Solving geometric problems with vectors

Using vectors to prove geometry: parallel vectors as scalar multiples, collinear points as parallel vectors through a shared point, dividing a line in a ratio, expressing figures in terms of base vectors, and comparing areas.

Pure · Vectors

Solving geometric problems with vectors

Why it works

Vectors turn geometry into algebra. Three ideas do most of the work.

Parallel. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other:

\vec{AB} = k\,\vec{CD}

. The number

k

also tells you the length ratio —

\vec{AB}

|k|

times as long as

\vec{CD}

.

Collinear. Three points

A

B

C

lie on one straight line exactly when

\vec{AB}

and

\vec{AC}

are parallel — i.e.

\vec{AB} = k\,\vec{AC}

. Parallel alone is not enough: the two vectors must also share a point (here

A

), or you have proved nothing about whether the lines are the same line. This is the workhorse for "show that

A

B

C

are collinear" and "find the value of

p

for which they lie on a straight line".

Dividing a line in a ratio. The point

P

with

AP:PB = m:n

has position vector

\mathbf{p} = \mathbf{a} + \dfrac{m}{m+n}(\mathbf{b} - \mathbf{a})

— travel the fraction

\frac{m}{m+n}

of the way from

A

B

.

To handle a figure, express every vector in terms of a couple of base vectors (say

\mathbf{a}

and

\mathbf{b}

) by reading off nose-to-tail routes, then compare. If

\vec{MN} = \lambda\,\vec{AB}

you have proved

MN \parallel AB

and found the length ratio in one line.

Areas. Two triangles that share the same apex and have their bases on the same straight line have the same height, so their areas are in the ratio of those bases (a length ratio, not its square). Only when two figures are similar — a genuine scale copy — does the area ratio become the square of the length ratio.