High school Β· MathematicsAbout 120 minutes
MathematicsHigh schoolAbout 120 minutes

Add vectors end-to-end, by the parallelogram rule and component-wise, and subtract by adding the additive inverse

Several sessions: instruction with grid drawings, modeled examples, independent practice with every magnitude and product re-computed, and a written explanation.

Ohio learning standards this draft targets

math-2017:N.VM.4math-2017:N.VM.4amath-2017:N.VM.4bmath-2017:N.VM.4c

Read the standard text

math-2017:N.VM.4 Add and subtract vectors.

math-2017:N.VM.4a Add vectors end-to-end, component-wise, and by the parallelogram rule. Understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes.

math-2017:N.VM.4b Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum.

math-2017:N.VM.4c Understand vector subtraction 𝒗𝒗 βˆ’ π’˜π’˜ as 𝒗𝒗 + (βˆ’π’˜π’˜), where βˆ’ π’˜π’˜ is the additive inverse of π’˜π’˜, with the same magnitude as π’˜π’˜ and pointing in the opposite direction. Represent vector subtraction graphically by connecting the tips in the appropriate order, and perform vector subtraction component-wise.

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 Β· LearnLearn the idea
Vectors u = ⟨4, 1⟩ and w = ⟨1, 3⟩ drawn from the origin; the parallelogram diagonal is the sum ⟨5, 4⟩, the same arrow found end-to-end. Magnitudes: |u| = √17 β‰ˆ 4.12, |w| = √10 β‰ˆ 3.16, |u + w| = √41 β‰ˆ 6.40, which is less than |u| + |w| = √17 + √10 β‰ˆ 7.29. Axis ticks are one unit apart.Vectors u = ⟨4, 1⟩ and w = ⟨1, 3⟩ drawn from the origin; the parallelogram diagonal is the sum ⟨5, 4⟩, the same arrow found end-to-end. Magnitudes: |u| = √17 β‰ˆ 4.12, |w| = √10 β‰ˆ 3.16, |u + w| = √41 β‰ˆ 6.40, which is less than |u| + |w| = √17 + √10 β‰ˆ 7.29. Axis ticks are one unit apart.123456123450u = ⟨4, 1⟩w = ⟨1, 3⟩u + w = ⟨5, 4⟩|u| = √17 β‰ˆ 4.12|w| = √10 β‰ˆ 3.16|u| + |w| β‰ˆ 7.29|u + w| = √41 β‰ˆ 6.406.40 < 7.29dashed sides: copies ofu and w placed end-to-end
Vectors u = ⟨4, 1⟩ and w = ⟨1, 3⟩ drawn from the origin; the parallelogram diagonal is the sum ⟨5, 4⟩, the same arrow found end-to-end. Magnitudes: |u| = √17 β‰ˆ 4.12, |w| = √10 β‰ˆ 3.16, |u + w| = √41 β‰ˆ 6.40, which is less than |u| + |w| = √17 + √10 β‰ˆ 7.29. Axis ticks are one unit apart.

Learning goal: Add vectors end-to-end, component-wise and by the parallelogram rule; understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes; given two vectors in magnitude and direction form, determine the magnitude and direction of their sum; and understand vector subtraction v βˆ’ w as v + (βˆ’w), where βˆ’w is the additive inverse of w, representing subtraction graphically by connecting the tips in the appropriate order and performing it component-wise.

Scope note. N.VM.4, with its parts a, b and c, is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

Three pictures of the same sum. Let u = ⟨4, 1⟩ and w = ⟨1, 3⟩. The sum u + w can be found three ways, and they must agree.

End-to-end (tip-to-tail). Draw u from the origin: it ends at (4, 1). Now draw w starting at the tip of u: from (4, 1) move 1 right and 3 up to (5, 4). The sum is the arrow from the tail of the first vector to the tip of the last, from (0, 0) to (5, 4). This picture matches the meaning of a sum: do one displacement, then the other, and ask where you ended up.

Parallelogram rule. Draw both u and w from the same initial point, the origin. Complete the parallelogram whose two sides are these arrows; its fourth vertex is at (5, 4). The sum is the diagonal from the common initial point to the opposite vertex. The parallelogram rule and the end-to-end rule are the same construction seen twice: the side of the parallelogram opposite w is a copy of w placed at the tip of u.

Component-wise. Add horizontal components together and vertical components together: ⟨4 + 1, 1 + 3⟩ = ⟨5, 4⟩. This is the version you compute with; the pictures are the version you check with.

A fully worked model, with magnitudes. |u| = √(4Β² + 1Β²) = √17 β‰ˆ 4.123. |w| = √(1Β² + 3Β²) = √10 β‰ˆ 3.162. Their sum of magnitudes is about 7.285. The sum vector ⟨5, 4⟩ has magnitude √(25 + 16) = √41 β‰ˆ 6.403. So |u + w| β‰ˆ 6.40 while |u| + |w| β‰ˆ 7.29. The magnitude of a sum of two vectors is typically not the sum of the magnitudes. Geometrically the three vectors form a triangle with sides √17, √10 and √41, and one side of a triangle is shorter than the other two put together. The general fact is |u + w| ≀ |u| + |w|, with equality only when u and w point in the same direction (or one of them is the zero vector). For instance ⟨2, 1⟩ and ⟨4, 2⟩ point the same way; their magnitudes are √5 and √20 = 2√5, and the sum ⟨6, 3⟩ has magnitude √45 = 3√5, exactly the sum. At the other extreme, ⟨3, 4⟩ + βŸ¨βˆ’3, βˆ’4⟩ = ⟨0, 0⟩, magnitude 0, though each part has magnitude 5.

Sums given in magnitude and direction form. Suppose two vectors are given by magnitude and direction: a has magnitude 10 at 0Β° (along the positive x-axis) and b has magnitude 10 at 60Β°. To determine the magnitude and direction of their sum, convert to components, add, and convert back. a = ⟨10 cos 0Β°, 10 sin 0°⟩ = ⟨10, 0⟩. b = ⟨10 cos 60Β°, 10 sin 60°⟩ = ⟨5, 5√3⟩ β‰ˆ ⟨5, 8.660⟩. Sum: ⟨15, 5√3⟩. Magnitude: √(15Β² + (5√3)Β²) = √(225 + 75) = √300 = 10√3 β‰ˆ 17.32. Direction: arctan(5√3/15) = arctan(√3/3) = 30Β° exactly (with the rounded 8.660, arctan(8.660/15) β‰ˆ 30.00Β°). The direction is exactly halfway between 0Β° and 60Β° because the two magnitudes are equal; and 17.32 is less than 10 + 10 = 20, as expected for vectors 60Β° apart. Check by converting back: ⟨17.32 cos 30Β°, 17.32 sin 30°⟩ = ⟨15.0, 8.66⟩.

The additive inverse. For any vector w, the vector βˆ’w is its additive inverse: it has the same magnitude as w, it points in the opposite direction whenever w is not the zero vector (βˆ’0 = 0, which has no direction), and w + (βˆ’w) = 0, the zero vector. In components, βˆ’βŸ¨1, 3⟩ = βŸ¨βˆ’1, βˆ’3⟩. Reversing an arrow does not change its length: |βˆ’w| = |w|.

Subtraction. Vector subtraction is defined by v βˆ’ w = v + (βˆ’w): to subtract w, add its additive inverse. Component-wise this means subtracting each component: ⟨v_x βˆ’ w_x, v_y βˆ’ w_y⟩. With v = ⟨2, 5⟩ and w = ⟨6, 1⟩: v βˆ’ w = ⟨2 βˆ’ 6, 5 βˆ’ 1⟩ = βŸ¨βˆ’4, 4⟩, and w βˆ’ v = ⟨4, βˆ’4⟩, the additive inverse of the first answer. Subtraction is not commutative, exactly as with numbers.

Subtraction graphically: connecting the tips. Draw v and w from the same initial point. The vector v βˆ’ w is the arrow connecting the tips in the appropriate order: from the tip of w to the tip of v. The order is the thing to remember: the arrow points toward the tip of the vector you are subtracting from. Why this works: starting at the tip of w, traveling along v βˆ’ w must land you at the tip of v, because w + (v βˆ’ w) = v. For v = ⟨2, 5⟩ and w = ⟨6, 1⟩ the tips are at (2, 5) and (6, 1), and the arrow from (6, 1) to (2, 5) is ⟨2 βˆ’ 6, 5 βˆ’ 1⟩ = βŸ¨βˆ’4, 4⟩, matching the component calculation. Drawing the arrow the wrong way round gives w βˆ’ v. The other graphical method is to draw βˆ’w (reverse w) and add it end-to-end to v; both pictures give the same arrow.

The misconception. Students who have just learned that vectors can be added often add magnitudes: "a 3 N force plus a 4 N force is a 7 N force." That is true only if the two forces point the same way. At right angles the resultant is 5 N; opposite, 1 N; at any other angle, somewhere between 1 N and 7 N. Two vectors' magnitudes tell you the possible range of |u + w|, from ||u| βˆ’ |w|| up to |u| + |w|; the angle between them decides where in that range the answer falls. A second error is subtracting in the graphical picture by drawing the arrow from the tip of v to the tip of w, which yields the inverse of the intended difference. A quick sign check on one component catches it.

Why all three additions agree. Placing w at the tip of u adds its horizontal move to u's horizontal move and its vertical move to u's vertical move; that is exactly what the component sum computes. The parallelogram is the same construction drawn twice, once with w first and once with u first, and both diagonals-from-the-origin meet at the same vertex, which is why vector addition is commutative: u + w = w + u.

Section 2 Β· ModelWork through the examples

Worked, end-to-end and component-wise: Add ⟨3, βˆ’2⟩ and βŸ¨βˆ’5, 4⟩. Component-wise: ⟨3 + (βˆ’5), βˆ’2 + 4⟩ = βŸ¨βˆ’2, 2⟩. End-to-end: from the origin go 3 right and 2 down to (3, βˆ’2), then 5 left and 4 up to (βˆ’2, 2); the sum arrow runs from (0, 0) to (βˆ’2, 2). Magnitudes: |⟨3, βˆ’2⟩| = √13 β‰ˆ 3.606, |βŸ¨βˆ’5, 4⟩| = √41 β‰ˆ 6.403, sum of magnitudes β‰ˆ 10.01, but |βŸ¨βˆ’2, 2⟩| = √8 β‰ˆ 2.828. The parts partly cancel because they point in nearly opposite directions.

Worked, magnitude and direction form: p has magnitude 12 at 90Β° and q has magnitude 5 at 0Β°. Components: p = ⟨0, 12⟩, q = ⟨5, 0⟩. Sum ⟨5, 12⟩, magnitude √(25 + 144) = 13, direction arctan(12/5) β‰ˆ 67.38Β°. Check: 13 lies between 12 βˆ’ 5 = 7 and 12 + 5 = 17, and 67.38Β° lies between 0Β° and 90Β°, closer to 90Β° because the 12-unit vector is larger.

Worked, equal magnitudes 60Β° apart: a = 10 at 0Β°, b = 10 at 60Β°. ⟨10, 0⟩ + ⟨5, 5√3⟩ = ⟨15, 5√3⟩ β‰ˆ ⟨15, 8.660⟩; magnitude √300 = 10√3 β‰ˆ 17.32 at exactly 30Β°, since tan ΞΈ = 5√3/15 = √3/3. Converting back: 17.32 cos 30Β° β‰ˆ 15.0 and 17.32 sin 30Β° = 8.66. The sum bisects the angle because the two magnitudes are equal.

Worked, subtraction two ways: v = ⟨2, 5⟩, w = ⟨6, 1⟩. Component-wise: v βˆ’ w = βŸ¨βˆ’4, 4⟩. Via the additive inverse: βˆ’w = βŸ¨βˆ’6, βˆ’1⟩, and v + (βˆ’w) = ⟨2 βˆ’ 6, 5 βˆ’ 1⟩ = βŸ¨βˆ’4, 4⟩. Graphically: draw both from the origin; the arrow from the tip of w at (6, 1) to the tip of v at (2, 5) is βŸ¨βˆ’4, 4⟩. |v βˆ’ w| = √32 β‰ˆ 5.657. Reversing the order gives w βˆ’ v = ⟨4, βˆ’4⟩, the same length pointing the opposite way.

Worked, when magnitudes do add: ⟨2, 1⟩ + ⟨4, 2⟩ = ⟨6, 3⟩. Magnitudes: √5 β‰ˆ 2.236, √20 β‰ˆ 4.472, and √45 β‰ˆ 6.708 = 2.236 + 4.472. The magnitudes add exactly because ⟨4, 2⟩ = 2⟨2, 1⟩ points in the same direction; the triangle has collapsed into a line.

Section 3 Β· DoPractice, demonstrate, and explain

Recording task: On grid paper draw u = ⟨4, 1⟩ and w = ⟨1, 3⟩ from the origin. Construct the sum three times: end-to-end (copy w to the tip of u), by the parallelogram rule (complete the parallelogram and draw the diagonal from the origin), and component-wise in a table. Confirm all three give ⟨5, 4⟩. Measure the three arrows with a ruler in grid units and record |u| β‰ˆ 4.12, |w| β‰ˆ 3.16 and |u + w| β‰ˆ 6.40, then write the inequality 6.40 < 4.12 + 3.16 in words. Finally draw u βˆ’ w by connecting the tip of w to the tip of u and check that it equals ⟨3, βˆ’2⟩ by components.

Independent performance: Choose two vectors of your own with different magnitudes, neither along an axis and not parallel. Present them first in component form and then, separately, in magnitude and direction form (magnitude to two decimals, angle to two decimals). Compute their sum by all three methods and their difference in both orders, both component-wise and graphically. Then design one pair of vectors for which the magnitude of the sum does equal the sum of the magnitudes and one pair for which the sum is the zero vector, and explain what is special about each. Success criteria: the three addition methods agree; the graphical difference arrow points from the tip of the subtracted vector to the tip of the other and matches the components; every magnitude is computed with the Pythagorean theorem and the inequality |u + w| ≀ |u| + |w| is stated numerically for each pair; and the magnitude-and-direction sum is converted back to components as a check.

Section 4 Β· Check8-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

Add u = ⟨4, 1⟩ and w = ⟨1, 3⟩ component-wise, then describe how the end-to-end picture and the parallelogram rule each produce the same arrow.

Show worked solution

Component-wise: ⟨4 + 1, 1 + 3⟩ = ⟨5, 4⟩. End-to-end: draw u from the origin to (4, 1), then w from (4, 1) to (5, 4); the sum runs from the origin to (5, 4). Parallelogram: draw both from the origin, complete the parallelogram with fourth vertex (5, 4), and take the diagonal from the origin. All three give the arrow to (5, 4), because placing w at the tip of u adds its components to those of u.

Problem 2

Show numerically that |u + w| is not |u| + |w| for u = ⟨4, 1⟩ and w = ⟨1, 3⟩. When would the two be equal?

Show worked solution

|u| = √17 β‰ˆ 4.12, |w| = √10 β‰ˆ 3.16, so |u| + |w| β‰ˆ 7.29. But u + w = ⟨5, 4⟩ has magnitude √41 β‰ˆ 6.40. The sum is shorter because the vectors form a triangle with the sum as the third side, and a side is shorter than the other two combined. Equality holds only when the two vectors point in the same direction (or one of them is the zero vector), for example ⟨2, 1⟩ and ⟨4, 2⟩, where the triangle collapses to a line.

Problem 3

Two vectors are given in magnitude and direction form: 10 units at 0Β° and 10 units at 60Β°. Determine the magnitude and direction of their sum.

Show worked solution

Convert to components: ⟨10, 0⟩ and ⟨10 cos 60Β°, 10 sin 60°⟩ = ⟨5, 5√3⟩ β‰ˆ ⟨5, 8.660⟩. Sum: ⟨15, 5√3⟩. Magnitude √(225 + 75) = √300 = 10√3 β‰ˆ 17.32; direction arctan(5√3/15) = arctan(√3/3) = 30Β°. The direction bisects the 60Β° angle because the magnitudes are equal, and 17.32 is less than 20 because the vectors are not aligned.

Problem 4

Explain what βˆ’w means for w = ⟨6, 1⟩, and use it to compute v βˆ’ w for v = ⟨2, 5⟩.

Show worked solution

βˆ’w is the additive inverse of w: the vector with the same magnitude as w, √37, pointing in the opposite direction, so βˆ’w = βŸ¨βˆ’6, βˆ’1⟩ and w + (βˆ’w) = ⟨0, 0⟩. Then v βˆ’ w = v + (βˆ’w) = ⟨2 + (βˆ’6), 5 + (βˆ’1)⟩ = βŸ¨βˆ’4, 4⟩, which is the same as subtracting component-wise: ⟨2 βˆ’ 6, 5 βˆ’ 1⟩.

Problem 5

v = ⟨2, 5⟩ and w = ⟨6, 1⟩ are drawn from the same point. Which arrow connecting their tips represents v βˆ’ w, and how do you know the direction is right?

Show worked solution

The arrow from the tip of w at (6, 1) to the tip of v at (2, 5). Its components are ⟨2 βˆ’ 6, 5 βˆ’ 1⟩ = βŸ¨βˆ’4, 4⟩, matching the component-wise subtraction. The direction is right because starting at the tip of w and traveling along v βˆ’ w must land at the tip of v, since w + (v βˆ’ w) = v. The arrow drawn the other way, from (2, 5) to (6, 1), is w βˆ’ v = ⟨4, βˆ’4⟩.

Problem 6

A student says a 3 N force and a 4 N force acting on the same object always combine to 7 N. Give the full range of possible resultant magnitudes and the cases at each end.

Show worked solution

The resultant can be anywhere from 4 βˆ’ 3 = 1 N to 3 + 4 = 7 N. It is 7 N only when both forces point the same way, 1 N only when they point in opposite directions, and 5 N when they are at right angles, since √(9 + 16) = 5. The angle between the forces decides where in the range the resultant falls; the magnitudes alone do not.

Problem 7

Compute ⟨3, βˆ’2⟩ + βŸ¨βˆ’5, 4⟩ and compare the magnitude of the sum with the sum of the magnitudes.

Show worked solution

Sum: ⟨3 βˆ’ 5, βˆ’2 + 4⟩ = βŸ¨βˆ’2, 2⟩ with magnitude √8 β‰ˆ 2.83. The parts have magnitudes √13 β‰ˆ 3.61 and √41 β‰ˆ 6.40, totalling about 10.01. The sum is much shorter than that because the two vectors point in nearly opposite directions and largely cancel.

Problem 8

Verify with components that u + w = w + u for u = ⟨4, 1⟩ and w = ⟨1, 3⟩, and explain what this says about the parallelogram picture.

Show worked solution

u + w = ⟨5, 4⟩ and w + u = ⟨1 + 4, 3 + 1⟩ = ⟨5, 4⟩; addition of the components is addition of numbers, which is commutative. In the parallelogram, going along u then along a copy of w, or along w then a copy of u, reaches the same opposite vertex, so both routes give the same diagonal.