assign2_soln

assign2_soln - Math 136 1 Determine Assignment 2 Solutions...

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Math 136 Assignment 2 Solutions 1. Determine a) proj (2 , 1 , 1) (3 , 2 , 1) and perp (2 , 1 , 1) (3 , 2 , 1). Solution: proj (2 , 1 , 1) (3 , 2 , 1) = (3 , 2 , 1) · (2 , 1 , 1) ± (2 , 1 , 1) ± 2 (2 , 1 , 1) = 9 6 (2 , 1 , 1) = (3 , 3 / 2 , 3 / 2) perp (2 , 1 , 1) (3 , 2 , 1) = (3 , 2 , 1) - proj (2 , 1 , 1) (3 , 2 , 1) = (3 , 2 , 1) - (3 , 3 / 2 , 3 / 2) = (0 , 1 / 2 , - 1 / 2) b) proj ( - 1 , 1 , 0 , 2) (2 , - 2 , 1 , 1) and perp ( - 1 , 1 , 0 , 2) (2 , - 2 , 1 , 1). Solution: proj ( - 1 , 1 , 0 , 2) (2 , - 2 , 1 , 1) = (2 , - 2 , 1 , 1) · ( - 1 , 1 , 0 , 2) ± ( - 1 , 1 , 0 , 2) ± 2 ( - 1 , 1 , 0 , 2) = - 2 6 ( - 1 , 1 , 0 , 2) = (1 / 3 , - 1 / 3 , 0 , - 2 / 3) perp ( - 1 , 1 , 0 , 2) (2 , - 2 , 1 , 1) = (2 , - 2 , 1 , 1) - proj ( - 1 , 1 , 0 , 2) (2 , - 2 , 1 , 1) = (2 , - 2 , 1 , 1) - (1 / 3 , - 1 / 3 , 0 , - 2 / 3) = (5 / 3 , - 5 / 3 , 1 , 5 / 3) 2. Use a projection to fnd the distance From the point to the plane. a) The point P (2 , - 3 , - 1) and the plane 2 x 1 - 3 x 2 - 5 x 3 = 7. Solution: Let A (1 , 0 , - 1) be a point in the plane (any point satisFying the equation oF the plane will do). Then, the distance d From P to the plane is the length oF the projection -→ AP = (1 , - 3 , 0) along a normal to the plane, ±n = (2 , - 3 , - 5): d = ± proj AP ± = ± ± ± ± ± (1 , - 3 , 0) · (2 , - 3 , - 5) ² 2 2 +( - 3) 2 - 5) 2 ± ± ± ± ± = ± ± ± ± 2 + 9 + 9 38 ± ± ± ± = 11 38 .

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assign2_soln - Math 136 1 Determine Assignment 2 Solutions...

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