Represent vector quantities by directed line segments and find their components from initial and terminal points

Learning goal: Recognize vector quantities as having both magnitude and direction, represent them by directed line segments, use appropriate symbols for vectors and for their magnitudes, and find the components of a vector by subtracting the coordinates of its initial point from the coordinates of its terminal point.

Scope note. N.VM.1 and N.VM.2 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

Scalars and vectors. Some quantities are completely described by one number with a unit: a temperature of 21 °C, a mass of 3 kg, a speed of 60 km/h. These are scalar quantities. Others are not described until you also say which way: a wind of 30 km/h from the northwest, a push of 50 newtons straight down, a displacement of 4 km toward the east. A vector quantity has both a magnitude (how much) and a direction (which way). Velocity, force, displacement and acceleration are vector quantities; speed, distance, mass and time are scalars. The distinction is not decoration. Two cars both traveling at 60 km/h have the same speed, but if one heads north and the other south their velocities are different vectors, and everything that depends on velocity, such as where they are a minute later, differs too.

Directed line segments. We represent a vector quantity by a directed line segment: an arrow. The length of the arrow, drawn to some chosen scale, shows the magnitude, and the way the arrow points shows the direction. The arrow starts at an initial point (its tail) and ends at a terminal point (its tip or head). Any arrow with the same length and the same direction represents the same vector, wherever it is drawn on the page. That is the key convention: a vector is not tied to a location. The arrow from (1, 2) to (5, 5) and the arrow from (−2, 0) to (2, 3) are the same vector, because each one goes 4 units right and 3 units up.

Symbols. In print a vector is usually a bold lowercase letter, v. In handwriting, where bold is hard to see, we write the letter with a small arrow drawn above it. These lesson pages are set as plain text and do not typeset an arrow over a letter, so in this lesson the same symbol is typed with the arrow immediately after the letter, v→, and you should draw it above the letter when you write by hand. A vector named by its endpoints is written with the arrow above both letters, typed here as PQ→ for the vector from P to Q, and the order matters: PQ→ and QP→ point opposite ways. The magnitude of v is written |v| or ‖v‖; both mean the same non-negative number, the length of the arrow. Write the symbol for the vector and the symbol for its magnitude differently, because they are different kinds of object: v has a direction, |v| is a plain number. "v = 5" is not a vector statement; "|v| = 5" is.

Components. On a coordinate grid a vector is described by how far it moves horizontally and how far it moves vertically. These two numbers are its components, written ⟨v_x, v_y⟩ with angle brackets, or as a column of two numbers, to keep them from being confused with the coordinates of a point. A point is a location; a vector is a displacement. The vector ⟨4, 3⟩ means "4 to the right and 3 up", starting from anywhere.

Finding components from two points. Suppose the initial point is P(x₁, y₁) and the terminal point is Q(x₂, y₂). To get from P to Q horizontally you move x₂ − x₁, and vertically you move y₂ − y₁. So the components of PQ→ are ⟨x₂ − x₁, y₂ − y₁⟩: subtract the coordinates of the initial point from the coordinates of the terminal point. Terminal minus initial, in that order. Subtracting the other way gives ⟨x₁ − x₂, y₁ − y₂⟩, which is QP→, the vector pointing the opposite way.

A fully worked model. Let P = (1, 2) be the initial point and Q = (5, 5) the terminal point. Horizontal component: 5 − 1 = 4. Vertical component: 5 − 2 = 3. So PQ→ = ⟨4, 3⟩. The arrow, the horizontal move of 4 and the vertical move of 3 form a right triangle, so by the Pythagorean theorem the magnitude is |PQ→| = √(4² + 3²) = √(16 + 9) = √25 = 5 units. The direction can be given as an angle measured counterclockwise from the positive x-axis: tan θ = 3/4, so θ = arctan(3/4) ≈ 36.87°. Now take R = (3, 0) and S = (7, 3), the second arrow in the diagram. RS→ = ⟨7 − 3, 3 − 0⟩ = ⟨4, 3⟩. (The arrow from (−2, 0) to (2, 3) gives the same result: ⟨2 − (−2), 3 − 0⟩ = ⟨4, 3⟩.) Same components, so the same vector, though it sits somewhere else on the grid. The magnitude formula in general is |⟨v_x, v_y⟩| = √(v_x² + v_y²), and it always returns a non-negative number because it is a length.

Direction angles need the quadrant. The calculator's arctan only returns angles between −90° and 90°, so it cannot tell ⟨−6, 8⟩ from ⟨6, −8⟩. For ⟨−6, 8⟩ the horizontal component is negative and the vertical is positive, so the arrow points up and to the left, into the second quadrant. Compute the reference angle arctan(8/6) ≈ 53.13° and then use 180° − 53.13° = 126.87°. For ⟨6, −8⟩ the arrow points down and right, in the fourth quadrant: 360° − 53.13° = 306.87°. Always sketch the arrow before naming its angle.

The misconception: a vector is a point. Because ⟨4, 3⟩ and (4, 3) use the same numbers, students often treat them as the same thing. They are not. The point (4, 3) is one place on the grid. The vector ⟨4, 3⟩ is a move that can be made from any place; drawn from the origin its tip does land on (4, 3), which is where the confusion comes from, but drawn from (1, 2) its tip lands on (5, 5). A second version of the same error is subtracting initial minus terminal. From P(1, 2) to Q(5, 5) that gives ⟨−4, −3⟩, an arrow pointing down-left, which describes traveling from Q back to P. The magnitude comes out the same, 5, which is why the error survives a magnitude check; the direction check catches it.

Going the other way: from components to the terminal point. If a vector ⟨−6, 8⟩ is drawn with its initial point at (3, −2), the terminal point is (3 + (−6), −2 + 8) = (−3, 6). Add the components to the initial point. Check by subtracting: ⟨−3 − 3, 6 − (−2)⟩ = ⟨−6, 8⟩.

From magnitude and direction to components. A wind of 20 km/h blowing toward a direction 30° above the positive x-axis has components ⟨20 cos 30°, 20 sin 30°⟩ = ⟨17.32, 10⟩ to two decimal places, because cos 30° = √3/2 ≈ 0.866 and sin 30° = 1/2. Check with the exact component 10√3: √((10√3)² + 10²) = √(300 + 100) = √400 = 20 (with the rounded 17.32, √(17.32² + 10²) ≈ 20.00). Converting between the two descriptions, magnitude-and-direction and components, is the everyday work of vector problems, and each conversion should be checked by converting back.

Worked, components and magnitude: Find the components and magnitude of AB→ with A(−3, 4) and B(2, −8). Terminal minus initial: horizontal 2 − (−3) = 5, vertical −8 − 4 = −12. AB→ = ⟨5, −12⟩. Magnitude: √(5² + (−12)²) = √(25 + 144) = √169 = 13. The negative vertical component tells you the arrow goes down; the magnitude is still positive. Direction: the arrow points right and down, fourth quadrant; reference angle arctan(12/5) ≈ 67.38°, so the direction angle is 360° − 67.38° = 292.62°.

Worked, a horizontal vector: C(6, 1) to D(2, 1). CD→ = ⟨2 − 6, 1 − 1⟩ = ⟨−4, 0⟩. Magnitude √(16 + 0) = 4. The vector points straight left, direction angle 180°. Notice |CD→| = 4 while the horizontal component is −4: the component carries a sign, the magnitude never does.

Worked, terminal point from a vector: The vector u = ⟨−6, 8⟩ is drawn with initial point (3, −2). Terminal point: (3 + (−6), −2 + 8) = (−3, 6). Magnitude |u| = √(36 + 64) = √100 = 10. Direction: left and up, second quadrant; reference angle arctan(8/6) ≈ 53.13°, direction angle 180° − 53.13° = 126.87°.

Worked, same vector in two places: Show that EF→ with E(0, 0), F(4, 3) and GH→ with G(1, 2), H(5, 5) are the same vector. EF→ = ⟨4 − 0, 3 − 0⟩ = ⟨4, 3⟩. GH→ = ⟨5 − 1, 5 − 2⟩ = ⟨4, 3⟩. Equal components mean equal magnitude (both 5) and equal direction (both about 36.87° above the x-axis), so they are the same vector even though F ≠ H.

Worked, from magnitude and direction: A displacement of 20 m at 30° above the positive x-axis has components ⟨20 cos 30°, 20 sin 30°⟩ = ⟨10√3, 10⟩ ≈ ⟨17.32, 10⟩. Check the magnitude: √(300 + 100) = √400 = 20 m.

Recording task: On grid paper draw the arrow from P(1, 2) to Q(5, 5) and, in a different color, the arrow from R(3, 0) to S(7, 3), as in the diagram. For each arrow record a three-line entry: the subtraction that gives the components (terminal minus initial, written out), the components in angle brackets, and the magnitude computed from the components with the Pythagorean theorem. Then draw the right triangle under each arrow with legs 4 and 3, and write a sentence explaining why the two arrows represent one vector although they touch different points.

Independent performance: Choose five pairs of points of your own, at least one in each of these situations: both coordinates of the terminal point smaller than those of the initial point; a horizontal vector; a vertical vector; a vector whose magnitude is a whole number (use a 5-12-13 or 8-15-17 triangle); and a vector given as a magnitude and a direction angle, such as 26 units at 120°, that you must convert to components. For every vector record the symbols correctly (an arrow over the endpoint letters for the vector, bars for its magnitude), the components, the magnitude to two decimal places, and the direction angle measured counterclockwise from the positive x-axis with the quadrant named, or, for a horizontal or vertical vector, the axis it lies along (its angle is then 0°, 90°, 180° or 270°). Success criteria: every component comes from terminal minus initial and is signed correctly; every magnitude is non-negative and checks against a sketch; every direction angle lies in the correct quadrant, or on the correct axis for a horizontal or vertical vector; and the converted vector, when its components are turned back into a magnitude, returns the magnitude you started with to within rounding.