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Let a-hat and b-hat be two unit vectors such that |(a-hat + b-hat) + 2*(a-hat x b-hat)| = 2. If theta in (0, pi) is the angle between a-hat and b-hat, then consider: (S1): 2|a-hat x b-hat| = |a-hat - b-hat| (S2): The projection of a-hat on (a-hat + b-hat) is 1/2.

1. Only (S1) is true
2. Only (S2) is true
3. Both (S1) and (S2) are true
4. Both (S1) and (S2) are false

Correct answer: Both (S1) and (S2) are true

Solution

Since a-hat x b-hat is perpendicular to both a-hat and b-hat (and hence to a-hat+b-hat), we have (a+b).(a x b)=0. So |(a+b)+2(a x b)|² = |a+b|² + 4|a x b|² = 4. Substituting: (2+2cos(theta)) + 4sin²(theta) = 4 => 2cos(theta) + 4(1-cos²(theta)) = 2 => 4cos²(theta) - 2cos(theta) - 2 = 0 => (2cos(theta)+1)(cos(theta)-1)=0 => cos(theta)=-1/2 (since theta≠0) => theta=2pi/3. Now check S1: 2|axb|=2sin(2pi/3)=sqrt(3). |a-b|=sqrt(2-2cos(2pi/3))=sqrt(2+1)=sqrt(3). Equal. S1 is TRUE. Check S2: projection of a-hat on (a+b) = a.(a+b)/|a+b| = (|a|²+a.b)/|a+b| = (1+cos(2pi/3))/sqrt(2+2cos(2pi/3)) = (1-1/2)/sqrt(1) = (1/2)/1 = 1/2. S2 is TRUE.

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