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# Exam1Solutions - EXAM I SOLUTIONS Math 51 Spring 2003 You...

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Unformatted text preview: EXAM I SOLUTIONS Math 51, Spring 2003. You have 2 hours. No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck! Name ID number 1. (/20 points) 2. (/15 points) 3. (/15 points) 4. (/20 points) 5. (/15 points) 6. (/15 points) Bonus (/10 points) Total (/100 points) “On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.” Signature: Circle your TA’s name: Byoung-du Kim (2 and 6) Ted Hwa (3 and 7) Jacob Shapiro (4 and 8) Ryan Vinroot (A02) Michel Grueneberg (A03) Circle your section meeting time: 11:00am 1:15pm 7pm 1 1. Consider the following three vectors:-→ u = 3 4 12 -→ v = 2 3 6 -→ w = 3 6 6 (a) Find the magnitude of each of the given vectors. Solution: bardbl-→ u bardbl = √ 3 2 + 4 2 + 12 2 = 13 bardbl-→ v bardbl = √ 2 2 + 3 2 + 6 2 = 7 bardbl-→ w bardbl = √ 3 2 + 6 2 + 6 2 = 9 (b) Find a parametric representation of the unique plane that contains all of the given vectors. Solution: For the parametric representation we need to have two independent vec- tors which are parallel to the plane in question, and one point which is in the plane in question. Of course, any of the three given vectors can be used as the point in the plane. To find the vectors parallel to the plane, we subtract (arbitrarily)-→ v from-→ u and-→ w . This gives us-→ u--→ v = 1 1 6 -→ w--→ v = 1 3 Our parametric representation of the plane is then 2 3 6 + c 1 1 1 6 + c 2 1 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle c 1 , c 2 ∈ R 2 (c) Find an equation representing the unique plane that contains all of the given vectors. Solution: To find the equation of the given plane, we need to find a normal vector. Since we already have two vectors parallel to the plane, their cross product must be perpendicular to the plane:-→ n = (-→ u--→ v ) × (-→ w--→ v ) = 1 1 6 × 1 3 = - 18 6 2 The equation of the plane is then-→ n ·-→ x =-→ n ·-→ v- 18 x + 6 y + 2 z =- 6 or 9 x- 3 y- z = 3 3 2. Find a parametric representation of the complete set of solutions to the following system of equations.- 2 x- y + 5 z =- 5 3 x + y- 7 z = 6 2 x- 4 z = 2 Solution: First we write the augmented matrix that corresponds to the given system: - 2- 1 5 | - 5 3 1- 7 | 6 2- 4 | 2 We then use row operations to reduce to the RREF: 2- 4 | 2 3 1- 7 | 6- 2- 1 5 | - 5 r 3 r 2 r 1 = ⇒ 1- 2 | 1 3 1- 7 | 6- 2- 1 5 | - 5 r 1 / 2 r 2 r 3 = ⇒ 1- 2 | 1 1- 1 | 3- 1 1 | - 3 r 1 r 2- 3 r 1 r 3 + 2 r 1 = ⇒ 1 0- 2 | 1 0 1- 1 | 3 0 0 | r 1 r 2 r 3 + r 2 We have two pivots, and thus two pivot variables (in this case...
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## This note was uploaded on 08/19/2008 for the course MATH 51A taught by Professor Staff during the Spring '06 term at Stanford.

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Exam1Solutions - EXAM I SOLUTIONS Math 51 Spring 2003 You...

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