23f11-ex1-review-solutions (1)

23f11-ex1-review-solutions (1) - MATH 23, FALL 2011 REVIEW...

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MATH 23, FALL 2011 REVIEW SHEET SOLUTIONS Instructions: These problems represent the types of questions you might nd on the rst 4:00 exam in this term's Math 23. They are taken directly from old exams. The best way to use this review is to choose a number of these problems to make up a (somewhat long) exam for yourself, then solve it under exam conditions, no notes, no calculator, and no looking at answers. Then, using your notes and text, grade your trial exam. (1) If A = 3 i - j + 3 k , and B = 2 i + 4 j - k , nd: (5 points per part) (a) 2 A - B , Solution: 2 A - B = 2(3 i - j + 3 k ) - (2 i + 4 j - k ) = 4 i - 6 j + 7 k . (b) The cosine of the angle between the vectors A and B . Solution: Since A · B = k A kk B k cos θ , where θ is the angle between the vectors, cos θ = A · B k A kk B k = 3 · 2 + ( - 1) · 4 + 3 · ( - 1) 9 + 1 + 9 4 + 16 + 1 = - 1 19 21 . (c) The number t so that A + t B is perpendicular to i + j . Solution: A + tB is perpendicular to i + j when ( A + tB ) · ( i + j ) = 0 , which is when: 0 = ( A + tB ) · ( i + j ) = ((3 i - j + 3 k ) + t (2 i + 4 j - k )) · ( i + j ) = (3 + 2 t ) + ( - 1 + 4 t ) = 2 + 6 t, or t = - 1 / 3 . (2) Find a vector perpendicular to 3 i - 2 j + k . (5 points) Solution: There are lots of right answers to this one. 2 i + 3 j , i - 3 k , or j + 2 k will all work. All you need to nd is a vector whose dot product with the given one is 0. (3) If A = 2 i + j - 3 k , and B = - 2 i + 5 j + k , nd: (5 points per part) (a) A - 3 B , Solution:: A - 3 B = 8 i - 14 j - 6 k . (b) A · B . Solution:: A · B = (2)( - 2) + (1)(5) + ( - 3)(1) = - 2 . (c) A × B . Solution:: A × B = i j k 2 1 - 3 - 2 5 1 = 16 i + 4 j + 12 k . (4) Find the equation of the line of intersection of the planes x + y = 2 and y - z = 4 . 1
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2 MATH 23, FALL 2011 REVIEW SHEET SOLUTIONS Solution: The standard way to do this is to nd a vector parallel to both planes, which is the same as being perpendicular to both normals. So, the cross-product of the two normals will work. That is, the vector A = N 1 × N 2 = ( i + j ) × ( j - k ) = k + j - i = - i + j + k will be in the direction of the line of intersection of both planes. In order to nd the equation of this line, you need a point on the line; a point on both planes. But, if y = 0 , then x = 2 and z = - 4 , so (2 , 0 , - 4) is on the line. Then, the equation for the line is r ( t ) = 2 i - 4 k + t ( - i + j + k ) . An alternate way to solve this problem is to note that you can set y = t in the equation for both planes, and solve for x in the rst plane, and z in the second, giving x = 2 - t, y = t, and z = t - 4 . (5) Find an equation of the plane P through the points (2 , 1 , 0) , (1 , 1 , 1) , and (0 , 3 , - 1) . Solution:
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23f11-ex1-review-solutions (1) - MATH 23, FALL 2011 REVIEW...

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