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Let vector a = i + 2j - 3k and vector b = 2i - 3j + 5k. If vector r satisfies r cross a = b cross r, r dot (i + 2j + k) = 3, and r dot (2i + 5j - alpha*k) = -1 where alpha is a real number, find alpha + |r|².

1. 9
2. 15
3. 13
4. 11

Correct answer: 15

Solution

r cross a = b cross r => r cross a + r cross b = 0 => r cross (a+b) = 0 => r is parallel to (a+b). a+b = (1+2)i + (2-3)j + (-3+5)k = 3i - j + 2k. So r = lambda*(3,-1,2). Using r dot (1,2,1) = 3: 3*lambda - 2*lambda + 2*lambda = 3*lambda = 3 => lambda = 1. So r = (3,-1,2). |r|² = 9+1+4 = 14. Using r dot (2,5,-alpha) = -1: 6 - 5 + 2*alpha = 1 + 2*alpha = -1 => 2*alpha = -2 => alpha = -1... wait: r dot (2i+5j-alpha*k) = 3*2 + (-1)*5 + 2*(-alpha) = 6-5-2*alpha = 1-2*alpha = -1 => -2*alpha = -2 => alpha = 1. alpha + |r|² = 1+14 = 15.

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