This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 5 Inner Product Let us start with the following problem. Given a point P R 2 and a line L j R 2 , how can we find the point on the line closest to P ? Answer : Draw a line segment from P meeting the line in a right angle. Then, the point of intersection is the point on the line closest to P . Let us now take a plane L j R 3 and a point outside the plane. How can we find the point u L closest to P ? The answer is the same as before, go from P so that you meet the plane in a right angle. Observation In each of the above examples we needed two things: A1 We have to be able to say what the length of a vector is. B1 Say what a right angle is. Both of these things can be done by using the dotproduct (or inner product) in R n . Definition. Let ( x 1 ,x 2 ,...,x n ) , ( y 1 ,x 2 ,...,y n ) R n . Then, the dotproduct of these vectors is given by the number: (( x 1 ,x 2 ,...,x n ) , ( y 1 ,x 2 ,...,y n )) = x 1 y 1 + x 2 y 2 + ... + x n y n . 27 28 LECTURE 5. INNER PRODUCT The norm (or length) of the vector ~u = ( x 1 ,x 2 ,...,x n ) R n is the non negative number: k u k = p ( u,u ) = q x 2 1 + x 2 2 + ... + x 2 n . Examples Example. (a) ((1 , 2 , 3) , (1 , 1 , 1)) = 1 + 2 3 = 0 (b) ((1 , 2 , 1) , (2 , 1 , 3)) = 2 + 2 + 3 = 7 Perpendicular Because,  x 1 y 1 + x 2 y 2 + ... + x n y n  q x 2 1 + x 2 2 + ... + x 2 n q y 2 1 + y 2 2 + ... + y 2 n or  ( u,v )  k u k k v k we have that (for u,v 6 = 0) 1 ( u,v ) k u k k v k 1 . Hence we can define: cos( ( u,v )) = ( u,v ) k u k k v k . In particular, u v ( u is perpendicular to v ) if and only if ( u,v ) = 0....
View Full
Document
 Spring '08
 Staff

Click to edit the document details