A single object can feel several pushes at once — wind on a plane, thrust from its engines, gravity pulling down. Because vectors carry direction, you can’t just add their sizes; you have to combine them geometrically. Do it right and any number of vectors collapse into one resultant that has the same overall effect. The reverse trick, splitting one vector back into two, is called resolving. Master both and mechanics gets a lot simpler.
📘 What you need to know
Vectors combine by addition or subtraction to give a single resultant (the “net” vector)
Two graphical methods: the triangle method (head-to-tail) and the parallelogram method (tail-to-tail)
To subtract, reverse the direction of the vector and add it: a − b = a + (−b)
A scalar × a vector is still a vector (e.g. F = ma, p = mv)
For perpendicular vectors, find the resultant’s magnitude with Pythagoras and its direction with trigonometry
Resolving is the opposite of adding: split one vector into a horizontal F cos θ and vertical F sin θ
Combining vectors: the triangle method
To add two vectors with the triangle method, draw the first one, then start the second from the head of the first. The resultant is the arrow that closes the triangle — from the tail of the first vector to the head of the second.
Draw a
→ head-to-tail →
Draw b from a’s head
→ close it →
c = a + b
The triangle method: link vectors head-to-tail, then the resultant closes the triangle from start to end.
The parallelogram method
The parallelogram method reaches the same answer a different way. Draw both vectors starting from the same point (tail-to-tail), complete the parallelogram they form, and the diagonal from the shared starting corner is the resultant. It’s especially handy when two forces act from a single point, like two ropes tugging one hook.
🧭 Choosing your method
Triangle method — best when vectors follow one after another, like successive displacements (walk here, then there)
Parallelogram method — best when vectors act together from the same point, like two forces on one object
Either way — if the two vectors are perpendicular, skip the drawing and go straight to Pythagoras + trig
Subtracting vectors
Subtracting is just addition in disguise. To work out a − b, first flip b so it points the opposite way — that gives you −b — then add it to a with the triangle method exactly as before.
To subtract, reverse b into −b, then add head-to-tail. The resultant is a − b.
Multiplying a vector by a scalar
Multiplying a vector by a scalar scales its length but keeps (or exactly reverses) its direction — the result is always another vector. Two staples of mechanics work this way: force is mass times acceleration, and momentum is mass times velocity.
Scalar × vector = vectorF = m × a · p = m × v
Resolving into components
Resolving runs the process backwards: one resultant vector is rebuilt as two perpendicular ones that together have the same effect. Because those two components form a right angle, the magnitude of the resultant comes straight from Pythagoras, and its direction from trigonometry.
Perpendicular resultantR = √(Fx2 + Fy2) · tan θ = Fy ÷ Fx
A neat sanity check: the resultant is always longer than either component but shorter than the two added end-to-end as plain numbers. If your Pythagoras answer comes out bigger than Fx + Fy, you’ve slipped a plus sign in somewhere.
Quick recap: add vectors head-to-tail (triangle) or tail-to-tail (parallelogram); subtract by reversing and adding; and for perpendicular vectors use Pythagoras for size and tan θ for direction.
WE 1
A delivery drone flies 8.0 km due east, then 6.0 km due north. Calculate the magnitude of its displacement and its direction measured from the east (horizontal).
The two legs are perpendicular, so draw the right-angled triangle and use Pythagoras for the magnitude, then tan for the angle.
The two perpendicular legs form a right-angled triangle; the diagonal is the resultant displacement.
Magnitude (Pythagoras)
R = √(8.0² + 6.0²) = √(64 + 36)
R = √100 = 10 kmdisplacement = 10 kmDirection (trig)
tan θ = opposite ÷ adjacent = 6.0 ÷ 8.0
θ = tan⁻¹(0.75)θ = 37° north of east
WE 2
A boat heads straight across a river at 12 m s⁻¹. The current flows at 5.0 m s⁻¹ perpendicular to the boat’s heading. Calculate the boat’s resultant velocity and its direction relative to its heading.
The boat’s velocity and the current are perpendicular, so the same Pythagoras-plus-trig routine applies.
Resultant speed
v = √(12² + 5.0²) = √(144 + 25)
v = √169 = 13 m s⁻¹resultant = 13 m s⁻¹Direction
tan θ = 5.0 ÷ 12
θ = tan⁻¹(0.417)θ = 23° from the headingThe current sweeps the boat 23° downstream, even though it points straight across.
💡 Top tips
Perpendicular? Calculate. For a right-angled vector triangle, Pythagoras and trig are faster and more accurate than a scale drawing
Always sketch the triangle first — it tells you which side is opposite and which is adjacent for the angle
State the direction fully — a magnitude alone is only half a vector answer; add “north of east”, “from the heading”, etc.
Reverse, don’t guess: for subtraction, physically flip the second arrow before adding
⚠ Common mistakes
Adding vector magnitudes directly — 8 + 6 is not 10; you must combine them geometrically
Forgetting the direction in the final answer, so a vector result reads like a scalar
Mixing up opposite and adjacent, so tan θ is inverted — sketch the triangle to keep them straight
Subtracting by reversing the wrong vector; it’s b that flips in a − b
Up next: Scale Diagrams — when two vectors aren’t at right angles, Pythagoras won’t help, so we draw them accurately to scale with a ruler and protractor and measure the resultant straight off the page.
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Book a free meeting and let’s work through the tricky bits together.
