Vectors in the Plane

Understanding vectors, vector operations, and magnitude and direction

Vectors in the Plane

What is a Vector?

A vector is a quantity that has both magnitude (size) and direction.

Examples:

  • Displacement: "5 miles east"
  • Velocity: "30 m/s at 45°"
  • Force: "10 N downward"

Notation

Vectors can be written as:

  • Bold: v\mathbf{v}
  • Arrow: v⃗\vec{v}
  • Component form: ⟨a,b⟩\langle a, b \rangle or ⟨a,b⟩\langle a, b \rangle
  • Unit vector form: ai+bja\mathbf{i} + b\mathbf{j}

Component Form

A vector from the origin to point (a,b)(a, b) is: v=⟨a,b⟩\mathbf{v} = \langle a, b \rangle

  • aa is the horizontal component (xx-component)
  • bb is the vertical component (yy-component)

Vector Between Two Points

The vector from point P1(x1,y1)P_1(x_1, y_1) to P2(x2,y2)P_2(x_2, y_2) is: P1P2⃗=⟨x2−x1,y2−y1⟩\vec{P_1P_2} = \langle x_2 - x_1, y_2 - y_1 \rangle

Magnitude of a Vector

The magnitude (or length) of vector v=⟨a,b⟩\mathbf{v} = \langle a, b \rangle is:

∣v∣=a2+b2|\mathbf{v}| = \sqrt{a^2 + b^2}

This is the distance formula!

Direction of a Vector

The direction angle θ\theta is measured counterclockwise from the positive xx-axis:

tan⁡(θ)=ba\tan(\theta) = \frac{b}{a}

θ=tan⁡−1(ba)\theta = \tan^{-1}\left(\frac{b}{a}\right)

⚠️ Be careful with quadrants when finding θ\theta!

Vector Operations

Scalar Multiplication

If v=⟨a,b⟩\mathbf{v} = \langle a, b \rangle and kk is a scalar: kv=⟨ka,kb⟩k\mathbf{v} = \langle ka, kb \rangle

  • If k>0k > 0: same direction, scaled by kk
  • If k<0k < 0: opposite direction, scaled by ∣k∣|k|

Vector Addition

If u=⟨a1,b1⟩\mathbf{u} = \langle a_1, b_1 \rangle and v=⟨a2,b2⟩\mathbf{v} = \langle a_2, b_2 \rangle: u+v=⟨a1+a2,b1+b2⟩\mathbf{u} + \mathbf{v} = \langle a_1 + a_2, b_1 + b_2 \rangle

Geometric interpretation: Tip-to-tail method or parallelogram rule

Vector Subtraction

u−v=⟨a1−a2,b1−b2⟩\mathbf{u} - \mathbf{v} = \langle a_1 - a_2, b_1 - b_2 \rangle

Unit Vectors

A unit vector has magnitude 1.

The standard unit vectors are:

  • i=⟨1,0⟩\mathbf{i} = \langle 1, 0 \rangle (horizontal)
  • j=⟨0,1⟩\mathbf{j} = \langle 0, 1 \rangle (vertical)

Any vector can be written as: v=ai+bj\mathbf{v} = a\mathbf{i} + b\mathbf{j}

Finding a Unit Vector

To find a unit vector in the direction of v\mathbf{v}: u=v∣v∣=1∣v∣v\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{|\mathbf{v}|}\mathbf{v}

Dot Product

The dot product of u=⟨a1,b1⟩\mathbf{u} = \langle a_1, b_1 \rangle and v=⟨a2,b2⟩\mathbf{v} = \langle a_2, b_2 \rangle is:

u⋅v=a1a2+b1b2\mathbf{u} \cdot \mathbf{v} = a_1a_2 + b_1b_2

Properties

  • u⋅v=∣u∣∣v∣cos⁡(θ)\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos(\theta) where θ\theta is the angle between them
  • If u⋅v=0\mathbf{u} \cdot \mathbf{v} = 0, the vectors are perpendicular
  • The dot product is a scalar, not a vector!

📚 Practice Problems

1Problem 1easy

❓ Question:

Find the magnitude and direction angle of vector v=⟨3,4⟩\mathbf{v} = \langle 3, 4 \rangle.

💡 Show Solution

Solution:

Find the magnitude:

∣v∣=a2+b2=32+42|\mathbf{v}| = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2}

=9+16=25=5= \sqrt{9 + 16} = \sqrt{25} = 5

Find the direction angle:

tan⁡(θ)=ba=43\tan(\theta) = \frac{b}{a} = \frac{4}{3}

θ=tan⁡−1(43)\theta = \tan^{-1}\left(\frac{4}{3}\right)

θ≈53.13°\theta \approx 53.13°

Since both components are positive, the vector is in Quadrant I, so this angle is correct.

Answers:

  • Magnitude: 55
  • Direction: θ≈53.13°\theta \approx 53.13° or 0.9270.927 radians

2Problem 2easy

❓ Question:

Given vectors u⃗=⟨3,−4⟩\vec{u} = \langle 3, -4 \rangle and v⃗=⟨−2,5⟩\vec{v} = \langle -2, 5 \rangle:

a) Find u⃗+v⃗\vec{u} + \vec{v} b) Find 2u⃗−3v⃗2\vec{u} - 3\vec{v} c) Find the magnitude of u⃗\vec{u}

💡 Show Solution

Solution:

Part (a): Add components:

u⃗+v⃗=⟨3,−4⟩+⟨−2,5⟩=⟨3+(−2),−4+5⟩=⟨1,1⟩\vec{u} + \vec{v} = \langle 3, -4 \rangle + \langle -2, 5 \rangle = \langle 3 + (-2), -4 + 5 \rangle = \langle 1, 1 \rangle

Part (b): Scalar multiplication and subtraction:

2u⃗=2⟨3,−4⟩=⟨6,−8⟩2\vec{u} = 2\langle 3, -4 \rangle = \langle 6, -8 \rangle

3v⃗=3⟨−2,5⟩=⟨−6,15⟩3\vec{v} = 3\langle -2, 5 \rangle = \langle -6, 15 \rangle

2u⃗−3v⃗=⟨6,−8⟩−⟨−6,15⟩=⟨6−(−6),−8−15⟩=⟨12,−23⟩2\vec{u} - 3\vec{v} = \langle 6, -8 \rangle - \langle -6, 15 \rangle = \langle 6-(-6), -8-15 \rangle = \langle 12, -23 \rangle

Part (c): Magnitude formula: ∣u⃗∣=u12+u22|\vec{u}| = \sqrt{u_1^2 + u_2^2}

∣u⃗∣=32+(−4)2=9+16=25=5|\vec{u}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

3Problem 3easy

❓ Question:

Given u=⟨2,−3⟩\mathbf{u} = \langle 2, -3 \rangle and v=⟨−1,4⟩\mathbf{v} = \langle -1, 4 \rangle, find: (a) u+v\mathbf{u} + \mathbf{v}, (b) 3u−2v3\mathbf{u} - 2\mathbf{v}

💡 Show Solution

Solution:

Part a) u+v\mathbf{u} + \mathbf{v}

Add corresponding components: u+v=⟨2,−3⟩+⟨−1,4⟩\mathbf{u} + \mathbf{v} = \langle 2, -3 \rangle + \langle -1, 4 \rangle

=⟨2+(−1),−3+4⟩= \langle 2 + (-1), -3 + 4 \rangle

=⟨1,1⟩= \langle 1, 1 \rangle

Part b) 3u−2v3\mathbf{u} - 2\mathbf{v}

Step 1: Scalar multiplication. 3u=3⟨2,−3⟩=⟨6,−9⟩3\mathbf{u} = 3\langle 2, -3 \rangle = \langle 6, -9 \rangle

2v=2⟨−1,4⟩=⟨−2,8⟩2\mathbf{v} = 2\langle -1, 4 \rangle = \langle -2, 8 \rangle

Step 2: Subtract. 3u−2v=⟨6,−9⟩−⟨−2,8⟩3\mathbf{u} - 2\mathbf{v} = \langle 6, -9 \rangle - \langle -2, 8 \rangle

=⟨6−(−2),−9−8⟩= \langle 6 - (-2), -9 - 8 \rangle

=⟨8,−17⟩= \langle 8, -17 \rangle

Answers:

  • a) ⟨1,1⟩\langle 1, 1 \rangle
  • b) ⟨8,−17⟩\langle 8, -17 \rangle

4Problem 4medium

❓ Question:

A vector w⃗\vec{w} has magnitude 10 and makes an angle of 120°120° with the positive xx-axis.

a) Write w⃗\vec{w} in component form. b) Find a unit vector in the direction of w⃗\vec{w}.

💡 Show Solution

Solution:

Part (a): For a vector with magnitude rr and angle θ\theta:

w⃗=⟨rcos⁡θ,rsin⁡θ⟩\vec{w} = \langle r\cos\theta, r\sin\theta \rangle

w⃗=⟨10cos⁡120°,10sin⁡120°⟩\vec{w} = \langle 10\cos 120°, 10\sin 120° \rangle

cos⁡120°=−12\cos 120° = -\frac{1}{2} and sin⁡120°=32\sin 120° = \frac{\sqrt{3}}{2}

w⃗=⟨10(−12),10(32)⟩=⟨−5,53⟩\vec{w} = \left\langle 10\left(-\frac{1}{2}\right), 10\left(\frac{\sqrt{3}}{2}\right) \right\rangle = \langle -5, 5\sqrt{3} \rangle

Part (b): A unit vector has magnitude 1. To find the unit vector in the direction of w⃗\vec{w}:

w^=w⃗∣w⃗∣=⟨−5,53⟩10=⟨−12,32⟩\hat{w} = \frac{\vec{w}}{|\vec{w}|} = \frac{\langle -5, 5\sqrt{3} \rangle}{10} = \left\langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle

Verify: ∣w^∣=(−12)2+(32)2=14+34=1=1\left|\hat{w}\right| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 ✓

5Problem 5easy

❓ Question:

Find the dot product of u=⟨5,2⟩\mathbf{u} = \langle 5, 2 \rangle and v=⟨−3,4⟩\mathbf{v} = \langle -3, 4 \rangle. Are the vectors perpendicular?

💡 Show Solution

Solution:

Find the dot product:

u⋅v=a1a2+b1b2\mathbf{u} \cdot \mathbf{v} = a_1a_2 + b_1b_2

=(5)(−3)+(2)(4)= (5)(-3) + (2)(4)

=−15+8= -15 + 8

=−7= -7

Are they perpendicular?

Two vectors are perpendicular if and only if their dot product equals zero.

Since u⋅v=−7≠0\mathbf{u} \cdot \mathbf{v} = -7 \neq 0, the vectors are not perpendicular.

Answers:

  • Dot product: −7-7
  • The vectors are not perpendicular
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