Preview

Vector Addition

Add vectors graphically and by components

Here is an arithmetic problem with three correct answers: what is 3 + 4? Seven — if the numbers are scalars. Five — if they are the legs of the famous 3-4-5 right triangle, pointing east and north. One — if they point in opposite directions. Same numbers, three sums: the difference is that these numbers carry DIRECTIONS, and direction changes the rules of addition itself. That is the whole idea of a vector. Forces, velocities, displacements all behave this way, and the universe keeps the books with a beautifully simple rule: add the east-west parts separately, add the north-south parts separately, then let Pythagoras reassemble the total. The geometric picture — arrows chained tip-to-tail, or two arrows spanning a parallelogram — is that same arithmetic wearing a costume, and this lab lets you flip between the costume and the calculation until they obviously agree. It also hands you a meter you may not have met: the dot product, which reads zero the instant two vectors are perpendicular and maxes out when they align. Your mission in the lab: a 10 N force pulls east, an 8 N force pulls north, and your intuition will whisper a confident wrong answer for their combined strength. The correct answer, 12.806 N, is shorter than your whisper and longer than either force alone — sandwiched by a ceiling (18 N, reached only when the pulls align) and a floor (2 N, when they oppose). Find where the real answer lives, and why.

What you'll be able to do

  • Add vectors analytically with the panel's own numbers: at the default state (A = 10 N at 0°, B = 8 N at 90°) the rows read Rx = 10.00, Ry = 8.00, |R| = 12.806, θR = 38.7° — 'we simply add them component by component' (OpenStax §2.3 verbatim), then Pythagoras √(10²+8²) = √164 and atan2(8,10) for the direction
  • Bound every vector sum by the triangle inequality |A|−|B| ≤ |R| ≤ |A|+|B| with the preset bookends: Parallel (5 N + 3 N both at 30°) is the ONLY case where magnitudes add — |R| = 8.000 exactly — and Opposing (5 N east, 3 N west) collapses the sum to |R| = 2.000, shorter than either vector
  • Read the dot product as a parallelism meter: sweeping Vector 2's angle at defaults walks A·B through +80.000 (0° apart, = |A||B|), 0.000 (exactly 90° apart — the quiz Q2 signature of perpendicularity), and −80.000 (180° apart), matching A·B = |A||B|cosθ at every stop

Formulas

R=A+B\vec{R} = \vec{A} + \vec{B}
Vector sum (resultant)
Rx=Ax+Bx=AcosθA+BcosθBR_x = A_x + B_x = A\cos\theta_A + B\cos\theta_B
x-components
R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}
Resultant magnitude

Make a prediction

Default state: Vector A is 10.0 N pointing east (0°), Vector B is 8.0 N pointing north (90°). The panel will compute the resultant R = A + B. What does the |R| row read?

No grading here — commit to a guess, then scroll down and test it yourself.

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Check

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Your prediction

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Answer: 12.806 N — components add separately (Rx = 10 + 0, Ry = 0 + 8), then Pythagoras: |R| = √(10² + 8²) = √164

Vectors add component by component — 'we simply add them component by component' (OpenStax §2.3, verbatim): the east pull has zero northward part and the north pull has zero eastward part, so Rx = 10.00 and Ry = 8.00, and the resultant's length is the hypotenuse √164 = 12.806 N at atan2(8,10) = 38.7° north of east. Option B is the scalar carryover — and the sim's own misconception card answers it verbatim: '|A + B| ≠ |A| + |B| in general… equality only when A and B point in exactly the same direction.' The 18 N figure is not nonsense; it is the CEILING — drag Vector B's angle from 90° down to 0° and |R| really does climb to 18.000, but only when the pulls align (the Parallel preset: 5 + 3 = 8.000 exactly). Option C picks the floor instead: 2 N is what you get when the vectors are ANTIPARALLEL (the Opposing preset: 5 N east vs 3 N west → 2.000 N), not when they are perpendicular — perpendicular vectors don't cancel at all, since neither has any component along the other. Every sum lives inside the triangle inequality |A|−|B| ≤ |R| ≤ |A|+|B|, and 12.806 is where a right angle puts it.

Quiz (0/3)

Add a 10N east vector and an 8N north vector. What is the resultant magnitude and angle?

Three forces balance (net = 0). If two are known, how do you find the third?

A river flows east at 3 m/s. A swimmer swims north at 4 m/s relative to the water. What is their speed relative to the ground?

You can now

  • Add vectors analytically with the panel's own numbers: at the default state (A = 10 N at 0°, B = 8 N at 90°) the rows read Rx = 10.00, Ry = 8.00, |R| = 12.806, θR = 38.7° — 'we simply add them component by component' (OpenStax §2.3 verbatim), then Pythagoras √(10²+8²) = √164 and atan2(8,10) for the direction
  • Bound every vector sum by the triangle inequality |A|−|B| ≤ |R| ≤ |A|+|B| with the preset bookends: Parallel (5 N + 3 N both at 30°) is the ONLY case where magnitudes add — |R| = 8.000 exactly — and Opposing (5 N east, 3 N west) collapses the sum to |R| = 2.000, shorter than either vector
  • Read the dot product as a parallelism meter: sweeping Vector 2's angle at defaults walks A·B through +80.000 (0° apart, = |A||B|), 0.000 (exactly 90° apart — the quiz Q2 signature of perpendicularity), and −80.000 (180° apart), matching A·B = |A||B|cosθ at every stop

Step-by-step

  1. Use the magnitude and angle sliders to set each vector.
  2. The resultant vector (amber) updates live.
  3. Switch between the Head-to-Tail and Parallelogram constructions — the resultant never moves.
  4. Verify against the data panel that the geometric construction gives the same answer as component addition.

Key formulas

  • R=A+B\vec{R} = \vec{A} + \vec{B}Vector sum (resultant)
  • Rx=Ax+Bx=AcosθA+BcosθBR_x = A_x + B_x = A\cos\theta_A + B\cos\theta_Bx-components
  • R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}Resultant magnitude

Frequently asked questions

Add a 10N east vector and an 8N north vector. What is the resultant magnitude and angle?
R = √(10²+8²) = √164 ≈ 12.8N; θ = arctan(8/10) ≈ 38.7° north of east.
Three forces balance (net = 0). If two are known, how do you find the third?
F₃ = −(F₁ + F₂); add F₁ and F₂ first, then negate the resultant.
A river flows east at 3 m/s. A swimmer swims north at 4 m/s relative to the water. What is their speed relative to the ground?
V_ground = √(3²+4²) = 5 m/s at arctan(3/4)=37° east of north.