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Math 311: Topics in Applied Math 1 5: Orthogonality 5.1: The Scalar Product in Rn Summary Let x, y ∈ Rn . Solution Since x and y be linearly independent, neither vector is a multiple of the other. Hence the angle θ between them is not a multiple of π. Hence |cos θ | < 1. Therefore, T x y = kxk kyk cos θ = kxk kyk |cos θ | < kxk kyk < (2) (3) or xT y < 6. n • The scalar product of x and y is xT y = ∑ xk yk . k=1 √ • The Euclidean length of x is kxk = xT x. • The distance between x and y is ky − xk. • The angle θ between two vectors x and y satisfies xT y = kxk kyk cos θ . uT v. • For unit vectors u and v, we have cos θ = • Cauchy-Schwarz Inequality: xT y ≤ kxk kyk. • Vectors x and y are orthogonal if xT y = 0. In this case, we write x ⊥ y. • For nonzero vectors x & y, the scalar projection α, vector projection p, and orthogonal projection q of x onto y are respectively given by xT y kyk T 1 x y y p = projy x = α y = kyk yT y q = orthy x = x − p. α = compy x = Let P1 and P2 be points in 3-dimensional real space. We identify them with their position vectors P1 and P2 in R3 . −−→ P1 P2 denotes the vector P2 − P1 ∈ R3 . • Given a nonzero vector n ∈ R3 and a fixed point P0 in 3-dimensional space, the set of points P in space −→ such that n ⊥ P0 P is a plane. Here nT (P − P0 ) = 0 or nT P = nT P0 We say the vector n and the plane are normal to each other; n is called a normal vector. 224/9 Find an equation of the plane that passes through the points P1 (2, 3, 1), P2 (5, 4, 3), and P3 (3, 4, 4). Solution 3 1 −−→ −−→ • Let v = P1 P2 = 1 and w = P1 P3 = 1 . 2 3 Back in Calc 3 we’d compute our normal vector as the cross product n = v × w. For grins, let’s alternatively directly obtain a vector that is orthogonal to both v and w. T • Let n = a b c . Then nT v = 0 and nT w = 0. This gives a linear system in the variables a, b, c, whose augmented system matrix and reduced row echelon form are 3 1 2 0 1 0 −1/2 0 → . 1 1 3 0 0 1 7/2 0 T • Thus a = 12 c and b = − 72 c. So n = 1 −7 2 is a normal vector. Hence an equation of the plane is x T y = nT P1 n z x − 7y + 2z = −17. 225/12+ Prove that if x, y, z ∈ Rn , then (a) xT x ≥ 0 (b) xT y = yT x (c) xT (y + z) = xT y + xT z Solution Examples n k=1 Also watch the calculator video for more examples. Let x and y be linearly independent vectors in R2 . Given kxk = 2 and kyk = 3, what can we conclude about the possible values of xT y? xT y is a scalar (1 × 1 matrix), we have T xT y = a = aT = xT y = yT x (b) Since a = 224/4 ∑ xk2 ≥ 0. (a) We have xT x = whence xT y = yT x. (c) We have xT (y + z) = xT y + xT z since left matrix multiplication distributes over matrix addition.