Vectors in Geometry

Components, magnitude and direction

For $A(-2,1)$ and $B(4,5)$, the displacement is $(6,4)$. Its magnitude is

The point $(6,4)$ is the endpoint of a copy of that vector drawn from the origin. It is not the original endpoint $B$ unless the original starting point was the origin.

A zero vector has components $(0,0)$ and magnitude zero, with no unique direction. Every non-zero vector has a direction, but an inverse-tangent calculation needs quadrant information. The ratio $y/x$ alone cannot distinguish a vector from its negative. Bearings, Heights and Distances handles directional conventions in applications.

Adding and subtracting vectors

For $\mathbf u=(3,1)$ and $\mathbf v=(-1,4)$,

To add geometrically, place a copy of $\mathbf v$ at the head of $\mathbf u$. The resultant runs from the first tail to the final head. This is a displacement relationship, not a requirement that the arrows remain attached to the origin.

The two-step route has length $\sqrt{10}+\sqrt{17}$, while the resultant magnitude is $\sqrt{29}$. Magnitudes do not generally add because changes of direction affect the net displacement. They add for vectors pointing in the same direction, and subtraction can occur for opposite directions.

Scalar multiplication and unit vectors

Multiplying by a scalar $k$ changes components to $k\mathbf u$. Its magnitude is $|k|\,|\mathbf u|$. A negative scalar reverses direction.

For $\mathbf u=(3,4)$, $|\mathbf u|=5$ and the unit vector in its direction is

Its squared magnitude is $9/25+16/25=1$. A unit vector supplies direction without the original magnitude. There is no unit vector obtained by normalizing the zero vector, because that would divide by zero.

Non-zero vectors are parallel when one is a scalar multiple of the other. For example, $(6,-9)=-3(-2,3)$, so these vectors are parallel with opposite directions. Equal magnitudes alone do not imply parallelism.

The dot product and perpendicular directions

The dot product of $\mathbf u=(u_1,u_2)$ and $\mathbf v=(v_1,v_2)$ is

For non-zero vectors it also equals $|\mathbf u||\mathbf v|\cos\theta$, where $0\le\theta\le\pi$ is the angle between them. Therefore a zero dot product establishes perpendicular directions.

For $(3,4)$ and $(4,-3)$, the product is $12-12=0$, so the vectors are perpendicular. The zero vector has zero dot product with every vector algebraically, but it has no direction; do not use it to infer a right angle between two directed segments.

For $(1,0)$ and $(1,\sqrt3)$, the dot product is $1$, and the magnitudes are $1$ and $2$. Thus $\cos\theta=1/2$ and the angle is $60^\circ$.

Points dividing a segment

A point on the line through distinct points $A,B$ can be written

When $0\le t\le1$, it lies on the segment. The midpoint uses $t=1/2$. Values outside that interval lie on extensions of the segment.

If $AP:PB=m:n$ for positive $m,n$, then $t=m/(m+n)$ and

For $A(1,2)$, $B(10,8)$ and ratio $AP:PB=2:1$, the point is

The first segment is twice the second. The weights in the alternative formula are opposite the corresponding endpoint labels; using the displacement form helps avoid reversing them.

A vector proof of the centroid ratio

Let a triangle’s vertex position vectors be $\mathbf a,\mathbf b,\mathbf c$. The midpoint of $BC$ is $\mathbf m=(\mathbf b+\mathbf c)/2$. Define

Then

So $G$ lies two thirds of the way from $A$ to the midpoint of $BC$. Repeating the same calculation from $B$ and $C$ places the same $G$ on all three medians. This proves concurrency and the $2:1$ ratio simultaneously.

For vertices $(0,0)$, $(6,0)$ and $(0,3)$, the centroid is $(2,1)$. The algebraic proof applies beyond this particular triangle because the vertex vectors were arbitrary.

A perpendicularity proof without slopes

Let adjacent side vectors of a parallelogram be $\mathbf u$ and $\mathbf v$. Its diagonals have directions $\mathbf u+\mathbf v$ and $\mathbf u-\mathbf v$, up to a reversal of direction. Their dot product is

In a rhombus, adjacent side lengths agree, so the product is zero and the diagonals are perpendicular. This proof handles vertical directions without undefined slopes and explains exactly where the equal-side hypothesis is used.

Related topics

Coordinate Proofs uses vector differences to establish general geometry results. Translation and Dilation become concise vector rules, while Coordinate Axes and Coordinate Planes in 3D Space extends the coordinate setting.