Student name Student ID TA s name and or section MATH 32A Butler Midterm I 16 April 2010 This test is closed book and closed notes No calculator is allowed for this test For full credit show all of your work legibly Each problem is worth 10 points a total of 50 points 1 Find the projection of h2s 1 s 1i onto the vector h 2t 5 t2 4ti The formula for projection of a vector a onto a vector b is a b b b b Applying it to our case with a h2s 1 s 1i and b h 2t 5 t2 4ti we have that the projection is h2s 1 s 1i h 2t 5 t2 4ti h 2t 5 t2 4ti h 2t 5 t2 4ti h 2t 5 t2 4ti 2s 2t 1 5 t2 s 1 4t h 2t 5 t2 4ti 2t 2 5 t2 2 4t 2 4st 5 t2 4st 4t 2 h 2t 5 t2 4ti 4t 25 10t2 t4 16t2 5 4t t2 h 2t 5 t2 4ti 4 t 10t2 25 5 4t t2 h 2t 5 t2 4ti 2 2 t 5 On a side note we see that this projection only depends on t even though the vector we were projecting involved s 2 Find the area of the quadrilateral in the plane with vertices located at 3 1 7 3 4 4 and 0 3 using vector techniques So we are looking for the area shown in the figure below The technique that we have learned for finding area of triangles using vectors is to take the magnitude of the cross product and divide by 2 But cross product only works in three dimensions That is easy to fix we will think of these points as 3 1 0 7 3 0 4 4 0 and 0 3 0 i e all the z coordinates are 0 So let us break the quadrilateral into two triangles and use cross products to find the area Namely we will use the triangle with corners at 3 1 0 7 3 0 and 4 4 0 so we will use the vectors h 4 2 0i and h 3 1 0i and the triangle with corners at 4 4 0 0 3 0 and 3 1 0 so we will use the vectors h4 1 0i and h3 2 0i So we have Area h 4 2 0i h 3 1 0i h4 1 0i h3 2 0i 2 2 h0 0 10i h0 0 11i 21 2 2 2 It is easy to directly find the area of the quadrilateral directly to see that this is the correct value For completeness we do the cross products We have h 4 2 0i h 3 1 0i h4 1 0i h3 2 0i i j k 4 2 0 3 1 0 i j k 4 1 0 3 2 0 4 2 k h0 0 10i 3 1 4 1 k h0 0 11i 3 2 3 For each of the following statements determine if the statement is true or false for any arbitrary P Q and R points in three dimensional space If it is true give a proof i e a brief explanation of why it always holds If it is false give an example of three points P Q and R that shows it does not hold a P Q P R QP QR This statement is always true Since P Q P R and QP QR are both equal to twice the area of the triangle with vertices at P Q and R they must also be equal to each other Note this also can be shown to be equivalent to the law of sines but the area argument seems much cleaner b P Q P R QP QR This statement is false in general For example if P 0 0 0 Q 1 0 0 and R 0 1 0 then P Q P R h1 0 0i h0 1 0i 0 0 QP QR h 1 0 0i h 1 1 0i 1 1 which are not equal 4 Show that the parametric curve x t y t z t 1 sin 2t 2 sin t cos t cos 2t lies on a sphere centered at the origin Also find the radius of the sphere 2 2 2 We need to check that x t y t z t is a fixed constant i e it does not depend on t So computing we have math symbols click the To print higher resolution Hi Res Fonts for Printing button on the jsMath control panel 2 2 2 2 2 2 x t y t z t 1 sin 2t 2 sin t cos t cos 2t puleshan Toggle Home 2 Published2 Log Settings 2 Sage cos tReport cos 2t Sign out 1 2 sin 2t The sin 2t 2 sin2 t 2 2 sin z t cos t 2Help a Problem Notebook sin 2t 2 Version 4 3 4 2 1 Untitled 2 sin 2t 2 sin 2t sin 2t cos 2t 2 sin2 t cos2 t Save z z Save quit last edited on April 09 2010 07 45 AM by puleshan File Action Data sage 1 Discard quit 1 Typeset Print Worksheet 1 1 2 4 22 Edit Text Undo Share Publish Therefore the points on this parametric curve lie on the sphere of radius 2 t var t parametric plot3d 1side sin 2 t sqrt 2 sin t cos t cos 2 t t 0 7 note the plot of the points is shown below For people On a who are very good at visualizing you can see that this curve is the intersection of x2 y 2 z 2 22 and the cylinder x 1 2 z 2 1 Get Image evaluate 5 Find the point on the plane x 2y 5 3z which is closest to the point 4 4 7 Hint you do not need calculus to find the answer Useful hint the line that connects the point 4 4 7 to the nearest point on the plane will be perpendicular to the plane Based on the hint the useful one we first find the line that connects the point that is given 4 4 7 to the nearest point in the plane Since this is perpendicular to the plane the directional vector of the line is the same as the normal vector of the plane rewriting the equation for the plane as x 2y 3z 5 we see that the desired vector is h1 2 3i With our point and direction we have that the line connecting these two points in parametric form is x 4 t y 4 2t z 7 3t We need to find where this line and the plane intersects This will occur when x 2y 3z 5 substituting in the above values for x y z we have that the correct value of t will be when 5 4 t 2 4 2t 3 7 3t 4 t 8 4t 21 9t 33 14t or rearranging 14t 5 …