MATH 6A Lecture Notes - Lecture 3: Anticommutativity, Cross Product, Parametric Equation
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The norm in the denominator should be squared. Theorem 1. 6: let be an orthogonal set in . tthenwe can write any vector in terms of : A similar statement holds in (for orthogonal vectors) Note: we will use the vector notation and component notation. We can also use the dot product to find the equation of a plane which contains a point and is orthogonal to. If is a point in the plane, then represents the directed line segment. Since lies in the plane, it has to be orthogonal to : The solutions to this equation are exactly the points that lie on the plane. The cross product of two vectors and in is. Takes two vectors in and outputs a vector in (anticommutativity: (distributive) Example: theorem 1. 10: the cross product is orthogonal to both and . Its length is , where is the angle between and.
