Turn endpoints into a directed quantity

Try

A delivery cart moves from A=(1,−2) metres to B=(4,2) metres on a floor plan. Its displacement is (3,4) metres: three right and four up. Its displacement magnitude is 5 metres by the Pythagorean theorem. The final position is (4,2), which is not the displacement.

Draw an arrow from A to B. You may instead list the labeled endpoints and state “arrow from A to B; horizontal change 3, vertical change 4.” Both formats preserve the directed geometry. An unlabelled segment or the number 5 alone omits essential information.

Predict

Predict the displacement from B back to A. Both component signs reverse, but the magnitude remains 5 metres. Predict another arrow that represents the original displacement while starting at C=(−2,0). Its endpoint must be C+(3,4)=(1,4).

Two equal vectors need not begin at the same point. Equality concerns their directed changes. Conversely, two arrows of the same length are not equal vectors when their directions differ.

Experiment

Check both arrows in the equal-scale diagram. Count horizontal and vertical changes rather than measuring the screen with a ruler.

Reverse one arrow and record the new components. Then translate both endpoints by (2,−1) and check that their difference is unchanged. These are separate operations: reversal changes the vector, translation of both endpoints does not.

Understand

For an initial point P=(p₁,p₂) and terminal point Q=(q₁,q₂), the vector from P to Q is Q−P=(q₁−p₁,q₂−p₂). Subtract initial from terminal, in that order. Its magnitude is √((q₁−p₁)²+(q₂−p₂)²). Translating both points by (h,k) leaves each difference unchanged, proving that the vector is independent of its chosen starting position.

A nonzero vector can also be described by magnitude r>0 and direction θ, measured here counterclockwise from the positive horizontal axis. Right-triangle trigonometry gives components (r cosθ,r sinθ). For magnitude 6 and direction 60°, these are (3,3√3). The identity cos²θ+sin²θ=1 verifies magnitude r. At zero magnitude there is no unique direction.

Velocity is a vector whose components have units of distance per time. Constant velocity v=(2,−1) metres per second for 4 seconds gives displacement 4v=(8,−4) metres. Starting from (3,2) metres gives final position (11,−2) metres. The speed is √5 metres per second: it records magnitude but not direction. This model assumes velocity is constant over the time interval; a changing velocity cannot generally be handled by multiplying its initial value by the total time.

Apply

When interpreting a vector problem, label whether each ordered pair is a position, displacement or velocity. Include units in each step and use the initial position only when a final position is requested.

Your construction should make reversing an arrow visibly different from translating it. If you provide a text alternative, specify the ordered endpoints; this includes the arrow direction and gives exact data against which the diagram can be checked.

Continue

This school coordinate work addresses N-VM.A.1–3 without requiring vector-space theory. The same geometry will support forces, velocities and later linear algebra, but those interpretations need their own physical assumptions.

Next, add two vectors with different directions. The magnitude of their resultant usually differs from the sum of the two magnitudes. Constructing the arrows before calculating helps reveal why.