A matrix problem :)

Question

If $l_i,m_i,n_i$ ; $i=1,2,3$ denote the direction cosines of three mutually perpendicular vectors in space, provided that $AA^T=I$ ,where $$A=\begin{bmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \\ \end{bmatrix} $$

I couldn't quite understand the question. Sorry for not posting 'how I tried to solve the answer'.

Please provide assistance.
Thank you. :)

it is not to prove but to denote the direction cosines. is it not? — chndn, May 22 '13 at 14:32

score 2 · Accepted Answer · answered May 22 '13 at 13:56

2

Hint: Consider first $$A^TA= \begin{bmatrix}l_1 & l_2 & l_3 \\m_1 & m_2 & m_3 \\n_1 & n_2 & n_3 \\ \end{bmatrix} \begin{bmatrix}l_1 & m_1 & n_1 \\l_2 & m_2 & n_2 \\l_3 & m_3 & n_3 \\ \end{bmatrix} = \cdots $$ by working out a few of the elements in terms of the components of $\mathbf {l,m,n}$. Then ask yourself: if $AB=I$, then what is $A^{-1}$? Therefore, what is $BA$?

answered May 22 '13 at 13:56

not all wrong

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what is meant by 'direction cosines of three mutually perpendicular vectors in space'? – chndn May 22 '13 at 14:18
1

Oh, it means $\mathbf{l\cdot m}=0$ etc. (mutually perpendicular) and $\lVert \mathbf l\rVert^2=1$ etc. (unit length, direction cosines = projections onto axes only). – not all wrong May 22 '13 at 14:31
can u elaborate further? I still couldnt understand... +1 – chndn May 22 '13 at 14:34
It just means they are three, unit length, orthogonal vectors. That's all. – not all wrong May 22 '13 at 14:43

A matrix problem :)

1 Answers1