AP = + r a a u P a r B ( ) ( ) ( x,y,z ) = ( a 1 , a - - PDF document

1 Vector Space

• Vector has a direction and a magnitude, not location
• Vectors can be defined by their I, j, k components
• Point (position vector) is a location in a coordinate system
• dot product of two vectors is v1.v2=|v1||v2|cosθ (scalar)
• cross product of two vectors is perpendicular to both vectors

and its magnitude is |v1 x v2| =|v1||v2|sinθ (scalar)

What is v2 (red) - v1 (blue)?

Do you know how to get ..

• Equation of a line passing through two points, A and B (an

equation which gives the position vector of any point on the line)

• Equation of a plane passing through three non-collinear points A, B

and C (an equation which gives the position vector of any point on a plane)

• Equation of a plane with normal vector N and passes through a

point A

• Distance of a point to a plane
• Intersection of two planes

Line Equation

• A, B are two known points on the line whose position

vectors are a and b

• u is a vector obtained by subtracting A and B (a and b)
• An arbitrary point P (position vector r) on the line is the sum
• f A (represented by position vector) and a scaled version of

u (a vector)

u A B P

• a

b r (x,y,z) = (a1, a2, a3) + λ(u1, u2, u3)

u a a r

AP

• λ

+ = + =

→

λ = − = − = −

3 3 2 2 1 1

u a x u a y u a x

Plane Equation

Given points A, B, C

AC AB AP µ λ

→ → →

+ =

( ) ( ) ( ) ( ) ( ) ( )

3 3 3 3 3 2 2 2 2 2 1 1 1 1 1

a c a b a z a c a b a y a c a b a x − + − + = − + − + = − + − + = µ λ µ λ µ λ

2

Projection of vector a on normal vector n is |a|cos0, or is a ‘dot product’ normalised n

a.n Given 3 points A, B, C on the plane, normal vector can be calculated as

Plane Equation

P

Given a point on the plane, and the plane’s normal vector n, then the plane equation can be obtained using the fact that every vector in the plane should be perpendicular to n

|n| nxx+nyy+nzz+d=0; where d = -n.a, n= (nx,ny,nz) So

( )

= − = −

• =
• →

a n r n a r n AP n

• →

→

× AC AB

Plane Equation

through

• rigin

not through

• rigin

Distance of Point A to Plane

• The distance of A to the plane is the projection of

AP on the plane’s normal vector n

• Given plane equation the distance is

n

p r n =

• (

)

n n a r

• −

Distance of Point A to Plane

3 Intersection of Planes

Line equation should be of the form The intersection line should be parallel to vector Given two plane equations:

So to determine the line equation we need to find a point (any point) on the line (or on both planes), e.g., let the planes intersect with the (x,y) plane (z=0) to reduce variables and solve simultaneous equations for x and y.

u a r

• λ

+ = p r m =

• q

r n =

• r

m

• ×