Fall 2007 - Cioaba's Class - Exam 1 (Version 1)

Fall 2007 - Cioaba's Class - Exam 1 (Version 1) - Name:...

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Unformatted text preview: Name: PID: Discussion Section- No: Time: TA’s name: Midterm 1, Math 20C - Lecture D (Fall 2007) Duration: 50 minutes Please close your books, turn your calculators off and put them away. You can use one page of handwritten notes. To get full credit you should explain your answers. 1. Consider the following two lines L 1 : x − 4 2 = y + 1 5 = z − 2 3 and L 2 : x + 2 3 = y + 5 2 = z − 4 − 1 a.(3 points) Find the coordinates of the intersection point of the lines L 1 and L 2 . Solution. The parametric equations of the two lines are x = 4 + 2 t y = − 1 + 5 t z = 2 + 3 t and x = − 2 + 3 s y = − 5 + 2 s z = 4 − s The intersection point ( x, y, z ) must satisfy both equations which means that 4 + 2 t = − 2 + 3 s − 1 + 5 t = − 5 + 2 s 2 + 3 t = 4 − s From the last equation we get s = 2 − 3 t and substituting into the second equation gives − 1 + 5 t = − 5 + 2 s = − 5 + 2(2 − 3 t ) = − 5 + 4 − 6 t = − 1 − 6 t Solving this equation, we get that 11 t = 0 which means t = 0 and s = 2 − 3 · 0 = 2. One also checks that s = 2 and t = 0 satisfy the first equation 4+2 t = 4+0 = 4 = − 2+6 = − 2+3 s . Thus, the intersection point ( x, y, z ) is obtained substituting t = 0 into the equation of L 1 (or substituting s = 2 into the equation of L 2 ) and is (4+2 · , − 1+5 · , 2+3 · 0) = (4 , − 1 , 2)....
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This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.

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Fall 2007 - Cioaba's Class - Exam 1 (Version 1) - Name:...

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