Before we start, here is the single formula we will lean on all page. Two vectors of size A and B, with angle θbetween them (tail-to-tail), combine into a resultant whose length is:
Look at the picture below so the symbols A, B, θ, R have a home before we compute anything. The lavender arrow is A, the coral arrow is B, they share a tail; the small arc between them is θ; the mint arrow along the diagonal is the resultant R. The dashed lines complete the parallelogram whose diagonal R is.
What we do: for each, ask "do I need a direction to state it fully — AND does it add by the parallelogram law?"
(a) mass → scalar (just kilograms).
(b) velocity → vector (speed and which way).
(c) temperature → scalar (just degrees).
(d) force → vector (magnitude and direction, adds by parallelogram).
(e) electric current → scalar — this is the trap; it has a direction along the wire but currents add arithmetically (Kirchhoff), not by parallelogram.
(f) displacement → vector.
Recall Solution 1.2
Answer: speed. Distance and speed are the scalar members; displacement and velocity are the vector members. Distance is total path length (no direction), speed is distance-over-time (no direction). See Distance vs Displacement and Speed vs Velocity.
(a) Distance (scalar). Distance ignores direction, so it is plain addition:
d=3+4=7km.(b) Displacement (vector). East and North are perpendicular, so θ=90∘ and cos90∘=0. The cross term2ABcosθ (defined above) vanishes:
∣R∣=32+42+2⋅3⋅4⋅0=9+16=25=5km.What it looks like: a right triangle with legs 3 and 4; the diagonal (the displacement) is the hypotenuse 5. That is why scalar gives 7 but vector gives 5.
Recall Solution 2.2
What we do: force is a vector, so use the resultant formula with θ=90∘, cos90∘=0:
∣R∣=62+82+2⋅6⋅8⋅0=36+64=100=10N.
Recall Solution 2.3
(a) Speed uses total path length (distance):
speed=100400=4m/s.(b) Velocity uses displacement. The car returns to start, so net displacement =0:
∣velocity∣=1000=0m/s.
A directed quantity can cancel itself; a path length cannot.
What we do:A=B=5, θ=120∘, and cos120∘=−21. The negative cosine shrinks the sum:
∣R∣=52+52+2⋅5⋅5⋅(−21)=25+25−25=25=5N.Why it looks right: three equal arrows of 5N at 120∘ close into an equilateral pattern; two of them combine to give the third's size. So the resultant equals each force.
Recall Solution 3.2
What we do: current is a scalar, so despite the different directions we use plain addition (charge conservation, Kirchhoff):
Iout=2+3=5A.
We do not use the parallelogram — if we wrongly did, at say 90∘ we'd get 4+9≈3.6A, which is physically nonsense (charge would vanish). This is exactly why "has direction ⇒ vector" fails.
Recall Solution 3.3
Put θ=0∘, so cos0∘=1:
∣R∣=A2+B2+2AB⋅1=A2+2AB+B2=(A+B)2=A+B.Interpretation: aligned arrows just stack head-to-tail, so lengths add. The "+2ABcosθ" term is precisely what makes vectors differ from scalars when the angle is not zero.
Step 1 — find the between-angle. First leg points East. Second leg turns 60∘away from East. Placed tail-to-tail, the angle between the two arrows is θ=60∘, and cos60∘=21.
Step 2 — apply the formula with A=B=4:
∣R∣=42+42+2⋅4⋅4⋅21=16+16+16=48=43≈6.93km.What it looks like: the two legs and the resultant form a triangle. Compare against the two extreme cases so 6.93 has context: if the arrows pointed opposite (θ=180∘) the answer would be the minimum ∣4−4∣=0; if they pointed the same way (θ=0∘) it would be the maximum 4+4=8. Because the turn is gentle (60∘), our 6.93 sits comfortably between those two limits, nearer the aligned maximum.
Recall Solution 4.2
Maximum, A+B: need cosθ=+1, so θ=0∘ — arrows point the same way, lengths stack.
Minimum, ∣A−B∣: need cosθ=−1, so θ=180∘ — arrows point opposite, they partly cancel:
A2+B2−2AB=(A−B)2=∣A−B∣.Takeaway: every possible resultant of two vectors lies in the band ∣A−B∣≤∣R∣≤A+B. That band is the fingerprint of vector addition.
Set up:A=B=F, want ∣R∣=F:
F=F2+F2+2F2cosθ.Square both sides (both sides positive, safe to square):
F2=2F2+2F2cosθ.Divide by F2 (allowed since F=0):
1=2+2cosθ⇒2cosθ=−1⇒cosθ=−21.Undo the cosine: the angle whose cosine is −21 is
θ=120∘.Check: this matches Exercise 3.1 exactly — two 5N forces at 120∘ gave 5N. The algebra confirms the picture.
Recall Solution 5.2
Assume the opposite: suppose current were a vector obeying the parallelogram law.
Then the "out" current would be
22+32+2⋅2⋅3cos90∘=4+9+0=13≈3.61A.The contradiction:2A of charge per second plus 3A of charge per second deliver 5 coulombs each second into the junction. Charge cannot be created or destroyed, so 5Amust leave. But the parallelogram gave 3.61A — a loss of charge, impossible.
Conclusion: the vector assumption breaks conservation of charge, so current is a scalar, adding arithmetically to 5A. The direction along a wire is real, but it does not obey the parallelogram law — which is the whole point of the parent note's "steel-man" trap.
Recall Solution 5.3
(a) The boat's velocity across and the river's velocity along are two vectors at 90∘. They add by the parallelogram:
∣v∣=32+42+2⋅3⋅4cos90∘=9+16=5m/s.(b)Velocities (boat, river, resultant) are vectors — that's why they combined to 5, not 7. The speeds3, 4, 5 (their magnitudes) are scalars. Time, if involved, would be scalar too. This is Speed vs Velocity in action and previews relative velocity.
Components of a Vector — the next tool for angled problems.
Dot and Cross Products — where cosθ reappears as the dot product.
Units and Dimensions — every answer above still carries a unit.
Recall Self-check (each line is
Question ::: Answer — reveal the part after the triple colon)
The band every two-vector resultant lives in ::: ∣A−B∣≤∣R∣≤A+B
Two equal forces give a resultant equal to one of them at what angle? ::: 120∘
Why is current a scalar despite having a direction? ::: it adds arithmetically (Kirchhoff), not by the parallelogram law
