C. Geometry and Algebra of Vectors ยท Vectors in Two-Space and Three-SpaceAdding Vectors Geometrically: Triangle and Parallelogram Laws
C2.1About 70 minutes
Learning goals- I can add vectors drawn as directed line segments using the triangle (tip-to-tail) law and the parallelogram law.
- I can simplify sums of vectors such as ๐๐ + ๐๐ + ๐๐.
- I can calculate the magnitude and direction of a resultant using the cosine and sine laws.
- I can explain why |๐ฎ + ๐ฏ| โค |๐ฎ| + |๐ฏ|.
Words to know
resultant
The single vector that has the same effect as two or more vectors combined: their sum.
triangle law
To add ๐ฎ + ๐ฏ, place the tail of ๐ฏ at the head of ๐ฎ; the sum runs from the tail of ๐ฎ to the head of ๐ฏ.
parallelogram law
To add ๐ฎ + ๐ฏ, place them tail to tail and complete the parallelogram; the sum is the diagonal from the shared tail.
angle between two vectors
The angle ฮธ (0ยฐ โค ฮธ โค 180ยฐ) between two vectors placed tail to tail.
triangle inequality
|๐ฎ + ๐ฏ| โค |๐ฎ| + |๐ฏ|.
Adding arrows, not numbers
A canoe is paddled across a river while the current pushes it downstream. A sign hangs from two cables that pull in different directions. In both cases two vectors act together, and the combined effect is their sum, called the resultant. You can't just add magnitudes: 3 km plus 4 km of displacement can give anything from 1 km to 7 km, depending on directions. In this lesson you add vectors geometrically (as directed line segments) and use the cosine and sine laws to get exact answers.
The triangle law (tip to tail)
To find ๐ฎ + ๐ฏ:
- Draw ๐ฎ.
- Slide ๐ฏ so its tail is at the head of ๐ฎ (sliding doesn't change a vector).
- The sum ๐ฎ + ๐ฏ is the arrow from the tail of ๐ฎ to the head of ๐ฏ.
With points, this is the very useful rule ๐๐ + ๐๐ = ๐๐: going from A to B and then from B to C has the same effect as going straight from A to C.
The parallelogram law (tail to tail)
If ๐ฎ and ๐ฏ start at the same point, complete the parallelogram. The sum ๐ฎ + ๐ฏ is the diagonal that starts at the shared tail. This gives the same arrow as the triangle law, because the opposite side of the parallelogram is a copy of ๐ฏ. It also shows at once that ๐ฎ + ๐ฏ = ๐ฏ + ๐ฎ.
Calculating the resultant
Let ฮธ be the angle between ๐ฎ and ๐ฏ when they are placed tail to tail (0ยฐ โค ฮธ โค 180ยฐ). In the tip-to-tail triangle, the angle at the joint is 180ยฐ โ ฮธ. So, by the cosine law from MCR3U:
|๐ฎ + ๐ฏ|ยฒ = |๐ฎ|ยฒ + |๐ฏ|ยฒ โ 2|๐ฎ||๐ฏ| cos(180ยฐ โ ฮธ)
Since cos(180ยฐ โ ฮธ) = โcos ฮธ, this can also be written |๐ฎ + ๐ฏ|ยฒ = |๐ฎ|ยฒ + |๐ฏ|ยฒ + 2|๐ฎ||๐ฏ| cos ฮธ.
Then use the sine law to find the angle that ๐ฎ + ๐ฏ makes with ๐ฎ (or with a north line). If ฮธ = 90ยฐ, the triangle is a right triangle and the Pythagorean theorem and tangent are quicker.
Find the right angle for the cosine law
The most common mistake is putting the angle between the vectors (tail to tail) into the cosine law instead of the angle inside the triangle (tip to tail). Always draw the tip-to-tail triangle and mark the angle at the joint. A check: if the vectors point in roughly the same direction, the resultant should be longer than either one.
Worked exampleExample 1: Two vectors at 60ยฐ
|๐ฎ| = 5 and |๐ฏ| = 8, and the angle between them (tail to tail) is 60ยฐ. Find |๐ฎ + ๐ฏ| and the angle between ๐ฎ + ๐ฏ and ๐ฎ.
- Draw ๐ฎ, then ๐ฏ tip to tail. The angle at the joint is 180ยฐ โ 60ยฐ = 120ยฐ.
- Cosine law: |๐ฎ + ๐ฏ|ยฒ = 5ยฒ + 8ยฒ โ 2(5)(8) cos 120ยฐ = 25 + 64 + 40 = 129, so |๐ฎ + ๐ฏ| = โ129 โ 11.36.
- Sine law for the angle ฮฑ between ๐ฎ and ๐ฎ + ๐ฏ (it is opposite the side of length 8): sin ฮฑ / 8 = sin 120ยฐ / โ129.
- sin ฮฑ = 8(0.8660)/11.358 โ 0.6100, so ฮฑ โ 37.6ยฐ. (ฮฑ must be acute because it is opposite a side shorter than the longest side.)
|๐ฎ + ๐ฏ| = โ129 โ 11.36, at about 37.6ยฐ from ๐ฎ (toward ๐ฏ).
Worked exampleExample 2: Adding displacements with bearings
A float plane flies 30 km north, then 40 km on a bearing of 060ยฐ. Find its resultant displacement from the start.
- Sketch with a north line at each point. The second leg makes 60ยฐ with north at the turning point B.
- The angle inside the triangle at B is between the leg back to the start (pointing south) and the second leg: 180ยฐ โ 60ยฐ = 120ยฐ.
- |๐|ยฒ = 30ยฒ + 40ยฒ โ 2(30)(40) cos 120ยฐ = 900 + 1600 + 1200 = 3700, so |๐| โ 60.8 km.
- Angle at the start, A: sin A / 40 = sin 120ยฐ / 60.83, so sin A โ 0.5695 and A โ 34.7ยฐ. This is measured from north toward east.
About 60.8 km on a bearing of 034.7ยฐ (N 34.7ยฐ E).
Draw a north line at every turning point. The 60ยฐ bearing at B makes a 120ยฐ angle inside the triangle.Worked exampleExample 3: Chains of vectors
Simplify (a) ๐๐ + ๐๐ + ๐๐ and (b) ๐๐ + ๐๐ + ๐๐.
- (a) Apply the triangle law twice: ๐๐ + ๐๐ = ๐๐, then ๐๐ + ๐๐ = ๐๐.
- (b) ๐๐ + ๐๐ = ๐๐, then ๐๐ + ๐๐ = ๐๐, which is the zero vector: you end up where you started.
When vectors are joined head to tail, the sum goes from the first tail to the last head: ๐๐ + ๐๐ + ๐๐ = ๐๐.Just for funVectors make great hikers. They never get lost: no matter how winding the trail, they can always tell you the straight-line arrow back to the parking lot.
Hands-on activityTip to tail, by hand
You need: a sheet of paper, a ruler, a protractor and a pencil.
- Draw ๐ฎ: 5 cm pointing right. From its head, draw ๐ฏ: 8 cm at 60ยฐ above the horizontal. Draw ๐ฎ + ๐ฏ from the tail of ๐ฎ to the head of ๐ฏ. Measure its length and its angle above the horizontal.
- On another part of the page, draw ๐ฏ first and then ๐ฎ from its head. Measure ๐ฏ + ๐ฎ.
- Draw ๐ฎ and ๐ฏ tail to tail, complete the parallelogram, and measure the diagonal.
- Compare all three measurements with Example 1.
What to look for: all three methods should give about 11.4 cm at about 37.6ยฐ above the horizontal (within a couple of millimetres and a degree or two). This shows that the order of adding doesn't matter, and that the triangle and parallelogram laws agree.
Your turn
1Simplify ๐๐ + ๐๐.
2Simplify ๐๐ + ๐๐ + ๐๐.
๐ (the path returns to A).
3๐ฎ and ๐ฏ are perpendicular, with |๐ฎ| = 6 and |๐ฏ| = 8. Find |๐ฎ + ๐ฏ| and the angle between ๐ฎ + ๐ฏ and ๐ฎ.
|๐ฎ + ๐ฏ| = โ(36 + 64) = 10. Angle: tanโปยน(8/6) โ 53.1ยฐ.
4|๐ฎ| = 10, |๐ฏ| = 7, and the angle between them is 40ยฐ. Find |๐ฎ + ๐ฏ|.
Angle in the triangle = 140ยฐ. |๐ฎ + ๐ฏ|ยฒ = 100 + 49 โ 140 cos 140ยฐ โ 149 + 107.2 = 256.2, so |๐ฎ + ๐ฏ| โ 16.0.
5For the vectors in the previous question, find the angle between ๐ฎ + ๐ฏ and ๐ฎ.
sin ฮฑ = 7 sin 140ยฐ / 16.01 โ 0.2811, so ฮฑ โ 16.3ยฐ.
6Two vectors each of magnitude 5 make an angle of 120ยฐ with each other. Find the magnitude of their sum.
The angle in the triangle is 60ยฐ, and the two sides are equal, so the triangle is equilateral: |๐ฎ + ๐ฏ| = 5. (Cosine law: 25 + 25 โ 50 cos 60ยฐ = 25.)
7A hiker walks 4 km on a bearing of 030ยฐ and then 6 km on a bearing of 120ยฐ. Find her displacement.
The legs differ by 90ยฐ, so the triangle has a right angle at the turn. |๐| = โ(16 + 36) = โ52 โ 7.2 km. The angle at the start is tanโปยน(6/4) โ 56.3ยฐ, so the bearing is 30ยฐ + 56.3ยฐ โ 086.3ยฐ.
More practice: warm-up
1Simplify ๐๐ + ๐๐.
2What is the resultant of 5 N east and 5 N west?
3Which law adds two vectors placed tail to tail?
5|๐| = 3 and |๐| = 4 and they are perpendicular. Find |๐ + ๐|.
More practice: on your own
1|๐| = 12, |๐| = 20 and the angle between them is 75ยฐ. Find |๐ + ๐| and the angle between ๐ + ๐ and ๐.
Angle in the triangle 105ยฐ: |๐ + ๐|ยฒ = 144 + 400 โ 480 cos 105ยฐ โ 544 + 124.2 = 668.2, so |๐ + ๐| โ 25.9. sin ฮฑ = 20 sin 105ยฐ / 25.85 โ 0.7473, ฮฑ โ 48.4ยฐ.
2A boat sails 8 km north, then 5 km on a bearing of 050ยฐ. Find its displacement.
Angle at the turn: 180ยฐ โ 50ยฐ = 130ยฐ. |๐|ยฒ = 64 + 25 โ 80 cos 130ยฐ โ 140.4, |๐| โ 11.9 km. sin A = 5 sin 130ยฐ / 11.85 โ 0.3232, A โ 18.9ยฐ. Bearing โ 018.9ยฐ.
3|๐ฎ| = 9 and |๐ฏ| = 4. What are the largest and smallest possible values of |๐ฎ + ๐ฏ|? When does each happen?
Largest 13, when ๐ฎ and ๐ฏ point the same way. Smallest 5, when they point in opposite directions.
4In regular hexagon ABCDEF, simplify ๐๐ + ๐๐ + ๐๐ and find its magnitude if each side is 3 cm.
๐๐. AD is a long diagonal through the centre, made of two radii, and each radius equals a side. So |๐๐| = 6 cm.
5Explain using the triangle law why |๐ฎ + ๐ฏ| โค |๐ฎ| + |๐ฏ|.
๐ฎ, ๐ฏ and ๐ฎ + ๐ฏ form the sides of a triangle (tip to tail). In any triangle, one side is never longer than the sum of the other two (the triangle inequality). Equality happens only when the "triangle" is flat: ๐ฎ and ๐ฏ point the same way.
Challenge
1|๐ฎ| = 7, |๐ฏ| = 8 and |๐ฎ + ๐ฏ| = 13. Find the angle between ๐ฎ and ๐ฏ.
Using |๐ฎ + ๐ฏ|ยฒ = |๐ฎ|ยฒ + |๐ฏ|ยฒ + 2|๐ฎ||๐ฏ| cos ฮธ: 169 = 49 + 64 + 112 cos ฮธ, so cos ฮธ = 56/112 = 0.5 and ฮธ = 60ยฐ.
2|๐ฎ| = 5, the angle between ๐ฎ and ๐ฏ is 60ยฐ, and |๐ฎ + ๐ฏ| = 7. Find |๐ฏ|.
Let |๐ฏ| = v. 49 = 25 + vยฒ + 2(5)(v) cos 60ยฐ = 25 + vยฒ + 5v. So vยฒ + 5v โ 24 = 0, (v + 8)(v โ 3) = 0, and since v > 0, |๐ฏ| = 3.
3A plane flies 200 km on a bearing of 040ยฐ and then 150 km on a bearing of 160ยฐ. On what bearing and for what distance must it fly to return straight to its start?
Angle at the turn: the back bearing of leg 1 is 220ยฐ, and 220ยฐ โ 160ยฐ = 60ยฐ. |๐|ยฒ = 200ยฒ + 150ยฒ โ 2(200)(150) cos 60ยฐ = 32 500, |๐| โ 180.3 km. Angle at the start: sin A = 150 sin 60ยฐ / 180.28 โ 0.7206, A โ 46.1ยฐ, so the plane is on a bearing of 040ยฐ + 46.1ยฐ = 086.1ยฐ from the start. To return: about 180.3 km on a bearing of 266.1ยฐ.
Quiz
1. What is ๐๐ + ๐๐ + ๐๐ equal to?
2. |๐ฎ| = 3 and |๐ฏ| = 4, and ๐ฎ is perpendicular to ๐ฏ. What is |๐ฎ + ๐ฏ|?
3. The angle between ๐ฎ and ๐ฏ (tail to tail) is 70ยฐ. Which angle goes in the cosine law to find |๐ฎ + ๐ฏ|?
4. |๐ฎ| = 10 and |๐ฏ| = 6. Which value is impossible for |๐ฎ + ๐ฏ|?
5. Which statement describes the parallelogram law?
6. Two vectors of magnitude 10 make an angle of 60ยฐ with each other. What is the magnitude of their sum?
Big idea (C2.1): vectors add tip to tail (or by completing a parallelogram), and the cosine and sine laws turn the drawing into exact numbers.
How to help: Insist on a sketch for every question, with north lines drawn at each turning point when bearings are involved. Ask, "Which angle is inside the triangle?" before your teen uses the cosine law.
Common difficulties:
- Putting the tail-to-tail angle into the cosine law instead of 180ยฐ minus it.
- Adding magnitudes directly (5 + 8 = 13) instead of adding vectors.
- Losing track of which reference line a direction is measured from.
- Rounding |๐ฎ + ๐ฏ| before using the sine law; keep full calculator values.