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Assessment of Learning Unit 7 1. a. u =( 2, 4 ) v =( 1, 3 ) 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 cos θ = u ⋅⃗ v |⃗ u || v | cos θ = 14 20 10 ¿ 14 200 ¿ 0.9899 θ = cos 1 ( 0.9899 )= 8.13 ° The angle is 8.13°. b.
p =(− 1,4,5 ) q =( 3, 1,3 ) 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 cos θ = p q | p || q | cos θ = 8 42 19 ¿ 8 798 ¿ 0.2832 θ = cos 1 ( 0.2832 )= 73.55 ° The angle is 73.55°. 2. 6 x 3 y + 2 = 0 Ax + By + C = 0 u =( A, B ) =( 6, 3 ) m = B A m = 3 6 m =− 1 2
The slope of the vector that is perpendicular to this scalar equation is 1 2 . 3. First, let’s find another point on the line. We can do this by substituting any real number into the equation. If t = 1: w =( 4, 1,3 )+ t (− 2,1,7 ) =( 4, 1,3 )+( 1 )(− 2,1,7 ) =( 4 2, 1 + 1,3 + 7 ) P =( 2,0 , 10 ) Now we change the direction of the vector. Do change the direction we multiply the direction (-2, 1, 7) by any real number (in this case 2): v = 2 (− 2,1,7 ) v =(− 4,2 , 14 ) Therefore an alternative vector equation for w =( 4, 1,3 )+ t (− 2,1,7 ) is w =( 2,0 , 10 )+ t (− 4,2 , 14 ) . 4. a. a =( 10 , 4,1 ) b =( 3, 2,2 ) a b =( 10 )( 3 )+(− 4 )(− 2 )+( 1 )( 2 )= 30 + 8 + 2 = 40 Since the dot product is positive, the angle between the vectors is acute. b. p =( 0,4, 3 ) q =( 7, 2,1 ) p ⋅⃗ q =( 0 )( 7 )+( 4 )(− 2 )+(− 3 )( 1 )= 0 +(− 8 )+(− 3 )=− 11 Since the dot product is negative, the angle between the vectors is obtuse. 5.
( x, y )=( 3,2 )+ t ( 2,4 ) u =( 2,4 ) A = 2 ,B = 4 Ax + By + C = 0 2 x + 4 y + C = 0 Insert the point ( 3,2 ) : 2 ( 3 )+ 4 ( 2 )+ C = 0 6 + 8 + C = 0 14 + C = 0 C =− 14 Therefore, the scalar equation for the line is: 2 x + 4 y 14 = 0 6. A direction vector that is parallel to the y-axis could have any y value, and an x and y value of zero. One example of a direction vector that is parallel to the y-axis is u = ( 0,3,0 ) . We know that a point on the line is P ( 1,0,3 ) , so the vector equation for this line could be: ( x, y ,z ) = ( 1,0,3 ) + t ( 0,3,0 ) . 7.
u =( 4, 1,1 ) v =( 1,3, 1 ) w =( 2, 5, 13 ) u ⋅⃗ v =( 4, 1,1 )⋅( 1,3, 1 ) =( 4 )( 1 )+(− 1 )( 3 )+( 1 )(− 1 ) = 4 +(− 3 )+(− 1 ) = 0 v ⋅⃗ w =( 1,3, 1 )⋅( 2, 5, 13 ) =( 1 )( 2 )+( 3 )(− 5 )+(− 1 )(− 13 ) = 2 +(− 15 )+ 13 = 0 w u =( 2, 5, 13 )⋅( 4, 1,1 ) =( 2 )( 4 )+(− 5 )(− 1 )+(− 13 )( 1 ) = 8 + 5 +(− 13 ) = 0 The dot products of all the paired vectors is zero, which means that they are all perpendicular to each other. 8. First find a direction vector: PQ = Q P =( 2,7,1 )−(− 3,5,2 )=( 2 + 3,7 5,1 2 )=( 5,2, 1 ) Use one point and the direction vector to write the vector equations: Using P (− 3,5,2 ) , the vector equation is: ( x, y ,z )=(− 3,5,2 )+ t ( 5,2, 1 ) Using Q ( 2 , 7 , 1 ) , the vector equation is: ( x, y ,z )=( 2,7,1 )+ t ( 5,2, 1 ) Now we use the vector equations to find the parametric equations: The parametric equations for ( x, y ,z )=(− 3,5,2 )+ t ( 5,2, 1 ) are:
x =− 3 + 5 t y = 5 + 2 t z = 2 t

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