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Lecturenotes7-9February2011

# Lecturenotes7-9February2011 - Theorem 3.2.1 If v is a...

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Theorem 3.2.1: If v is a vector in R n , and if k is any scalar, then (a) || v || ≥ 0 (b) || v || = 0 if and only if v=0 (c) || kv || = | k ||| v || Proof: Exercise, (Hint: express u as u = ( u 1 , u 2 , . . . , u n ) and use the definition of norm) Applying the definition of length to a vector between two points, one can find the distance between two points. Let P 1 = ( x 1 , y 1 ) , P 2 = ( x 2 , y 2 ) then d ( P 1 , P 2 ) = || ~ P 1 P 2 || = p ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 . Definition: If P = ( u 1 , u 2 , . . . , u n ) and Q = ( v 1 , v 2 , . . . , v n ) are two points in R n then d ( P, Q ) = || ~ PQ || = p ( v 1 - u 1 ) 2 + . . . + ( v n - u n ) 2 Unit vector: unit vector= vector with length 1 A way to obtain a unit vector: Let v be a vector, then v || v || will be a unit vector. Proof: Consider || v || v || || = || 1 || v || v || . Note that 1 || v || is a scalar and by Theorem 3.2.1 (c) || 1 || v || v || = | 1 || v || ||| v || = 1 || v || || v || by Theorem 3.2.1 (a). Hence || v || v || || = 1 in R n we have standard unit vectors e 1 = (1 , 0 , 0 , . . . , 0) , e 2 = (0 , 1 , 0 , 0 , . . . , 0) , e 3 = (0 , 0 , 1 , 0 , . . . , 0) , . . . , e n = (0 , 0 , . . . , 0 , 1) . If v = ( v 1 , v 2 , . . . , v n ) is a vector in R n then v = v 1 e 1 + v 2 e 2 + . . . + v n e n . That is any vector in R n can be written as a linear combination of standard unit vectors. Relation of dot product to length(= norm) of a vector Recall the definition of dot product: Given two vectors u = ( u 1 , . . . , u n ) , v = ( v 1 , . . . , v n ) in R n u · v = u 1 v 1 + u 2 v 2 + . . . + u n v n Consider v · v = v 2 1 + v 2 2 + . . . + v 2 n and compare with || v || = p v 2 1 + v 2 2 + . . . + v 2 n , hence the relation is || v || = v · v Theorem 3.2.2:

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