• Here's a link to the course syllabus.
• Our first paper homework set had two problems. In the first problem, the task was to find the point nearest (a,b,c) on the line through (1,0,0) and (2,1,1), where (a,b,c) are the last three digits of your own UIN. The answer is found by observing that the vector v parallel to the line is (1,1,1) and the vector u from (1,0,0) to (a,b,c) is (a-1,b,c). So the projection p of u onto v is ((a-1,b,c)dot(1,1,1))/3 times (1,1,1), which works out to (a+b+c-1)/3 times (1,1,1). Add that to (1,0,0) to get the answer. For my UIN, p=(14/3,14/3,14/3) and the nearest point is (17/3,14/3,14/3). Different UINs will result in different details, of course. The second problem was to find the point in a certain plane and nearest to (2,3,4). The plane is the plane through (0,0,0), (1,1,1), and (1,3,2). Two vectors parallel to the plane are (1,1,1) and (1,3,2), and if we needed a third we could take the vector from (1,1,1) to (1,3,2) and get (0,2,1). The cross product of any two of these will give a vector n perpendicular to the plane. So taking the first two of our vectors, we'd get n=(-1,-1,2). The vector from (0,0,0) to (2,3,4) is just (2,3,4) and the projection p of (2,3,4) onto n is (3/6)(-1,-1,2)=(-1/2,-1/2,1). The point we seek, call it q, is also the vector v from (0,0,0) to q. Now v has two properties: v+p=(2,3,4), and v.n=0. The first property gives us v: v=(2,3,4)-p=(5/2,7/2,3). The second property is a check: is (5/2,7/2,3) dot (-1,-1,2) really 0? And yes, it is. So we are happy and (5/2,7/2,3) is both calculated and checked to be the answer to the question.
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