https://pau.um.es/examenes/pdf/2008_1_67.pdf

https://pau.um.es/examenes/pdf/2007_1_67.pdf

https://pau.um.es/examenes/pdf/2006_1_67.pdf

https://pau.um.es/examenes/pdf/2005_1_67.pdf

https://pau.um.es/examenes/pdf/2004_1_67.pdf

https://pau.um.es/examenes/pdf/2003_1_37.pdf

https://pau.um.es/examenes/pdf/2002_1_37.pdf

https://pau.um.es/examenes/pdf/2001_1_37.pdf

https://pau.um.es/examenes/pdf/2000_1_37.pdf

https://pau.um.es/examenes/pdf/1999_1_37.pdf

https://pau.um.es/examenes/pdf/1998_1_37.pdf

https://pau.um.es/examenes/pdf/1997_1_37.pdf

https://pau.um.es/examenes/pdf/1996_1_37.pdf

https://pau.um.es/examenes/pdf/2009_1_58.pdf

https://pau.um.es/examenes/pdf/2008_1_58.pdf

https://pau.um.es/examenes/pdf/2007_1_58.pdf

https://pau.um.es/examenes/pdf/2006_1_58.pdf

https://pau.um.es/examenes/pdf/2005_1_58.pdf

https://pau.um.es/examenes/pdf/2004_1_58.pdf

https://pau.um.es/examenes/pdf/2003_1_28.pdf

https://pau.um.es/examenes/pdf/2002_1_28.pdf

https://pau.um.es/examenes/pdf/2001_1_28.pdf

https://pau.um.es/examenes/pdf/2000_1_28.pdf

https://pau.um.es/examenes/pdf/1999_1_28.pdf

https://pau.um.es/examenes/pdf/1998_1_28.pdf

https://pau.um.es/examenes/pdf/1997_1_28.pdf

https://pau.um.es/examenes/pdf/1996_1_28.pdf

In this tutorial, we will use vector methods to represent lines and planes in 3-space.

Displacement Vector

The displacement vector v with initial point (x₁,y₁,z₁) and terminal point (x₂,y₂,z₂) is

v = (x₂-x₁,y₂-y₁,z₂-z₁)

That is, if vector v were positioned with its initial point at the origin, then its terminal point would be at (x₂-x₁,y₂-y₁,z₂-z₁).

Example

The vector v with initial point (-1,4,5) and final point (4,-3,2) is

v = ( 4-(-1),-3-4,2-5 ) = (5,-7,-3)

Parametric Equations for a Line in 3-space

The line through the point (x₀,y₀,z₀) and parallel to the non-zero vector v = (a,b,c) has parametric equations

x = x₀ + at

y = y₀ + bt

z = z₀ + ct

Example

The line through (2,-1,3) and parallel to the vector v = (3,-7,4) has parametric equations

x = 2+3t

y = -1-7t

z = 3+4t

Notice that when t = 0, we are at the point (2,-1,3). As t increases or decreases from 0, we move away from this point parallel to the direction indicated by (3,-7,4).

If you know two points p₁ = (x₁,y₁,z₁) and p₂ = (x₂,y₂,z₂) that a line passes through, you can find a parameterization for the line. First, find the displacement vector v = (x₂-x₁,y₂-y₁,z₂-z₁). then write down parametric equations for the line through either p₁ or p₂ and parallel to v.

Equation of a Plane in 3-space

The equation of the plane containing the point (x₀,y₀,z₀) with normal vector n = (a,b,c) is

a(x-x₀)+ b(y-y₀)+c(z-z₀) = 0

Thus, the graph of the equation

ax+by+cz = d

is a plane with normal vector (a,b,c).

Example

The equation of the plane containing (2,4,-1) and normal to the vector n = (3,5,-2) is

3(x-2)+5(y-4)-2(z-(-1)) = 0

Simplifying,

3x+5y-2z = 28.

With a little extra work, we can use this procedure to find the equation of the plane defined by any three points. First, compute displacement vectors u and v between two pairs of these points. Then n = u × v is normal to the plane. Now, use one of the points and the vector n = u × v to obtain the equation of the plane.

Key Concepts

• Displacement Vector

The displacement vector v with initial point (x₁,y₁,z₁) and terminal point (x₂,y₂,z₂) is v = (x₂-x₁,y₂-y₁,z₂-z₁).

• Parametric Equations for a line in 3-space

The line through the point (x₀,y₀,z₀) and parallel to the non-zero vector v = (a,b,c) has parametric equations

x = x₀ + at

y = y₀ + bt

z = z₀ + ct

• Equation of a plane in 3-space

The equation of the plane containing the point (x₀,y₀,z₀) with normal vector n = (a,b,c) is

a(x-x₀)+ b(y-y₀)+c(z-z₀) = 0