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Vector Space. What is v2 (red) - v1 (blue)?. 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|cos q (scalar)
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Vector Space What is v2 (red) - v1 (blue)? • 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|cosq (scalar) • cross product of two vectors is perpendicular to both vectors • and its magnitude is |v1 x v2| =|v1||v2|sinq (scalar)
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
A u P a r B o b 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 of A (represented by position vector) and a scaled version of u (a vector) (x,y,z) = (a1, a2, a3) + l(u1, u2, u3)
Plane Equation Given points A, B, C
Plane Equation 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 Projection of vector a on normal vector n is |a|cosq, or is a ‘dot product’ normalised n a.n P r |n| nxx+nyy+nzz+d=0; where d = -n.a, n= (nx,ny,nz) So Given 3 points A, B, C on the plane, normal vector can be calculated as
Plane Equation not through origin through origin
Intersection of Planes Line equation should be of the form Given two plane equations: The intersection line should be parallel to vector 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.