Plane vectors and geometric arguments

A vector has magnitude and direction

The column vector (3; -2) describes a displacement three units right and two down. Translating any point by it gives the same directed displacement. Its magnitude is square root of (3² + (-2)²) = square root of 13.

For A(1,2) and B(5,3), the position vectors are a = vector OA and b = vector OB. The displacement is vector AB = b-a = (4; 1). Reversing the journey gives vector BA = - vector AB, with the same magnitude and opposite direction.

Join vectors head to tail

vector AB + vector BC = vector AC. The intermediate position cancels: (b-a) + (c-b) = c-a.

Multiplying a vector by positive 3 triples its magnitude and preserves direction. Multiplying by −2 doubles magnitude and reverses direction. Two non-zero vectors are parallel when one is a scalar multiple of the other. Equal magnitudes alone do not establish parallelism.

A point on a segment needs a position condition

If vector AP = t vector AB, then p = (1-t)a + tb. For 0 < t < 1, P lies strictly between A and B; for t > 1 it lies beyond B; for t < 0 it lies beyond A in the opposite direction. Collinearity alone does not prove that P is on the bounded segment.

The midpoint has t = 1/2, so position vector (a + b)/2. For A(1,2), B(5,3), its coordinates are (3,2.5).

Check your understanding 1 Use the stated vertex order

A, B, C, D are consecutive vertices of a parallelogram, with position vectors a, b, c, d. Express d in terms of a, b and c, then check A(1,2), B(5,3), C(7,6).

Your answer or drawing on paper

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vector AD = vector BC, so d-a = c-b and d = a + c-b. The coordinates are D(3,5). Both AD and BC are (2,3); both AB and DC are (4,1).

Magnitude is not component addition

For displacement (−3,4), magnitude is 5, not 1 or 7. The component signs describe direction. Squaring them in the Pythagorean length removes direction but preserves the correct non-negative distance.