07spr-m1sols - Math 51 Spring 2007 Exam 1 Solutions Page 1...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 51, Spring 2007 Exam 1 Solutions — April 24, 2007 Page 1 of 9 1. (10 points) Complete each of the following sentences. (a) A collection of vectors -→ v 1 , . . . , -→ v k is defined to be linearly independent if (5 points) . . . the equation c 1 -→ v 1 + c 2 -→ v 2 + · · · + c k -→ v k = -→ 0 for scalars c 1 , . . . , c k implies c 1 = · · · = c k = 0. OR: . . . no vector in the collection can be written as a linear combination of the other vectors. (b) A basis for a subspace V is defined to be (5 points) . . . a linearly independent set (or collection) of vectors whose span is V . (Note: saying only the portion “linearly independent set” earned 3 points; saying only the portion “set that spans V ” also earned 3 points.) OR: . . . a collection C of vectors such that any vector in V can be uniquely expressed as a linear combination of vectors in C .
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Math 51, Spring 2007 Exam 1 Solutions — April 24, 2007 Page 2 of 9 2. (10 points) Let Q be the set of all vectors in R 3 that are orthogonal to -→ w = (1 , 3 , - 1). (a) Find a parametrization for the set Q . (Hint: Q forms a plane in R 3 .) (5 points) If -→ v = ( x, y, z ) is orthogonal to -→ w , then -→ v · -→ w = 0, i.e., x + 3 y - z = 0. This is a “system” with two free variables y and z , so that x y z = - 3 y + z y z = - 3 y y 0 + z 0 z = y - 3 1 0 + z 1 0 1 . Equivalently, Q = s - 3 1 0 + t 1 0 1 s, t R = x y z x = - 3 s + t, y = s, z = t . (b) Suppose P is a plane in R 3 , parallel to Q , such that P passes through the point (2 , 0 , - 1). Find an equation for P , written in the form ax + by + cz = d . (5 points) Solution 1: The plane Q has normal vector -→ w , since every vector in Q is orthogonal to -→ w . Since P is parallel to Q , it also has normal vector -→ w . Now if ( x, y, z ) is any point in P , then since (2 , 0 , - 1) also lies in P , we know that the vector -→ w is orthogonal to the vector -→ u = x - 2 y z + 1 from ( x, y, z ) to (2 , 0 , - 1). Thus, 0 = -→ w · -→ u = ( x - 2) + 3 y - ( z + 1) = x + 3 y - z - 3 , so that x + 3 y - z = 3 is the equation for P . (Note: In this instance, we were given -→ w as a normal vector to Q , but in other situations we might need to construct such a vector using Q ’s parametrization. In such a case, remember this can be done by taking the cross product of any two non-collinear vectors that span Q .) Solution 2: Since P is parallel to Q , the plane P in parametric form is 2 0 - 1 + s - 3 1 0 + t 1 0 1 s, t R , or equivalently x y z x = 2 - 3 s + t, y = s, z = - 1 + t . We can combine the above equations to eliminate s and t , and we find x = 2 - 3 y + ( z + 1) , which simplifies to x + 3 y - z = 3 , as before.
Image of page 2
Math 51, Spring 2007 Exam 1 Solutions — April 24, 2007 Page 3 of 9 3. (10 points) Compute, showing all steps, the reduced row echelon form of the matrix 1 0 2 1 0 - 1 0 - 2 0 - 1 2 2 2 3 7 - 2 2 - 6 0 6 .
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern