07spr-m1sols

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

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Unformatted text preview: 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 =-→ 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 . 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 + z z = y - 3 1 + z 1 1 . Equivalently, Q = s - 3 1 + t 1 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 , ,- 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 , ,- 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 , ,- 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- 1 + s - 3 1 + t 1 1 s,t ∈ R , or equivalently x y z x = 2- 3 s + t, y = s, z =- 1 + t ....
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This note was uploaded on 01/12/2010 for the course MATH 51 at Stanford.

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

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