By Arthur Wouk
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Extra info for A course of applied functional analysis
Here is an example. 1 Find a parametric equation for the line between the points (1, 2, 3) and (2, −3, 1) . A direction vector is (1, −5, −2) because this is the vector from the ﬁrst to the second of these. Then an equation of the line is (x, y, z) = (1, 2, 3) + t (1, −5, −2) , t ∈ R The example shows how to do this in general. If you have two points in Rn , a, b, then a parametric equation for the line containing these points is of the form x = a + t (b − a) . Note that when t = 0 you get the point a and when t = 1, you get the point b.
I 1 3 j k −1 2 −2 1 = −1 −2 2 1 i− 1 3 2 1 j+ 1 3 −1 k = 3i + 5j + k. 5 Find the area of the parallelogram determined by the vectors, (i − j + 2k) , (3i − 2j + k) . 4. 4. Thus the area is 9 + 25 + 1 = 35. 6 Find the area of the triangle determined by (1, 2, 3) , (0, 2, 5) , (5, 1, 2) . 4. THE CROSS PRODUCT 37 This triangle is obtained by connecting the three points with lines. Picking (1, 2, 3) as a starting point, there are two displacement vectors, (−1, 0, 2) and (4, −1, −1) such that the given vector added to these displacement vectors gives the other two vectors.
What was just described would be called a force vector. It has two essential ingredients, its magnitude and its direction. Note there are n special vectors which point along the coordinate axes. These are ei ≡ (0, · · · , 0, 1, 0, · · · , 0) th where the 1 is in the i of R3 . slot and there are zeros in all the other spaces. See the picture in the case z e3 ✻ e1✠ ✲ e2 y x The direction of ei is referred to as the ith direction. Given a vector v = (a1 , · · · , an ) , it follows that n ∑ v = a1 e1 + · · · + an en = ai e i .
A course of applied functional analysis by Arthur Wouk