\text{-}17$ makes magnitudes fast to compute.

1) $6$ right, $8$ up: $\lVert\vec{v}\rVert = \,?$ 2) $5$ right, $12$ up: $\lVert\vec{v}\rVert = \,?$ 3) $9$ right, $12$ up: $\lVert\vec{v}\rVert = \,?$

$a$ is the horizontal component (how far right, if positive),
$b$ is the vertical component (how far up, if positive).

This is the "standard position" of a vector: tail at the origin $(0,0)$ , head at the point $(a, b)$ .

⚠️ Brackets, not parentheses. We write $\langle a, b\rangle$ for a vector and $(a, b)$ for a point. They look similar but mean different things — a point is a location, a vector is a displacement. AP graders care about this distinction.

$\vec{v} = \langle 3, -2 \rangle$ means: starting from the tail, move $3$ right and $2$ down. Its head, placed in standard position, sits at the point $(3, -2)$ .

Read the Components 🔽

Finding Components From Two Points

To find the components of $\vec{PQ}$ (from $P$ to $Q$ ), subtract the tail from the head — that is, terminal minus initial:

$\vec{PQ} = \langle\, x_Q - x_P,\;\; y_Q - y_P \,\rangle$

Worked Example: $P(1, 2)$ and $Q(4, 7)$

$\vec{PQ} = \langle 4 - 1,\; 7 - 2 \rangle = \langle 3,\, 5 \rangle$

So going from $P$ to $Q$ means moving $3$ right and $5$ up. ✓

⚠️ Head minus tail, every time. A common mistake is subtracting in the wrong order. $\vec{PQ}$ uses $Q$ (the head) minus $P$ (the tail). Reversing it gives , the opposite vector.

Worked Example: $A(-3, 5)$ and $B(2, -1)$

$\vec{AB} = \langle 2 - (-3),\; -1 - 5 \rangle = \langle 5,\, -6 \rangle$

And its magnitude:

$\lVert\vec{AB}\rVert = \sqrt{5^2 + (-6)^2} = \sqrt{25 + 36} = \sqrt{61}$

Since $\sqrt{61}$ is not a perfect square, we leave it exact (or approximate: $\sqrt{61} \approx 7.81$ ).

💡 Watch the signs. Subtracting a negative coordinate flips it to addition: $2 - (-3) = 2 + 3 = 5$ . Careful sign work is where most component errors hide.

Components From Points 🧮

Find $\vec{PQ} = \langle x_Q - x_P,\; y_Q - y_P\rangle$ . Enter the horizontal component, then the vertical component.

1) $P(2, 1),\; Q(7, 4)$ : $\vec{PQ} = \langle\,?,\; ?\,\rangle$ :

Putting It Together 🎯

\langle a_1, b_1\rangle - \langle a_2, b_2\rangle = \langle a_1 - a_2,\; b_1 - b_2\rangle

$\langle 3, 5\rangle + \langle 4, -2\rangle = \langle 3+4,\; 5+(-2)\rangle = \langle 7,\, 3\rangle$ $\langle 3, 5\rangle - \langle 4, -2\rangle = \langle 3-4,\; 5-(-2)\rangle = \langle -1,\, 7\rangle$

💡 Tip-to-tail. Geometrically, $\vec{u} + \vec{v}$ means: draw $\vec{u}$ , then start $\vec{v}$ at the head of $\vec{u}$ . The sum (called the resultant) runs from the original tail to the final head. The component rule is the shortcut for that picture.

Scalar Multiplication

Multiplying a vector by a scalar $k$ multiplies each component by $k$ :

$k\langle a, b\rangle = \langle ka,\; kb\rangle$

This scales the length by $|k|$ . The direction:

stays the same if $k > 0$ ,
reverses if $k < 0$ (the arrow flips around).

Examples

$3\langle 2, -1\rangle = \langle 6,\, -3\rangle \quad\text{(3× longer, same direction)}$ $-2\langle 2, -1\rangle = \langle -4,\, 2\rangle \quad\text{(2× longer, reversed direction)}$

🔑 Magnitude scales by $|k|$ : $\;\lVert k\vec{v}\rVert = |k|\,\lVert\vec{v}\rVert$ . So is three times as long, and is twice as long but points the opposite way.

Combine the Vectors 🧮

Let $\vec{u} = \langle 4, -1\rangle$ and $\vec{v} = \langle -2, 3\rangle$ . Enter each result as $\langle\,?,\;?\,\rangle$ (horizontal, then vertical).

1) $\vec{u} + \vec{v} = \langle\,?,\;?\,\rangle$

Concept Check 🎯

The Zero Vector & Opposite Vectors

The zero vector $\vec{0} = \langle 0, 0\rangle$ has magnitude $0$ and no direction. It is the additive identity: $\vec{v} + \vec{0} = \vec{v}$ .
The opposite of $\vec{v}$ is $-\vec{v} = (-1)\vec{v}$ — same length, reversed direction. Adding a vector to its opposite gives zero:

$\vec{v} + (-\vec{v}) = \vec{0}$

In fact, subtraction is just adding the opposite: $\;\vec{u} - \vec{v} = \vec{u} + (-\vec{v})$ .

💡 This is exactly like numbers: $7 - 3 = 7 + (-3)$ . Vectors follow the same rules — they just do it in two components at once.

Zero & Opposite Vectors 🔽

A unit vector in the direction of $\vec{v}$ is found by dividing $\vec{v}$ by its own magnitude:

$\hat{u} = \frac{\vec{v}}{\lVert\vec{v}\rVert} = \left\langle \frac{a}{\lVert\vec{v}\rVert},\; \frac{b}{\lVert\vec{v}\rVert} \right\rangle$

This is called normalizing the vector. The result always has magnitude $1$ .

Worked Example: $\vec{v} = \langle 3, 4\rangle$

First the magnitude: $\lVert\vec{v}\rVert = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ .

$\hat{u} = \left\langle \frac{3}{5},\; \frac{4}{5} \right\rangle$

✅ Check the length: $\sqrt{\left(\tfrac{3}{5}\right)^2 + \left(\tfrac{4}{5}\right)^2} = \sqrt{\tfrac{9}{25} + \tfrac{16}{25}} = \sqrt{\tfrac{25}{25}} = \sqrt{1} = 1$ ✓

Normalize It 🧮

Find the unit vector in the direction of each vector. Enter each component as a fraction (e.g. 5/13 or -12/13).

1) $\vec{v} = \langle 6, 8\rangle$ (first $\lVert\vec{v}\rVert$ , then divide): $\hat{u} = \langle\,?,\;?\,\rangle$ 2) $\vec{v} = \langle -5, 12\rangle$ : $\hat{u} = \langle\,?,\;?\,\rangle$

The $\hat{\imath},\,\hat{\jmath}$ (Standard Unit Vector) Form

Two special unit vectors point along the axes:

$\hat{\imath} = \langle 1, 0\rangle \quad\text{(one unit right)}, \qquad \hat{\jmath} = \langle 0, 1\rangle \quad\text{(one unit up)}$

Every vector is a combination of these:

$\langle a, b\rangle = a\,\hat{\imath} + b\,\hat{\jmath}$

Examples

Component form	$\hat{\imath},\,\hat{\jmath}$ form
$\langle 3, 5\rangle$

💡 The two forms are completely interchangeable. $\langle a, b\rangle$ and $a\hat{\imath} + b\hat{\jmath}$ are the same vector written two ways. AP problems use both, so get comfortable switching.

Switch Between Forms 🔽

Building a Vector of a Given Length

A unit vector is a "direction tool." To get a vector of length $L$ pointing the same way as $\vec{v}$ , scale its unit vector by $L$ :

$L\,\hat{u} = L\cdot\frac{\vec{v}}{\lVert\vec{v}\rVert}$

Example

Find a vector of length $15$ in the direction of $\langle 3, 4\rangle$ .

The unit vector is $\left\langle \tfrac{3}{5}, \tfrac{4}{5}\right\rangle$ , so multiply by :

$15\left\langle \tfrac{3}{5}, \tfrac{4}{5}\right\rangle = \left\langle \tfrac{45}{5}, \tfrac{60}{5}\right\rangle = \langle 9, 12\rangle$

✅ Check: $\lVert\langle 9, 12\rangle\rVert = \sqrt{81 + 144} = \sqrt{225} = 15$ ✓ — exactly the length we wanted, same direction.

Concept Check 🎯

positive $x$ -axis

$\lVert\vec{v}\rVert = \sqrt{a^2 + b^2}, \qquad \tan\theta = \frac{b}{a}$

⚠️ The calculator's $\arctan$ only returns angles in $(-90^\circ, 90^\circ)$ . You must check which quadrant the vector is in and adjust. If $a < 0$ , add $180^\circ$ to the arctangent result.

Worked Example: $\vec{v} = \langle 1, \sqrt{3}\rangle$

$\lVert\vec{v}\rVert = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$ $\tan\theta = \frac{\sqrt{3}}{1} = \sqrt{3} \;\Rightarrow\; \theta = 60^\circ \quad(\text{Quadrant I, no adjustment})$

So $\vec{v}$ has magnitude $2$ at a direction angle of $60^\circ$ .

The Quadrant Adjustment

Because $\arctan(b/a)$ can't tell which way the vector actually points, use this guide:

Quadrant of $\langle a, b\rangle$	Sign of $a$ , $b$	Adjustment to $\arctan\!\left(\tfrac{b}{a}\right)$
I	$+,\ +$	none (answer is $0^\circ$ – $90^\circ$ )
II	$-$

Worked Example: $\vec{v} = \langle -3, 3\rangle$ (Quadrant II)

$\arctan\!\left(\frac{3}{-3}\right) = \arctan(-1) = -45^\circ$

Since $a = -3 < 0$ , add $180^\circ$ : $\;\theta = -45^\circ + 180^\circ = 135^\circ$ .

✅ Sanity check: $\langle -3, 3\rangle$ points up and to the left — that should be in Quadrant II, between $90^\circ$ and $180^\circ$ . fits. ✓

Find the Direction Angle 🧮

Find $\theta$ (in degrees, $0^\circ \le \theta < 360^\circ$ ) for each vector. These all land on familiar reference angles.

1) $\langle 0, 5\rangle$ (straight up): $\theta = \,?$ 2) $\langle -2, 0\rangle$ (straight left): $\theta = \,?$ 3) $\langle -4, 4\rangle$ (Quadrant II): $\theta = \,?$

From Magnitude & Direction to Components

Going the other way, if a vector has magnitude $r$ and direction angle $\theta$ , its components come from right-triangle trigonometry:

$\vec{v} = \langle\, r\cos\theta,\; r\sin\theta \,\rangle$

This is the workhorse formula for word problems — you are given a speed/force and a heading, and you need components.

Worked Example: magnitude $10$ , direction $30^\circ$

$a = 10\cos 30^\circ = 10\cdot\frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$

So $\vec{v} = \langle 5\sqrt{3},\, 5\rangle \approx \langle 8.66,\, 5\rangle$ .

💡 Memory aid: $\cos$ goes with the horizontal (cosine ↔ $x$ ), $\sin$ with the vertical (sine ↔ $y$ ). Mixing them up is the #1 error in vector word problems.

Concept Check 🎯

Components From Magnitude & Direction 🧮

Use $\vec{v} = \langle r\cos\theta,\; r\sin\theta\rangle$ . Enter exact decimals where the angle is "nice."

1) $r = 4,\;\theta = 0^\circ$ : $\langle\,?,\;?\,\rangle$ 2) :

\langle 0, 40\rangle

Add the components:

$\vec{r} = \langle 200, 0\rangle + \langle 0, 40\rangle = \langle 200,\, 40\rangle$

Magnitude (ground speed):

$\lVert\vec{r}\rVert = \sqrt{200^2 + 40^2} = \sqrt{40000 + 1600} = \sqrt{41600} \approx 203.96 \text{ mph}$

Direction (north of east):

$\theta = \arctan\!\left(\frac{40}{200}\right) = \arctan(0.2) \approx 11.3^\circ \text{ north of east}$

💡 The plane's ground speed ( $\approx 204$ mph) is more than its airspeed, and it drifts about $11.3^\circ$ off course — exactly why pilots correct for wind.

Find the Resultant 🧮

A boat motors east at $9$ mph; a current pushes it south at $12$ mph.

1) Resultant components $\langle\,?,\;?\,\rangle$ (east is $+x$ , north is $+y$ ) 2) Resultant speed (magnitude) $= \,?$ mph

Forces & Components of a Single Vector

Often you are given a vector by its magnitude and direction and must break it into components first.

Worked Example: Pulling a Wagon

You pull a wagon with a force of $50$ N at an angle of $30^\circ$ above the horizontal. How much force acts horizontally vs. vertically?

$F_x = 50\cos 30^\circ = 50\cdot\frac{\sqrt{3}}{2} = 25\sqrt{3} \approx 43.3 \text{ N (horizontal)}$ $F_y = 50\sin 30^\circ = 50\cdot\frac{1}{2} = 25 \text{ N (vertical)}$

💡 Most of your $50$ N pull ( $\approx 43.3$ N) moves the wagon forward; only $25$ N lifts it. Resolving a force into components tells you which part of your effort actually does the job.

A Note on Bearings

Navigation often uses bearings — angles measured clockwise from due north, not counterclockwise from east.

Compass direction	Bearing	Standard direction angle $\theta$
North	$000^\circ$	$90^\circ$
East	$090^\circ$	$0^\circ$
South	$180^\circ$	$270^\circ$
West	$270^\circ$	$180^\circ$

⚠️ Don't mix conventions. A bearing of $90^\circ$ means east, but a standard direction angle of $90^\circ$ means north. Read each problem carefully and convert before plugging into $\langle r\cos\theta, r\sin\theta\rangle$ .

Set Up the Components 🔽

A car drives at $60$ mph in various directions. Choose the correct component setup (east $= +x$ , north $= +y$ ).

Application Check 🎯

⚠️ Top traps: subtract head $-$ tail (not tail $-$ head); add $180^\circ$ when $a<0$ ; magnitude is never negative; $\cos$ ↔ horizontal, $\sin$ ↔ vertical.

Mixed Practice 🔽

Let $\vec{u} = \langle 3, -4\rangle$ and $\vec{v} = \langle -1, 2\rangle$ .

Mixed Practice 🧮

1) Magnitude of $\vec{PQ}$ for $P(-2, 1),\; Q(1, 5)$ : $\lVert\vec{PQ}\rVert = \,?$ 2) Components of a vector with magnitude $6$ at direction angle $90^\circ$ : $\langle\,?,\;?\,\rangle$

Exit Quiz ✅

Answer all three to finish the lesson.

P(-1, 3),\; Q(4, -2)

\vec{PQ} = \langle\,?,\; ?\,\rangle

⟨−4,2⟩(2× longer, reversed direction)

\vec{u} - \vec{v} = \langle\,?,\;?\,\rangle

3\vec{u} = \langle\,?,\;?\,\rangle

3\hat{\imath} + 5\hat{\jmath}

60∘(Quadrant I, no adjustment)

b = 10\sin 30^\circ = 10\cdot\frac{1}{2} = 5

r = 8,\;\theta = 180^\circ

r = 8, θ = 18 0^{\circ}

\langle\,?,\;?\,\rangle

Vectors in the Plane

Vectors in the Plane - Complete Interactive Lesson

Part 1: What Is a Vector?

➡️ Vectors in the Plane

Topics in This Part

Scalars vs. Vectors

Drawing & Naming Vectors

Notation

Equal Vectors

Magnitude (Length)

Part 2: Component Form

➡️ Vectors in the Plane

Component Form

Part 3: Adding, Subtracting & Scaling Vectors

➡️ Vectors in the Plane

Adding & Subtracting

Part 4: Unit Vectors & the i, j Form

➡️ Vectors in the Plane

Finding a Unit Vector

Part 5: Direction Angle & Magnitude–Direction Form

➡️ Vectors in the Plane

From Components to Magnitude & Direction

Part 6: Applications: Velocity, Force & Bearings

➡️ Vectors in the Plane

The Resultant

Worked Example: A Plane in Wind

Part 7: Mixed Practice & Mastery Check

➡️ Vectors in the Plane

Quick Reference

Example

Example

Finding Components From Two Points

Worked Example: P(1,2)P(1, 2)P(1,2) and Q(4,7)Q(4, 7)Q(4,7)

Worked Example: A(−3,5)A(-3, 5)A(−3,5) and B(2,−1)B(2, -1)B(2,−1)

Examples

Scalar Multiplication

Examples

The Zero Vector & Opposite Vectors

Worked Example: v⃗=⟨3,4⟩\vec{v} = \langle 3, 4\ranglev=⟨3,4⟩

The ı^, ȷ^\hat{\imath},\,\hat{\jmath}^,^​ (Standard Unit Vector) Form

Examples

Building a Vector of a Given Length

Example

Worked Example: v⃗=⟨1,3⟩\vec{v} = \langle 1, \sqrt{3}\ranglev=⟨1,3​⟩

The Quadrant Adjustment

Worked Example: v⃗=⟨−3,3⟩\vec{v} = \langle -3, 3\ranglev=⟨−3,3⟩ (Quadrant II)

From Magnitude & Direction to Components

Worked Example: magnitude 101010, direction 30∘30^\circ30∘

Forces & Components of a Single Vector

Worked Example: Pulling a Wagon

A Note on Bearings

Worked Example: $P(1, 2)$ and $Q(4, 7)$

Worked Example: $A(-3, 5)$ and $B(2, -1)$

Worked Example: $\vec{v} = \langle 3, 4\rangle$

The $\hat{\imath},\,\hat{\jmath}$ (Standard Unit Vector) Form

Worked Example: $\vec{v} = \langle 1, \sqrt{3}\rangle$

Worked Example: $\vec{v} = \langle -3, 3\rangle$ (Quadrant II)

Worked Example: magnitude $10$ , direction $30^\circ$