# Task 3 11 in your own words describe the types of

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Task 3 11. In your own words, describe the types of intersections that can be formed by two lines in 3-space. Define each type of intersection and provide a diagram for each. 12. Given u = ( 3,7,2 ) , v = ( 6,2,1 ) w =(− 2,3,8 ) , verify the following arithmetic properties of vectors u
v =[ ( 7 ) ( 1 ) ( 2 ) ( 2 ) , ( 2 ) ( 6 ) ( 1 ) ( 3 ) , ( 3 ) ( 2 ) ( 6 ) ( 7 ) ] v =[ ( 7 4 ) , ( 12 ( 3 ) ) , ( 6 42 ) ] v = ( 3,15 , 48 ) w =(− 3,7,2 ) × (− 2,3,8 ) w = u 2 w 3 w 2 u 3 ,u 3 w 1 w 3 u 1 ,u 1 w 2 w 1 u 2 w =[ ( 7 ) ( 8 ) ( 3 ) ( 2 ) , ( 2 ) ( 2 ) ( 8 ) ( 3 ) , ( 3 ) ( 3 ) ( 2 ) ( 7 ) ] w =[ ( 56 6 ) , ( 4 ( 24 ) ) , ( 9 ( 14 ) ) ] w =( 50,20,5 ) v + w = ( 3,15 , 48 ) +( 50,20,5 ) v + w =( 53,35 , 43 ) ( v + w ) = ( 53,35 , 43 ) v + w =( 53,35 , 43 ) left side is equal to right side of equation and the equation can be verified as correct. b. u∙ ( v + w ) = u∙ v + v ∙ w ` Left side u∙ ( v + w ) = ( 3,7,2 ) [ ( 6,2,1 ) + ( 2,3,8 ) ] u∙ ( v + w ) =(− 3,7,2 ) ( 4,5,9 ) u∙ ( v + w ) =(− 12 + 35 + 18 ) u∙ ( v + w ) = 41 Right side u∙ v =(− 3,7,2 ) ( 6,2,1 ) u∙ v =(− 18,14,2 ) u∙ w =(− 3,7,2 ) (− 2,3,8 ) u∙ w =( 6,21,16 ) u∙ v + u∙ w = ( 18,14,2 ) +( 6,21,16 ) u∙ v + u∙ w = 14 + 2 + 6 + 21 + 16 18 u∙ v + u∙ w = 41 u∙ ( v + w ) = 41 u∙ v + u∙ w = 41 meaning left side is equal to right side of equation and the equation can be verified as correct. c. ( u + v ) ( u + v ) = ¿ u ¿ 2 + ¿ v ¿ 2 + 2 ( u∙ v ) Left side
( u + v ) =[ ( 3,7,2 ) + ( 6,2,1 ) ] ( u + v ) =( 3,9,3 ) ( u + v ) ( u + v ) =( 3,9,3 ) ( 3,9,3 ) ( u + v ) ( u + v ) =( 9 + 81 + 9 ) ( u + v ) ( u + v ) = 99 Right side ¿ u ¿ 2 + ¿ v ¿ 2 = ¿ 13. Describe how the dot product can be used to determine whether two vectors are perpendicular. Create a question with vectors in 3-space to illustrate this property. Be sure to solve the question as well. Task 4 2 ) ) 1 2
System of equations 6 + 3 t = 4 s [1] 1 + t = s [2] 4 t = 5 + 5 s [3] s =− 3 t 2 [4] s = t 1 [5] 5 s + 4 t =− 5 [6] [4]=[5] t 1 =− 3 t 2 4 t =− 1 t = 1 4 Sub t = 1 4 into [4] s =− 3 t 2 s =− 3 ( 1 4 ) 2 s = 3 4 2 s = 5 4 Sub t = 1 4 and s = 5 4 into equation [3] Left side 4 t ¿ 4 ( 1 4 ) ¿ 1 Right side 5 + 5 s ¿ 5 + 5 ( 5 4 ) ¿ 5 6.25 ¿ 1.25
1 1.25 Values for s and t do not satisfy equation [3] therefore there is no point of intersection and the lines are skew.