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This booklet introduces the $64000 recommendations of finite-dimensional vector areas during the cautious research of Euclidean geometry. In flip, tools of linear algebra are then utilized in the examine of coordinate adjustments wherein an entire class of conic sections and quadric surfaces is bought.
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Extra resources for Vectors, Matrices and Geometry
1, we get st(l + r)x + (1 + s)y - s(l + t)z =st(b + re) + (c + sa) - sea + tb) =0 . 13, the point X, Y and Z will be collinear if we can prove that st(l + r) + (1 + s) - and that the scalars st(l+r), l+s and former follows from the condition rst to show that 1 + s :/; O. Indeed if s which is impossible. The proof is now = s(l + t) = 0 s(l+t) are not all equal to zero. The = -1. For the latter it is sufficient -1 then IT = -YA; hence C A complete. = EXERCISES 1. Find the value of k so that the points (2,5), (0, k), and (-1,1) are collinear.
Given x and a "# o. It is required to find a scalar t such that x - ta =b is orthogonal a" Taking dot product with a on both sides, we get (x "a) - tea "a) = (b "a) = 0 " Since (a" a) "# 0, we obtain solution to the problem as t = (x" a)/(a "a) and b =x - ta " Graphically the condition that (x-ta) be orthogonal to a means that the scalar multiple ta is the perpendicular projection of x on a. o Fig 1-11 Let us put the result of our discussion into the following statement. 7 THEOREM AND DEFINITION Let a be a non-zero vector.
C) Find the area of 6ABC if A = (1,2), B = (-3,-1)' and C = (3, -3). 26. Consider 6ABC where position vectors of the vertices are a, b, and c respectively. Suppose that the perpendicular bisectors of BC and AB meet at the origin. (a) Show that lal2 = Ibl 2 = Ic1 2. 31 Vectors, Matrices and Geometry (b) Hence deduce that the perpendicular bisectors of the sides of 6ABC are concurrent. 27. Let M be the midpoint of BC of 6ABC. If AM = x, AB = y and Aa = z, show that (a) Ixl 2= ~IYI2 + ~(y. z) + ~lzI2, (b) AB2 and + AC 2 = 2(AM2 + BM2).