Represent and combine plane vectors exactly

A vector describes magnitude and direction

A directed segment from A to B represents the vector AB. Translating the entire segment does not change its vector. Equal vectors have equal components, hence equal magnitude and direction when nonzero. A position vector OA depends on the chosen origin; the displacement from A to B does not. The zero vector has magnitude zero and no defined direction

Use coordinate columns or unit-basis notation

The basis vectors give the horizontal and vertical components. For A=(−2,1) and B=(3,−1), reversing the displacement changes both component signs. The two displacements are opposite vectors

For A=(−2,1) and B=(3,−1), find the horizontal component of AB

For A=(−2,1) and B=(3,−1), find the vertical component of BA

What happens to AB if both endpoints are translated by the same vector t?

Add head to tail or by components

To add u and v, place the tail of v at the head of u. The sum runs from the first tail to the final head. With common tails it is the diagonal of the parallelogram. Components add independently. Subtraction u−v means add −v; with common tails it points from the head of v to the head of u.

Let u=(2,−3), v=(−1,4). Find the second component of 3u−2v

Scalar multiplication and magnitude

For scalar λ, λ(x,y)=(λx,λy). If λ>0, direction is unchanged; if λ<0, it reverses. Length is multiplied by |λ|. If λ=0 the result has no direction. Algebraic rules such as u+v=v+u and λ(u+v)=λu+λv follow component by component

Magnitude is a nonnegative length

Magnitude is not the sum of the components or the sum of their absolute values. A unit vector in the same direction exists only for a nonzero vector. The two unit vectors parallel to a nonzero vector have opposite signs

Find the magnitude of v=(−6,8)

For v=(−6,8), find the first component of the unit vector in the same direction

Exact irrational magnitudes

For v=(2,2), magnitude is 2√2 and the unit vector is (√2/2,√2/2). In a structured answer, write magnitude a√d with d square-free and use d=1 for rational lengths. If |v|=a√d then v/|v|=(x√d/(ad),y√d/(ad)). This also handles one zero component

For v=(2,2), give the coefficient of √2 in the first unit-vector component

Can the zero vector be normalised to a unit vector in its direction?

Convert between direction and components

For magnitude r>0 and direction θ measured anticlockwise from the positive x-axis, v=(r cosθ,r sinθ). Conversely both signed ratios cosθ=x/r and sinθ=y/r identify θ modulo 2π. Use an appropriate quadrant or an atan2 function; a plain inverse tangent of y/x can choose the wrong quadrant and fails when x=0.

For v=(−3,4), let θ be its direction measured anticlockwise from the positive x-axis. Find cosθ

For v=(0,−7), let θ be its direction measured anticlockwise from the positive x-axis. Give θ/π for 0≤θ<2π

Parallel vectors and scalar multiples

Nonzero u,v are parallel exactly when v=λu for a nonzero scalar λ. The sign distinguishes same versus opposite direction. Algebraically the zero vector is a scalar multiple of every vector, but it has no direction. Use a nonzero coordinate to recover λ and verify the other coordinate; do not divide by zero

For u=(0,3), v=(0,−6), find λ with v=λu.

Linear combinations and independent directions

A linear combination αu+βv scales and adds the two vectors. If u,v are nonparallel, every plane vector has a unique such representation. Equating both coordinates gives a simultaneous system. Parallel vectors cannot form a basis of the plane: they generate at most one line

For u=(1,2), v=(2,1) and w=(7,8), solve w=αu+βv. Find α

For u=(1,2), v=(2,1) and w=(7,8), solve w=αu+βv. Find β

Solve a vector equation component by component

For nonzero scalar k, kz+a=b gives z=(b−a)/k. Subtract both known components before dividing. This is scalar division, not a general operation of dividing one vector by another

Solve 2z+(1,−3)=(7,5). Find the second component of z.

Resultant displacement differs from path distance

A journey of 3 km east followed by 4 km north has displacement (3,4)km, magnitude 5 km and travelled distance 7 km. Reversing a leg changes its vector sign but never makes its path length negative. Use consistent units and state the chosen axes

A traveller goes 5 km east and then 2 km west. Find the total path distance in kilometres

Prove vector identities and draw them, ungraded

Prove AB+BC=AC using position vectors, then justify the head-to-tail and parallelogram rules. Explain why shifting the coordinate origin does not change a displacement. For A=(−2,1),B=(3,−1),C=(4,2), sketch and label the two-leg path and its resultant. This proof and sketch are ungraded

Model: intermediate positions cancel

AB+BC=(b−a)+(c−b)=c−a=AC. Here (5,−2)+(1,3)=(6,1). Head-to-tail construction follows these endpoint differences. For common tails, the fourth parallelogram vertex is u+v; either order reaches it. Replacing every position by its value minus the new-origin position leaves each difference unchanged

Prove uniqueness and assess a displacement model, ungraded

Explain why nonparallel u,v give unique coefficients in αu+βv=w. Then assess a model that adds two travel vectors on a flat map: state the information needed, distinguish path length from displacement and describe how you would check it. Independent proof and modelling are ungraded

Model: subtract representations and check assumptions

Let u and v be nonzero, nonparallel vectors. If αu+βv=α′u+β′v, then (α−α′)u=−(β−β′)v. A nonzero coefficient would make the nonzero vectors parallel, contradicting the hypothesis; hence both coefficient differences vanish. A travel model needs one coordinate frame, units, directions and leg lengths. Sum the displacement components but add leg magnitudes for total distance. Check the predicted endpoint against observations; curved terrain, inaccurate headings or unrecorded detours can invalidate the planar model