Vector Notes: Magnitude, Direction, and Components

Vector Fundamentals

• A vector is a quantity that has both magnitude and direction. Examples discussed: velocity, force. A scalar has only magnitude (no direction).
• Magnitude: the size or length of the vector. Examples: A long arrow represents a larger magnitude; a short arrow a smaller magnitude.
• Direction: where the vector points.
• To describe a vector in a plane, we use components along a coordinate system (x, y): a vector a at angle θ has components along the x- and y-axes.

Coordinate System, Origin, and Units

• To describe a vector in space, we need:
  • a coordinate system (typically x and y in 2D),
  • an origin, and
  • consistent units (meters, meters per second, etc.).

Decomposing Vectors into Components

• Any vector a with magnitude |a| and direction θ can be written as
  • $a<em>x = |a| \cos\theta, \quad a</em>y = |a| \sin\theta.$
• This uses the idea that cosine gives the adjacent component (along the direction of θ) and sine gives the opposite component.
• Note on signs: the sign of ax and ay depends on the quadrant in which the vector points.
• Quick intuition trick used in class: if the angle is close to the x-axis (adjacent side), use cosine for the x-component; if the angle is close to the y-axis (opposite side), use sine for that component.

Worked Examples: Components

• Example 1: velocity vector v = 200 m/s at θ = 20°
  • $v_x = 200 \cos(20^{\circ}) \approx 200 \cdot 0.9397 \approx 1.88\times 10^{2}\ \mathrm{m/s}$
  • $v_y = 200 \sin(20^{\circ}) \approx 200 \cdot 0.3420 \approx 6.84\times 10^{1}\ \mathrm{m/s}$
  • So, v ≈ (188, 68.4) m/s in (x, y) components.
• Example 2: vector a = 50 m at θ = 30°
  • $a_x = 50 \cos(30^{\circ}) \approx 50 \cdot 0.8660 \approx 43.3\ \mathrm{m}$
  • $a_y = 50 \sin(30^{\circ}) = 50 \cdot 0.5 = 25\ \mathrm{m}$
• Example 3: vector b = 50 m at θ = 45° pointing to the southwest (i.e., negative x and negative y)
  • If the vector lies in the III quadrant (both x and y negative):
  • $b_x = -50 \cos(45^{\circ}) = -50 \cdot 0.7071 \approx -35.4\ \mathrm{m}$
  • $b_y = -50 \sin(45^{\circ}) = -50 \cdot 0.7071 \approx -35.4\ \mathrm{m}$
  • (In the transcript, signs were discussed carefully; the key is identifying the quadrant to assign signs.)
• Important reminder: for a vector defined by magnitude and angle, use the standard formulas with the angle measured from the +x axis and assign signs according to the quadrant.

Adding Vectors via Components

• If two vectors a and b are given by components: a = (ax, ay), b = (bx, by), then the sum c = a + b has components:
  • $c<em>x = a</em>x + b<em>x, \quad c</em>y = a<em>y + b</em>y.$
• The resultant magnitude and direction are:
  • $|c| = \sqrt{c<em>x^2 + c</em>y^2},$
  • $\theta<em>c = \tan^{-1}\left(\frac{c</em>y}{c_x}\right)$ (use proper quadrant, or use atan2).
• Worked walkthrough based on class example:
  • Let a = (ax, ay) with a = 50 at 30°: ax ≈ 43.3, ay = 25.
  • Let b = 50 at 45° in the southwest: bx ≈ -35.4, by ≈ -35.4.
  • Then c = a + b = (43.3 - 35.4, 25 - 35.4) ≈ (7.9, -10.4).
  • Magnitude: $|c| \approx \sqrt{7.9^2 + (-10.4)^2} \approx 13.1\ \mathrm{m}.$
  • Direction: $\theta_c \approx \tan^{-1}\left(-10.4/7.9\right) \approx -52.7^{\circ}.$
  • Interpretation: about 52.7° below the +x axis (in the fourth quadrant).

A Practical Example: Time Across a Ravine (Velocity Components and Time)

• If you know the velocity vector and a horizontal distance across a ravine, the time to cross depends on the velocity component in the x-direction.
  • Distance across along x: $D_x$.
  • Time to cross along x: $t = \frac{D<em>x}{v</em>x}.$
  • Why not simply use total distance divided by total speed? Because the motion is at an angle; only the x-component of velocity contributes to crossing along x, while the y-component changes the vertical position.
  • If the path has a y-component as well, you still use the same time t for both x and y components; time is scalar and is the same for the motion as it evolves.
• Quick intuition: if vx = |v| cos θ and vy = |v| sin θ, then crossing a distance Dx requires t = Dx / (|v| cos θ).

Multiple Vectors: A Step-by-Step Demonstration

• Scenario: Walk 50 m at 30°, then walk 50 m at 45° in a different direction; find the displacement from the start.
  • Decompose each vector:
  • a = 50 m at 30°: ax ≈ 50 cos 30° ≈ 43.3; ay ≈ 50 sin 30° = 25.
  • b = 50 m at 45° in the southwest: bx ≈ -50 sin 45° ≈ -35.4; by ≈ -50 cos 45° ≈ -35.4.
  • Sum: c = a + b ≈ (43.3 - 35.4, 25 - 35.4) ≈ (7.9, -10.4).
  • Magnitude: $|c| = \sqrt{7.9^2 + (-10.4)^2} \approx 13.1\ \mathrm{m}.$
  • Direction: $\theta_c = \tan^{-1}\left(\frac{-10.4}{7.9}\right) \approx -52.7^{\circ}.$
• This demonstrates the core rule: to add vectors, break them into components, add components, then recombine to get magnitude and direction.

A More Complex Two-Vector Addition (Geowalk problem)

• Problem setup (as in the GeoWalk example from the transcript):
  • Vector a: 50 m west along Tenth Street. So a = (-50, 0).
  • Vector b: 60° southeast, 5 m/s for 2 s, so distance |b| = 10 m, direction +60° toward southeast.
  • Decompose b: bx = +10 cos 60° = +5, by = -10 sin 60° = -8.66.
  • Resultant after a and b: c = a + b = (-50 + 5, 0 - 8.66) = (-45, -8.66).
• Then a third movement: d = 20 m at 60° west of south.
  • Components: dx = -20 sin 60° = -10√3 ≈ -17.32, dy = -20 cos 60° = -10.
  • Total displacement g = c + d = (-45 - 17.32, -8.66 - 10) ≈ (-62.32, -18.66).
• Final magnitude and direction:
  • Magnitude: $|g| = \sqrt{(-62.32)^2 + (-18.66)^2} \approx 65.0\ \mathrm{m}.$
  • Direction: since both components are negative, the vector points southwest. The angle south of west is
    $\theta<em>{g} = \tan^{-1}\left(\frac{|g</em>y|}{|g_x|}\right) = \tan^{-1}\left(\frac{18.66}{62.32}\right) \approx 16.7^{\circ}.$
  • So g is about 65 m, 16.7° south of west.
• Takeaway: for problems with multiple vectors, always break into x and y components, sum components, then compute magnitude and angle from the resulting components.

Scalar vs Vector: Class Activity Highlights

• Scalars: quantities with only magnitude (no direction). Examples from the discussion: 25 mph (without direction) is a scalar. 35 pounds (without direction) is a scalar.
• Vectors: quantities with both magnitude and direction. Examples: 10 miles south, 10 m/s at 30° north of east, 2 light years toward the North Star.
• Edge questions discussed:
  • Temperature: a scalar quantity (e.g., 10°C, 20°C). Temperature can have a gradient, which is a vector (direction of changing temperature), but the temperature itself is a scalar.
  • Pronunciation and interpretation of “below zero” can be a scalar quantity with a negative value; it’s not a direction per se, though the value is negative.
• One revolution equals 360°:
  • Useful for angular problems and converting between units of angle.

Real-World Contexts: Motion Tracking and Biology

• Example: tracking motion of biological cells (macrophages) is naturally expressed with position vectors that evolve over time.
  • Researchers use machine learning to study patterns of cell movement, using the vector positions to classify cell types without looking at microscopic images.
  • This illustrates how vector concepts apply beyond physics to biology and data science, where tracking displacement, velocity fields, and trajectories is essential.
• The overarching message: vectors provide a language to model motion, forces, and distributions in real-world systems.

Quick Practice Problems (Mixed Review)

• Problem: Decide whether each statement is a vector or a scalar. (From the in-class activity.)
  • 35 pounds east -> Vector (has magnitude and direction).
  • 25 miles per hour -> Scalar (magnitude only unless a direction is specified).
  • 10 miles south of a point -> Vector.
  • 2 light years away toward a star -> Vector (magnitude + direction).
• Problem: A temperature value (e.g., 10°C) is a scalar. A temperature gradient (temperature change with position) is a vector.
• Problem: If given a velocity vector and a straight-line distance, find the crossing time along the x-direction: t = Dx / vx, where v_x = |v| cos θ.

Practice Problem Walkthrough (Summary of steps)

• Step 1: Identify magnitudes and directions; decide orientation relative to +x axis.
• Step 2: Break each vector into components:
  • For a vector with magnitude m and angle θ from +x axis: $v<em>x = m \cos\theta, \quad v</em>y = m \sin\theta.$
• Step 3: Sum components independently to get resultant components: $R<em>x = \sum v</em>{i,x}, \quad R<em>y = \sum v</em>{i,y}.$
• Step 4: Compute resultant magnitude and direction:
  • Magnitude: $|R| = \sqrt{R<em>x^2 + R</em>y^2},$
  • Direction: $\theta<em>R = \tan^{-1}\left(\frac{R</em>y}{R_x}\right)$ (take into account the quadrant).
• Step 5: Interpret the result in the given context, e.g., how long it takes to cross a distance along a particular axis or what the overall displacement is after several moves.

Tips for Exam Prep

• Always start by drawing a diagram and breaking vectors into components along x and y.

• Keep track of signs carefully based on the quadrant.

• Use the standard convention: angles are measured from the +x axis, counterclockwise is positive.

• When adding several vectors, add all x-components together and all y-components together first, then compute the magnitude and direction of the resultant.

• For problems involving time and motion with angled velocity, use the velocity component in the direction of the motion you’re analyzing (e.g., x-direction for horizontal travel) to compute time or distance.

• If you want more practice, try creating additional problems by combining different magnitudes and angles, then verify with a graphing tool or Desmos to visualize the resultant vector.