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A unit vector c-hat is inclined at angle theta to each of two mutually perpendicular unit vectors a-hat and b-hat. If c-hat = lambda*(a-hat + b-hat) + mu*(a-hat x b-hat) where lambda and mu are real numbers, then theta belongs to:

1. (-pi/4, 0)
2. [0, pi/4)
3. [3*pi/4, pi)
4. [pi/4, 3*pi/4]

Correct answer: [pi/4, 3*pi/4]

Solution

c-hat. a-hat = lambda*(1+0) + mu*(a-hat x b-hat).a-hat = lambda (since (axb).a=0). So lambda = cos(theta). Similarly lambda = cos(theta) from b-hat. Now |c-hat|² = lambda²*(|a|²+|b|²+2a.b) + mu²*|axb|² = 2*lambda² + mu² = 1. So 2cos²(theta) + mu² = 1, giving mu² = 1 - 2cos²(theta) >= 0. Thus cos²(theta) <= 1/2, meaning |cos(theta)| <= 1/sqrt(2), so theta in [pi/4, 3pi/4].

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