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Lecturenotes9-11February2011

Lecturenotes9-11February2011 - To determine the direction...

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To determine the direction of u × v we have the right hand rule; Fingers point the first vector, Palm points the second vector, tumb gives the direction of the cross product. Theorem 3.5.1 (Relationship involving Cross Product and Dot Prod- uct:) If u,v and w are vectors in 3-space, then (a) u · ( u × v ) = 0 (that is u × v and u are orthogonal) (b) v · ( u × v ) = 0 (that is u × v and v are orthogonal) (c) || u × v || 2 = || u || 2 || v || 2 - ( u · v ) 2 (Lagrange’s Identity) (d) u × ( v × w ) = ( u · w ) v - ( u · v ) w (e) ( u × v ) × w = ( u · w ) v - ( v · w ) u Proof: Write u = ( u 1 , u 2 , u 3 ) , v = ( v 1 , v 2 , v 3 ) and apply the definitions of dot prod- uct and cross product. Importance: From part (a) and (b) we see that if one needs to find a vector that is orthogonal to both vectors u and v then u × v will do the work. We will see that this property can be used to write a plane equation. Example: Let u = (1 , 0 , 2) , v = ( - 1 , 1 , 3) then u × v = i j k 1 0 2 - 1 1 3 = - 2 i - 5 j + k = ( - 2 , - 5 , 1) ( u × v ) · u = ( - 2 , - 5 , 1) · (1 , 0 , 2) = 0 , ( u × v ) · v = ( - 2 , - 5 , 1) · ( - 1 , 1 , 3) = 0 . Theorem 3.5.2 (Properties of Cross Product): If u,v and w are any vectors in 3-space and k is any scalar, then (a) u × v = - ( v × u ) (b) u × ( v + w ) = ( u × v ) + ( u × w ) (c) ( u + v ) × w = ( u × w ) + ( v × w ) (d) k ( u × v ) = ( ku ) × v = u × ( kv ) 1

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(e) u × 0 = 0 × u = 0 (f) u × u = 0 Geometric Interpretation of Cross Product and Determinants: Theorem 3.5.3 (Area of a Parallelogram) If u and v are vectors in 3-space, then || u × v || is equal to the area of the parallelogram determined by u and v. Why || u × v || = || u |||| v || sin ( θ ) ? From Theorem 3.5.1 we have Lagrange’s Identity || u × v || 2 = || u || 2 || v || 2 - ( u · v ) 2 and we have cos ( θ ) = u · v || u |||| v || u · v = || u |||| v || cos ( θ ) . Then || u × v || 2 = || u || 2 || v || 2 - ( u · v ) 2 = || u || 2 || v || 2 - || u || 2 || v || 2 cos 2 ( θ ) = || u || 2 || v || 2 (1 - cos 2 ( θ )) = || u || 2 || v || 2 sin 2 ( θ ) ⇒ || u × v || = || u |||| v || sin ( θ ) Theorem 3.5.4: (a) The absolute value of the determinant det u 1 u 2 v 1 v 2 is equal to the area of the par- allelogram in 2-space determined by the vectors u = ( u 1 , u 2 ) and v = ( v 1 , v 2 ) .
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