Evaluating magnitude of vector

  • Thread starter Thread starter Raghav Gupta
  • Start date Start date
  • Tags Tags
    Magnitude Vector
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
12 replies · 2K views
Raghav Gupta
Messages
1,010
Reaction score
76

Homework Statement



Let a and b be two unit vectors such that | a + b| = √3. If
c = a+ 2b + 3(a x b), then 2|c| is equal to:
√55
√51
√43
√37

Homework Equations


a x b = |a| |b| sinθ n where n is a unit vector
## | a + b | = \sqrt{a^2 + b^2 + 2abcosθ} ##

The Attempt at a Solution


Found cosθ = 1/2 and θ = π/3
Then 3 a x b = 3√3/2 n
Now how to find c?
We don't know angle between n and a or n and b
 
Physics news on Phys.org
Raghav Gupta said:

Homework Statement



Let a and b be two unit vectors such that | a + b| = √3. If
c = a+ 2b + 3(a x b), then 2|c| is equal to:
√55
√51
√43
√37

Homework Equations


a x b = |a| |b| sinθ n where n is a unit vector
## | a + b | = \sqrt{a^2 + b^2 + 2abcosθ} ##

The Attempt at a Solution


Found cosθ = 1/2 and θ = π/3
Then 3 a x b = 3√3/2 n
Now how to find c?
We don't know angle between n and a or n and b

Yes, you do know the angle. Go back and review the definition and properties of axb.
 
Ray Vickson said:
Yes, you do know the angle. Go back and review the definition and properties of axb.
I know angle is π/3
What to do next?
 
I have also written value of a x b in attempt in post 1
 
Raghav Gupta said:
I know angle is π/3
What to do next?

No. The angle between a and b is π/3, but that is not the angle you asked about. You asked about the angle between a and axb or between b and axb.
 
Oh that is 90°. But how I will evaluate c?
 
Raghav Gupta said:
Oh that is 90°. But how I will evaluate c?

What is preventing you from writing out all the terms of |c|^2 and evaluating them one-by-one? In other words, use the fact that
$$ |\vec{c}|^2 = \vec{c} \cdot \vec{c} \\
= (\vec{a} + 2\vec{b} +3\, \vec{a} \times \vec{b}) \cdot (\vec{a} + 2\vec{b} +3\, \vec{a} \times \vec{b}) $$
and just expand it all out.
 
Last edited:
  • Like
Likes   Reactions: Raghav Gupta
Ray Vickson said:
What is preventing you from writing out all the terms of |c|^2 and evaluating them one-by-one?
## |c|^2 = (a+2b)^2 + [3( a X b )]^2 + 2( a+2b)3( a X b ) cos θ ##
What is the angle between (a+ 2b) and a x b ?
Between a and a x b it is 90°.
 
It yields 0, so the angle is 90°?
 
Raghav Gupta said:
## |c|^2 = (a+2b)^2 + [3( a X b )]^2 + 2( a+2b)3( a X b ) cos θ ##
What is the angle between (a+ 2b) and a x b ?
Between a and a x b it is 90°.

You have not "expanded it all out"; you should be getting 6 terms, not just the 3 you have written.
 
Got it on solving.
Thanks to both of you.