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for unit vectors a,b and c, if |a-b|^2+|b-c|^2+|c-a|^2 = 9 then what is |2a+5b+5c|?

for unit vectors a,b and c, if |a-b|^2+|b-c|^2+|c-a|^2 = 9 then what is |2a+5b+5c|?

Grade:

2 Answers

Y RAJYALAKSHMI
45 Points
6 years ago
|a – b|2 + |b – c|2 + |c – a|2 = 9 
=> 2(|a|2 + |b|2 + |c|2) – 2(a.b + b.c + c.a) = 9 
=> 2 * 3 – 2(a.b + b.c + c.a) = 9  (Since a, b, c are unit vectors)
=> (cos (a, b)  + cos (b, c)  + cos(c, a) = – 3/2  [Since a, b, c are unit vectors & a.b = |a||b| cos(a, b) where cos (a, b) ,  cos (b, c)  , cos(c, a) are cosine of angles between these vecotrs ] 
This gives us cos (a, b)  =  cos (b, c)  = cos(c, a) = –1/2
=> a.b = b.c = c.a = –1/2
|2a + 5b + 5c|2 = 4|a|2 + 25|b|2 + 25|c|2 + 20 a. b + 50 b. c +  20 c.a
= 4 + 25 +25 + 20 * (–1/2) + 50 * (–1/2) + 20 * (–1/2) = 9
Therefore, |2a + 5b + 5c| = 3
 
 
 
Kushagra Madhukar
askIITians Faculty 629 Points
8 months ago
Dear student,
Please find the solution to your problem.
 
|a – b|2 + |b – c|2 + |c – a|2 = 9 
=> 2(|a|2 + |b|2 + |c|2) – 2(a.b + b.c + c.a) = 9 
=> 2 . 3 – 2(a.b + b.c + c.a) = 9  (Since a, b, c are unit vectors)
=> (cos (a, b)  + cos (b, c)  + cos(c, a) = – 3/2  [Since a, b, c are unit vectors & a.b = |a||b| cos(a, b) where cos (a, b) ,  cos (b, c)  , cos(c, a) are cosine of angles between these vecotrs ] 
This gives us cos (a, b)  =  cos (b, c)  = cos(c, a) = –1/2
=> a.b = b.c = c.a = –1/2
|2a + 5b + 5c|2 = 4|a|2 + 25|b|2 + 25|c|2 + 20 a. b + 50 b. c +  20 c.a
= 4 + 25 +25 + 20 * (–1/2) + 50 * (–1/2) + 20 * (–1/2) = 9
Therefore, |2a + 5b + 5c| = 3
 
Thanks and regards,
Kushagra

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