5.1 The Scalar Product in Rn

TT Definition 1: Let x(, x1 x 2 , , xnn ) and y (, y 1 y 2 , , y ), then the scalar product of xy and are defined

T x y x1 y 1  x 2 y 2   xnn y

Definition 2: Given a vector x in Rn , its Euclidean length is defined in terms of the scalar product 1 T 2 2 2 2 x x() x12  x   xn , and denoted by ||x || .

Definition 2: Let be vectors in . The distance between is defined as ||xy || .

First, we would like to derive some properties when the vector spaces are either R2 or R3 which we are familiar with, then we extend those properties to Rn.

Property 1: If are non-zero vectors in either or and  is the angle between them, then

xT y|| x || || y || cos( )

θ

Proof: We will prove the result for . The proof for R2 is similar. By the law of cosines,

||x y ||2  || x || 2  || y || 2  2 || x |||| y || cos  ||xy || || || cos

1 2 2 2 (||x ||  || y ||  || x  y || ) 2 1 (xxxyyy2  2 2 2 2 2 (( xy ) 2 ( xy ) 2 ( xy ) 2 )) 2 123123 11 22 33 1 (xxxyyyxxxyyy222222222222  ( 2 xyx  2 y 2 x y )) 2 123123123123 112 2 3 3

x y  x y  x y 1 1 2 2 3 3  xyT Cauchy-Schwarz Inequality: If xy and be vectors in either R2 or R3 , then

|xT y | || x || || y ||

T Since | cos( ) | 1, |x y | || x ||  || y ||  | cos( ) |  || x ||  || y ||

This is valid when x , y Rn . We will prove it later.

Definition 3: The vectors are non-zero vectors in Rn are called orthogonal if xyT  0 .

11        Example 1: Prove vectors xy2  ,    4  are orthogonal.     33   

1 T  Prove: xy1 2 3  4   1  8  9  0 , so they are orthogonal.  3

Property: If vectors are orthogonal non-zero vectors in , then

||x y ||2  || x || 2  || y || 2 .

Proof: ||xy ||2 ( xyxy )TTTTTTT (  ) ( x yxyxxxyyxyyx )(  ) || || 2 || y || 2

We call the above Pythagorean Law is a generalization of the Pythagorean Theorem.

Exercise: From Example 1, are orthogonal. Please verify that .

||xy ||2  (1  1) 2  (2  4) 2  (3  3) 2  40 ||xy ||22 || ||  1  4  9  1  16  9  40 Thus, ||x y ||2  || x || 2  || y || 2

Scalar and Vector projections in either or .

The scalar product can be used to find the component of one vector in the direction of another. Let x and y be nonzero vectors in either or .

1 Let uy a unit vector in the direction of y . We wish to find out vector projection x onto y , vector ||y || 1 p . We have xp orthogonal to . Let p u y . If we can find  , then find . ||y ||

yTT|| y ||2 x y (xpy )TTTTT  0 xypy 0 xy y  0 xy   0  ||y || || y || || y ||

xTT y1 x y p y y ||y || || y || || y ||2

xyT Scalar projection x onto y is   ||y ||

xyT Vector projection onto is py ||y ||2

Example 2: Find the distance from the point A  (1,2,1)to the plane 4x 2 y  z  1.

Solution: Find a point B in plane . Let x0, y  0  z   1, so point B (0,0, 1) is in the given plane. The normal vector n of the plane is n (4,2, 1)T . The distance from the point to the plane is the absolute of the scalar projection of AB (  1,  2,  2)T onto .

T AB n ( 1,  2,  2)(4,2,  1)T  4  4  2  6      ||n || 16 4 1 21 21

6 Thus the distance from the point to the plane is . 21

HW: 1,3,5,9,13,15.