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

Vector Space Do you know how to get .. o Equation of a line passing through two points, A and B (an equation which gives the position vector of any point on the line) o 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) What is v2 (red) - v1 (blue)? o Equation of a plane with normal vector N and passes through a o Vector has a direction and a magnitude, not location point A o Vectors can be defined by their I, j, k components o Distance of a point to a plane o Point (position vector) is a location in a coordinate system o Intersection of two planes o dot product of two vectors is v1.v2=|v1||v2|cos θ (scalar) o cross product of two vectors is perpendicular to both vectors and its magnitude is |v1 x v2| =|v1||v2|sin θ (scalar) Line Equation Plane Equation o A, B are two known points on the line whose position vectors are a and b Given points A, B, C o u is a vector obtained by subtracting A and B (a and b) → → → µ λ AC AP = AB + o An arbitrary point P (position vector r) on the line is the sum of A (represented by position vector) and a scaled version of u (a vector) A u � → � � � = + AP = + λ r a a u P a r B ( ) ( ) ( x,y,z ) = ( a 1 , a 2 , a 3 ) + λ ( u 1 , u 2 , u 3 ) = + λ − + µ − o x a b a c a b 1 1 1 1 1 ( ) ( ) = + λ − + µ − − − − y a b a c a x a y a x a 2 2 2 2 2 = = = λ 1 2 3 ( ) ( ) = + λ − + µ − z a b a c a 3 3 3 3 3 u u u 1 2 3 1

Plane Equation Plane Equation Given a point on the not through through plane, and the plane’s origin origin Projection of normal vector n , then P vector a on the plane equation normal vector n can be obtained using is |a|cos0, or is the fact that every a ‘dot product’ vector in the plane should be normalised n perpendicular to n a.n | n | � → � � � � � � � ( ) • = • − = − = n AP n r a n r n a 0 n x x + n y y + n z z + d =0; where d = -n.a , n = ( n x , n y , n z ) So Given 3 points A, B, C on the plane, normal vector → → can be calculated as × AC AB Distance of Point A to Plane Distance of Point A to Plane o The distance of A to the plane is the projection of AP on the plane’s normal vector n n � � n = r p o Given plane equation the distance is � − � � ( ) r � a n n 2

Intersection of Planes � � � Line equation should be of the form = + λ r a u � Given two plane equations: � � � m = n = r p r q � × � The intersection line should be parallel to vector m r 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. 3