Lecture5

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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 dot-product (or inner product) in R n . Definition. Let ( x 1 ,x 2 ,...,x n ) , ( y 1 ,x 2 ,...,y n ) ∈ R n . Then, the dot-product 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....
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