Question

Using Wolfram Mathematica to solve the problem

(1) Given the two vectors u = <6, -2, 1> and v = <1, 8, -4> Find u x v, and find u V a) b) Find angle between vectors u and v. c) Graph both u and V on the same system. d) Now, graph vectors u, V and on the same set of axes and give u x V a different color than vectors u and V. Rotate graph from part d and show two different views of the cross e) product. f) Find the normal vector to vector U. 6, 2, -1} u v:= {1, 8, -4}

Answer 1

u = 6, - 2, 1 v = 1, 8, - 4 u = 6i - 2j + k v = i + 8j - 4k a u times v = mod i 6 1 j - 2 8 k 1 - 4 = I (8 - 8) - j (- 24 - 1 + k (48 + 2) u times v = 25j + 50k u v = (6i - 2j + k) (i + 8j - 4k) = 6 - 16 - 4 = -14 b) angle between u and v Theta = cos^-1 [uv/mod u mod v] = cos^-1 [- 14/ squareroot 41 times squareroot 79] Theta = 104.24 4 u = 6i - 2j + k u = (6, - 2,1) p = 6 q = -2, R = 1 vector.n perpendicular vector.PQ vector.nnotequalto PR vector n = vector PQ times vector PR vector PQ = vector Q - vector