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Junior Member
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Apr 8, 2011, 01:07 PM
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Q.If O and o' are circumcentre and orthocentre of triangle ABC , than OAvector +OBvector +OCvector equals
1. 2oo'vector
2. oo'vector
3. o'ovector
4. 2o'ovector
please give your choice with correct explanation
Q. If a,b,c are non-coplanar unit vectors such that avector cross (bvector cross cvector) = (bvector+cvector)/sqrtroot2,
then the angle between avector and bvector is
1. 3pi/4
2. pi/4
3. pi/2
4. pi
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Senior Member
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Apr 9, 2011, 09:32 PM
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If you look at the attached picture where I've drawn it out, you can see that the answer to the first question is the OO' vector.
The circumcenter, O, is constructed with blue lines. The orthocenter, O', is constructed with green lines.
You can see when the OA, OB, and OC vectors (shown in red) are added head to tail starting at O, the finish point is exactly O'.
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Junior Member
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Apr 13, 2011, 02:48 PM
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Comment on jcaron2's post
Thanks a lot for your answer
Sir please also answer the 2nd question
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Senior Member
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Apr 13, 2011, 08:23 PM
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Here's how I approach the second question without putting too, too much thought into it:
The problem suggests that there's only one answer. That means that the angle between A and B is always the same, no matter what B and C vectors you choose. Hence if we can find the answer for some B and C combination, that must be the correct answer for all B's and C's.
Therefore, I choose to assume B and C are orthogonal unit vectors directed in the x and y directions, respectively. (If you write the vectors in terms of the typical I, j, and k unit vectors then B = I and C = j).
That means that the right side of the equation is equal to a unit vector directed at a pi/2 angle from B and C, but in the same plane.
Meanwhile, the cross-product in parentheses on the left side of the equation is equal to a unit vector in the z direction (a.k.a. the k vector).
So what vector A, crossed with k, results in the right side of the equation? One which is directed in the same plane as B and C directed along the -x and y directions. That puts it at an angle of 3pi/4 from B.
See the figure below for a better illustration of what I'm trying to say. The unit circle is just so you can see that the vectors all end up being unit length.
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Junior Member
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Apr 14, 2011, 03:48 AM
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Comment on jcaron2's post
sir thanks for giving your answer but I am having trouble to understand that how did u say that the angle is 3pi/4
sir the solution of this question( from the book from where I had given this question) is
avector.bvector =-1/sqrt2
cos@ = -1/sqrt2
@=3pi/4
where @ is the angle b/w A and B
in this solution I am not able to understand that how did we get avector.bvector=1/sqrt2
please help me
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Senior Member
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Apr 14, 2011, 08:39 AM
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ApoorvGoel, sorry, there was a much easier way to approach this problem. :o
Are you familiar with Lagrange's formula (also known as a triple product expansion)?
The formula says:
The means that the right-hand side of your equation is equal to the right-hand side of Lagrange's equation:
Since the B and C vectors are independent, you can separate the above into two separate equations.
Therefore
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Junior Member
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Apr 22, 2011, 08:38 PM
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Comment on jcaron2's post
Thanks a lot for your answer
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Senior Member
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Apr 22, 2011, 09:16 PM
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Comment on jcaron2's post
You're welcome!
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