Quiz #2. A = 1 2 bh 1 2 (6.1)(5.3)

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1 Math 50C Multivariable Calculus September 11, 2015 Quiz #2 Name: Answer Ke David Arnold Instructions. (20 points) This quiz is open notes, open book. This includes an supplementar tets or online documents. You must answer all of the eercises on our own. You are not allowed to work in groups or pairs on the quiz. You are not allowed to enlist the aid of a tutor, classmate, or friend to help with the quiz. All work must be our own. Place our solution to each problem in the space provided. Honor Pledge: I promise that the solution submitted was done entirel b me. I received no help from colleagues, friends, or tutors. I also did not share an parts of m solution with an of m classmates. Signature: (5 pts ea. ) 1. Perform each of the following tasks. (a) n the grid that follows, plot the points A(2, 1), B(8, 2), and C(6, 7), the connect them to draw the triangle ABC. Use our drawing to estimate the area of the triangle. Show the work used to determine the estimate of the area. Solution: Plot the points A(2, 1), B(8, 2), and C(6, 7), then draw the triangle ABC. Net, construct CD perpendicular to AB. This is the altitude of the triangle. To measure CD, take a compass and place the point at C, open the pencil of the compass to the point D, then draw an arc. We can now estimate that the length of CD, equal to that of CG, is approimatel 5.3 units. Net, place the needle of our compass at the point B, open the tip of the pencil to point A, then draw another arc. We can now find the length of AB, equal to the length of BG, as approimatel 6.1 units. C(6, 7) C(6, 7) A(2, 1) G D B(8, 2) G A(2, 1) B(8, 2) Hence, the area of the triangle is approimatel: A = 1 2 bh 1 2 (6.1)(5.3) (b) Use the cross product to determine the area of the triangle ABC. Solution: The cross product onl lives in three space, so we ll set the z-coordinate of each of the vertices of the triangle to zero:

2 Math 50C Multivariable Calculus/Quiz #2 Page 2 of 6 Name: Answer Ke 1. A = (2, 1, 0) 2. B = (8, 2, 0) 3. C = (6, 7, 0) You could just as easil set the z-coordinate of each point equal to 117 and get the same result. Now, compute AB and AC. AB = 8, 2, 0 2, 1, 0 = 6, 1, 0 Net, compute the cross product. AB AC = î ĵ ˆk = 32ˆk The area of the parallelogram spanned b the vectors AB and AC is: AB AC = 32 Therefore, the area of the triangle is one-half of this amount. Area ABC = 16 AB = 6, 7, 0 2, 1, 0 = 4, 6, 0 Note the nice agreement between this eact result and our approimation from part (a). (5 pts ea. ) 2. Perform each of the following tasks. (a) n the grid that follows, sketch the graph of the line = 12. Net, plot the point P (4, 6), then use our figure to estimate distance from point P (4, 6) to the line = 12. Provide some eplanator prose as to how our found our estimate. Solution: First, sketch the line = 12, which has -intercept (4, 0) and -intercept (0, 6). Net, plot the point P (4, 6), then drop a perpendicular from P to the line = 6. Take out our compass again, place the needle at P, place our pencil point at Q, then draw an arc. The length of P R equals the distance between P and Q, and is approimatel equal to 3.3 units.

3 Math 50C Multivariable Calculus/Quiz #2 Page 3 of 6 Name: Answer Ke P (4, 6) Q R = 12 (b) Use vectors and the cross product to determine the eact distance between the point P (4, 6) and the line = 12. No credit will be awarded for solutions that simpl quote the formula from the eercise set, then substitute the vectors. You will onl receive full credit if ou draw the appropriate vectors, sketch the appropriate parallelogram, then determine the distance using the cross product. Your solution must impress me that ou know where the formula comes from. Solution: First, sketch the line 3+2 = 12, using the -intercept A(0, 6) and the -intercept B(4, 0). Net, draw the parallelogram spanned b the vectors AP and AB. Then indicate the height h of the parallelogram b dropping a perpendicular from P to the side AB.

4 Math 50C Multivariable Calculus/Quiz #2 Page 4 of 6 Name: Answer Ke A(0, 6) P (4, 6) h Q B(4, 0) C(8, 0) = 12 Now, the area of the parallelogram is found b taking the magnitude of the cross product of AP and AB. If we divide this result b the length of the base, we get the height. Thus: h = AP AB AB The cross product does not eist in two-space, so we ll assign an arbitrar z-value of zero to each of the points A(0, 6, 0), B(4, 0, 0), and P (4, 6, 0). Then: 1. AP = 4, 0, 0 0, 6, 0 = 4, 0, 0, and 2. AB = 4, 0, 0 0, 6, 0 = 4, 6, 0. Hence: AP î ĵ ˆk AB = Hence, AP AB = 24. Second, note that: = 24 ˆk AB = 42 + ( 6) = 52 = 2 13

5 Math 50C Multivariable Calculus/Quiz #2 Page 5 of 6 Name: Answer Ke Thus, the height h of the parallelogram is: h = AP AB AB = = If we use a calculator to estimate this result, we see that 12/ , which agrees nicel with our approimation in part (a). Alternate solution using dot product. First, note that the vector AB = 4, 6. Thus, a vector perpendicular (orthogonal) to this vector is v = 6, 4. This is easil checked as their dot product equals zero. Now, we need a unit vector in the direction of v. û = v v 6, 4 = , 4 = =, A(0, 6) P (4, 6) Q û B(4, 0) = 12 Now, if we project the vector AP onto the unit vector û, then take the absolute value, we get the length of the projection of AP onto the vector û, which is the distance from point P to the line = 12.

6 Math 50C Multivariable Calculus/Quiz #2 Page 6 of 6 Name: Answer Ke d = 3 4, 0 2, = If we use a calculator to estimate this result, we see that 12/ , which agrees nicel with our approimation in part (a).