exam_1_20C_solutions

# exam_1_20C_solutions - Name Student ID TA/Section (circle):...

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Unformatted text preview: Name Student ID TA/Section (circle): Michele 4pm-D03 5pm-D04 Andy 6pm-D05 7pm-D06 8pm-D07 9pm-D08 Math 20C, Winter 2010, Midterm Exam 1 Solutions • Show all of your work to receive full credit. • Simplify your answers. You will not be penalized for a mistake in simplifying if a less simplified answer is sufficient for full credit. • Write your answers and work clearly and legibly; no credit will be given for illegible solutions. • Go back and check your answers if you finish early. # Points Score 1 5 2 5 3 5 4 5 5 5 6 5 Σ 30 1. (5 points) Determine if the lines given by the vector parametrizations r 1 ( t ) = ( 5 , − 16 , 19 ) + t ( 1 , − 3 , 4 ) and r 2 ( t ) = ( 5 , − 1 , − 11 ) + t (− 2 , 1 , 2 ) intersect and, if so, find the point of intersection. Answer: We need to see if there are values of s and t such that r 1 ( s ) = r 2 ( t ), so write r 1 ( s ) = ( s + 5 , − 3 s − 16 , 4 s + 19 ) and r 2 ( t ) = (− 2 t + 5 , t − 1 , 2 t − 11 ) and equate the components of each vector. s + 5 = − 2 t + 5 − 3 s − 16 = t − 1 4 s + 19 = 2 t − 11 First pick two of these equations and solve for s and t . From the second equation we get t = − 3 s − 15, and substituting this into the first equation gives s + 5 = − 2( − 3 s − 15) + 5 s + 5 = 6 s + 35 − 30 = 5 s s = − 6 Now, t = − 3 s − 15 = − 3( − 6) − 15 = 3. This tells us that the x and y components of the lines are equal at r 1 ( s = − 6) and r 2 ( t = 3). We can now see that r 1 ( s = − 6) = (− 1 , 2 , − 5 ) = r 2 ( t = 3) so the lines intersect at the point ( − 1 , 2 , − 5)....
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## This note was uploaded on 04/15/2010 for the course MATH 20C taught by Professor Lit during the Spring '10 term at UCLA.

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exam_1_20C_solutions - Name Student ID TA/Section (circle):...

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