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Let a = (3, -1, 2) and b = (1, 5, 1). There is a unique vector r…
Question
Let a = (3, -1, 2) and b = (1, 5, 1). There is a unique vector r satisfying both r × a = b and r·a = 14. The value of |r|^2 equals m/n in lowest terms with m, n positive integers. Find m + n.
✓ Verified answer: 237checked by our engine — not a guess
Step-by-step solution
For r × a = b to be solvable we need b ⊥ a; check a·b = 3 - 5 + 2 = 0, so it is consistent.
Taking the cross product of both sides with a: a × (r × a) = a × b. Using the triple product expansion, a × (r × a) = (a·a)r - (a·r)a = |a|^2 r - 14a.
Thus |a|^2 r = a × b + 14a. Here |a|^2 = 9 + 1 + 4 = 14.
a × b = ((-1)(1) - 2(5), 2(1) - 3(1), 3(5) - (-1)(1)) = (-11, -1, 16).
So 14r = (-11, -1, 16) + 14(3, -1, 2) = (31, -15, 44), giving r = (31/14, -15/14, 22/7).
|r|^2 = (31^2 + 15^2 + 44^2)/14^2 = (961 + 225 + 1936)/196 = 3122/196 = 223/14.
Then m + n = 223 + 14 = 237.
Final answer237
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