Exam 1 Review Solutions

# Exam 1 Review Solutions - Solutions To Review Problems 3...

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Unformatted text preview: Solutions To Review Problems 3. Let ~a = (6 , 2 , 3), ~ b = (- 1 , 5 , 2) and ~ c = (1 , , 1). Find (1) ~a + 3 ~ b- 2 ~ c ; (2) ~a Â· ~ b ; (3) ~a Ã— ~ b ; (4) | ~ c | ; (5) Proj ~ c ~ b ; (6) the area of the parallelogram determined by ~a and ~ b . (7) the volume of the parallelepiped determined by ~a , ~ b and ~ c . Solution: (1) ~a + 3 ~ b- 2 ~ c = (6 , 2 , 3) + 3(- 1 , 5 , 2)- 2(1 , , 1) = (1 , 17 , 7) . (2) ~a Â· ~ b = (6 , 2 , 3) Â· (- 1 , 5 , 2) =- 6 + 10 + 6 = 10 . (3) ~a Ã— ~ b = (6 , 2 , 3) Ã— (- 1 , 5 , 2) = fl fl fl fl fl fl fl ~ i ~ j ~ k 6 2 3- 1 5 2 fl fl fl fl fl fl fl = (- 11 ,- 15 , 32) . (4) | ~ c | = âˆš 1 + 1 = âˆš 2. (5) Proj ~ c ~ b = ~ c Â· ~ b | ~ c | 2 ~ c = 1 2 (1 , , 1) = ( 1 2 , , 1 2 ) . (6) The area = | ~a Ã— ~ b | = | (- 11 ,- 15 , 32) | = âˆš 11 2 + 15 2 + 32 2 = âˆš 1370. (7) The volume = | ( ~a Ã— ~ b ) Â· ~ c ) | = | (- 11 ,- 15 , 32) Â· (1 , , 1) | = 21 . 4. Find parametric equations for the line through the points (3 , 1 ,- 1) and (3 , 2 ,- 6). Solution: ~v = (3 , 2 ,- 6)- (3 , 1 ,- 1) = (0 , 1 ,- 5) is a direction vector along the line. x = 3 y = t + 1 z =- 5 t- 1 . 5. Find an equation for the plane through the points (3 , 1 ,- 1), (3 , 2 ,- 6) and (1 , 1 , 0). Solution: ~v 1 = (1 , 1 , 0)- (3 , 1 ,- 1) = (- 2 , , 1) and ~v 2 = (3 , 2 ,- 6)- (3 , 1 ,- 1) = (0 , 1 ,- 5) are two vectors in the plane. Their cross product ~v 1 Ã— ~v 2 = fl fl fl fl fl fl fl ~ i ~ j ~ k- 2 0 1 1- 5 fl fl fl fl fl fl fl = (- 1 ,- 10 ,- 2) is a normal vector for the plane. The plane equation is- ( x- 3)- 10( y- 1)- 2( z + 1) = 0 i.e. x + 10 y + 2 z- 11 = 0 . 1 2 6. Find an equation for the plane that passes the point (1 , 1 ,- 1) and perpendicular to the line x = 2 t + 1 ,y = 3 t- 2 ,z =- 4 t + 1....
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Exam 1 Review Solutions - Solutions To Review Problems 3...

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