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Vector applications

Cross Product

Given u = $u_1$i + $u_2$j + $u_3$k, and v = $v_1$i + $v_2$j + $v_3$k, the vector cross product is:

u × v = ($u_2v_3 - u_3v_2$)i + ($u_3v_1 - u_1v_3$)j + ($u_1v_2 - u_2v_1$)k

The magnitude of the vector product: |u × v| = |u| |v| sinθ


A point on a plane with position vector r satisfies the vector equation of a plane: $r↖{→} = a↖{→}+αu↖{→}+βv↖{→}$, where $a↖{→}$ is the position vector of a known point on that plane, and $u↖{→}$ and $v↖{→}$ are two non-collinear vectors lying on that plane.
A different, yet equivalent way of defining a vector equation of a plane is based on its normal vector. A vector is normal to a plane if it is perpendicular to all vectors on that plane. Hence, if we know the positional vector $a↖{→}$ of a point on our plane $π$ and a normal vector to $π$, we know that if a point of positional vector $r↖{→}=(\table x;y;z)$ lies on $π$ it will satisfy the condition: $$n↖{→}⋅(r↖{→}-a↖{→})=0$$ $$n↖{→}⋅r↖{→} = n↖{→}⋅a↖{→}$$ Another equation (called the Cartesian equation of the plane) for $π$ can therefore be: $$n_1x+n_2y+n_3z=n↖{→}⋅a↖{→}$$


The angle between two lines is: $θ = arccos(|{u⋅v}/{|u|⋅|v|}|)$

The angle between a line parallel to u and a plane with normal vector n is given by:

$θ = arcsin(|{u⋅n}/{|u|⋅|n|}|)$, where 0 ≤ θ ≤ 90°

The angle between a plane with normal vector m and a plane with normal vector n is equal to the angle between the lines in the direction of the vectors m and n, and is given by:

$θ = arccos(|{m⋅n}/{|m|⋅|n|}|)$, where 0 ≤ θ ≤ 90°


The shortest distance from a point to a plane is measured along a line perpendicular to the plane which contains the point.

Example: distance from a point to a plane

Point $A(1, -2, 1)$ lies on a perpendicular to a plane $π$ with Cartesian equation $x + y - z + 2 = 0$


Step 1: Find the vector equation of the line, $L$, which is perpendicular to the plane and passes through $A$.

r = (1 + λ)i + (-2 + λ)j + (1 - λ)k

Step 2: Find the point of intersection $P$ of the line $L$ and the plane $π$.

$P$ ∈ $L$ ⇒ $P(1 + λ, -2 + λ, 1 - λ)$

$1 + λ - 2 + λ + 1 - λ = -2$, ∴ $λ = -2$

$L$ intersects π at $P$(1 - 2, -2 - 2, 1 - -2) = $P$(-1, -4, 3)

Step 3: Find the distance from $A$ to $P$:

The distance between two points is: $√{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AP = √{(1--1)+(-2--4)+(1-3)} = √{(2)+(2)+(-2)} = √2$


All lines will intersect with any plane at a single point, provided the line is not parallel to the plane or on the plane.

The point of intersection can be found by simultaneous equations, to find the three Cartesian coordinates in common between the vector equation which describes the line and the Cartesian equation which describes the plane.

Example: intersection of line and plane

The line $l$ with vector equation r = i + j + 3k + λ(2i + j + k) and the plane π with Cartesian equation $2x + y - z + 3 = 0$ intersect at a point. Find the coordinates of the point of intersection.


P ∈ $l$ ⇒ P(1 + 2λ, 1 + λ, 3 + λ)

P ∈ $π$ ⇒ 2(1 + 2λ) + (1 + λ) - (3 + λ) + 3 = 0

= (2 + 1 - 3 + 3) + (4 + 1 - 3)λ = 0

$λ = {-3}/2$ ⇒ $P(1 - 3, 1 - 3/2, 3 - 3/2)$

The point of intersection is: $P(-2, -1/2, 3/2)$

Intersection of two lines

Lines in 3D space may not be parallel, and yet have no point in common. These are skew lines.

Skew lines do not define a plane, but lie in parallel planes.

Example: intersection of 2 lines

Do the lines $x+3=y=2z-1$ and $3x+6=2y-2=z+4$ intersect?

Solve simultaneously for x and y:

$3x+6=2(x+3)-2$, so $x=-2$, $y=1$

Solve for z in both equations: $2z - 1=y=1$, so $z = 1$

$z+4=2y-2=0$, so $z=-4$

Since $1≠-4$, the lines do not intersect.

Vector Art

Vector art, or vector graphics, is a way of drawing and colouring images with computers. All the points, lines, curves and shapes (polygons) are described by mathematical expressions.

Vector graphics consist of precise control points, with a path (a vector). These paths can be assigned colours, shapes, and thickness.

An important advantage of vector art over bitmaps (images created by a series of pixels) is the ability of the computer to use the vector expressions to zoom the image in or out without losing any resolution.

Vector graphics illustration
3D landscapes based on vector graphics can be developed and manipulated by computing software

Vector graphics are widely used in 2D image creation, and in 3D applications, such as CAD (computer aided design).

Content © Andrew Bone. All rights reserved. Created : April 1, 2015

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