MCV4UB Unit 7 Key Questions - Unit 7 Lesson 20 Assessment...

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Unit 7: Lesson 20 Assessment TASK 1: KNOWLEDGE & UNDERSTANDING 1. a) u∙ v = ( 2, 4 ) ( 1, 3 ) = ( 2 ) ( 1 ) + ( 4 ) ( 3 ) = 2 + 12 = 14 | u | = ( 2 ) 2 + ( 4 ) 2 = 20 | v | = ( 1 ) 2 + ( 3 ) 2 = 10 ¿ ( 14 20 10 ) = 8.13 ° cos θ = u∙ v | u || v | = 14 20 10 →θ = cos ¿ b) p∙ q = ( 1,4,5 ) ( 3, 1,3 ) = ( 1 ) ( 3 ) + ( 4 ) ( 1 ) + ( 5 ) ( 3 ) =− 3 4 + 15 = 8 | p | = ( 1 ) 2 + ( 4 ) 2 + ( 5 ) 2 = 42 | q | = ( 3 ) 2 + ( 1 ) 2 + ( 3 ) 2 = 19 ¿ ( 8 42 19 ) = 73.55 ° cos θ = p∙ q | p || q | = 8 42 19 →θ = cos ¿ 2. Let vector u represent perpendicular vector to 6 x 3 y + 2 = 0 u =( 6, 3 ) slopeof uis m = 3 6 = 1 2 . ∴ The slope of the vector that is perpendicular to 6 x 37 + 2 = 0 is 1 2 . 3. To find another point, substitute any real number for t : If t = 2 then w = ( 4, 1,3 ) + 2 ( 2,1,7 ) = ( 4, 1,3 ) + ( 4,2,14 ) = ( ( 4 4 ) , ( 1 + 2 ) , ( 3 + 14 ) ) = ( 0,1,17 ) →alternate point To find an alternate direction vector use scalar multiple of current direction vector: Let t = 2: 2 ( 2,1,7 ) = ( 4,2,14 ) →alternate directionvector For this alternate vector equation let it be z and let the parameter ¿ s . z = ( 0,1,17 ) + s (− 4,2,14 ) 4. a) a∙ b = ( 10, 4,1 ) ( 3, 2,2 ) = ( 10 ) ( 3 ) + ( 4 ) ( 2 ) + ( 1 ) ( 2 ) = 30 + 8 + 2 = 40 | a | = ( 10 ) 2 + ( 4 ) 2 + ( 1 ) 2 = 117 | b | = ( 3 ) 2 + ( 2 ) 2 + ( 2 ) 2 = 17
( 40 117 17 ) = 26.25 ° ´ ¿ ¿¿ cos θ = a∙ b | a || b | = 40 117 17 →θ = cos ¿ b) p∙ q = ( 0,4, 3 ) ( 7, 2,1 ) = ( 0 ) ( 7 ) + ( 4 ) ( 2 ) + ( 3 ) ( 1 ) = 0 8 3 =− 11 | p | = ( 0 ) 2 + ( 4 ) 2 + ( 3 ) 2 = 25 = 5 | q | = ( 7 ) 2 + ( 2 ) 2 + ( 1 ) 2 = 54 ¿ ( 11 5 54 ) = 107.42 ° OBTUSE cos θ = p∙ q | p || q | = 11 5 54 →θ = cos ¿ 5. ( x, y ) = ( 3,2 ) + t ( 2,4 ) We know a point on the line is (3, 2) . We can find the slope of the line using the direction vector, let that be u . The slope of u will then be m . u = ( 2, 4 ) m = 4 2 = 2 We can now use the formula y = m ( x x 1 ) + y 1 to find the equation of the line. y = m ( x x 1 ) + y 1 y = 2 ( x 3 ) + 2 y = 2 x 6 + 2 y = 2 x 4 Rearrange equation to get the scalar equation: 2 x y 4 = 0 6. A direction vector parallel to the y-axis is u =( 0,1,0 ) . Parallel to the y-axis because x, and z coordinates are 0. ∴Vector equation is: ( x, y ,z ) = ( 1,0,3 ) + t ( 0,1,0 ) TAKS 2: THINKING QUESTIONS 7. To find if a pair of vectors are perpendicular, the dot product must ¿ 0 . u∙ v = ( 4, 1,1 ) ( 1,3, 1 ) = ( 4 1 ) + ( 1 3 ) + ( 1 1 ) = 4 3 1 = 0 u v u∙ w = ( 4, 1,1 ) ( 2, 5, 13 ) = ( 4 2 ) + ( 1 5 ) + ( 1 13 ) = 8 + 5 13 = 0 u w v ∙ w = ( 1,3, 1 ) ( 2, 5, 13 ) = ( 1 2 ) + ( 3 5 ) + ( 1 13 ) = 2 15 + 13 = 0 v w ∴All the vectors are perpendicular to one another.
8. Find components of direction vector PQ :

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