Prelim_1_Fall_2007

Prelim_1_Fall_2007 - Math 192 Prelim 1 Solutions 7:30-9:00...

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Unformatted text preview: Math 192, Prelim 1 - Solutions September 27, 2007. 7:30-9:00 1) a) v ( t ) = i + 2 t j + 2 k , a ( t ) = 2 j b) Let θ ( t ) denote the angle. Then cos θ ( t ) = v ( t ) · a ( t ) | v ( t ) || a ( t ) | = 4 t √ 1+4 t 2 +4 √ 4 = 2 t √ 5+4 t 2 2) a)--→ AB = i- j- k is a vector parallel to the line through A and B and (2 , , 0) is a point on the line so the parametric equation for the line through A and B is x = 2 + t , y =- t , z =- t . Plug this into the equation for P 2 to get (2 + t )- 2(- t ) = 0 ⇒ t =- 2 3 and use the parametrization for the line through A and B to see that the point of intersection is ( 4 3 , 2 3 , 2 3 ) b) The normals for the given planes are h 1 , 2 , 3 i and h 1 ,- 2 , i so h 1 , 2 , 3 i × h 1 ,- 2 , i = h 6 , 3 ,- 4 i is parallel to the line of intersection. The point (0 , , 1 3 ) is in both planes and thus on the line of intersection. Hence the line of intersection is x = 6 t , y = 3 t , z = 1 3- 4 t 3) a)--→ AB = h- 3 , 1 ,- 3 i ,-→ AC = h , 1...
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This note was uploaded on 02/28/2008 for the course MATH 1920 taught by Professor Pantano during the Fall '06 term at Cornell.

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