Finding the angle of unit vectors

[rbnphlp]

If $\vec{OA}$ is the unit vector $l_1i+m_1j+n_1k$ and $\vec{OB}$ is the unit vector $l_2i+m_2j+n_2k$, by using the cosine formula in triangle OAB find the angle between $\vec{OA}$ &$\vec{OB}$..

I have tried expressing them as direction cosines , but none of that is working..can anyone point me in the right direction..
Also I have not been introduced to scalar product ..is there an easy way of go about doing this without using the scalar product
Thanks

 

[D H]

Hint: Use the law of cosines.

 

[rbnphlp]

Hint: Use the law of cosines.

the question does state to use that ..a further hint maybe:(

 

[D H]

What is the law of cosines?

 

[rbnphlp]

What is the law of cosines?

a^2=b^2+c^2-2bccosA ..

and since the length is 1 b=c

but what do I do with a unit vector?

 

[D H]

a^2=b^2+c^2-2bccosA ..

You are trying to solve for cos(A), not a^2. You already know (or can know) a^2. Why don't you re-arrange the above in terms of solving for cos(A)?

 

[rbnphlp]

You are trying to solve for cos(A), not a^2. You already know (or can know) a^2. Why don't you re-arrange the above in terms of solving for cos(A)?

thanks ...I have solved it now