Question

Show that the direction cosine L,M,N of vectors Ax,Ay,Az is given by L=Ax/|A| , M=Ay/|A|, N=Az/|A| and hence L^2+M^2+N^2=1

Answer 1

Answer on Question #54673, Physics Mechanics Kinematics Dynamics

Show that the direction cosine $\mathrm{L}, \mathrm{M}, \mathrm{N}$ of vectors $\mathrm{Ax}$ , $\mathrm{Ay}$ , $\mathrm{Az}$ is given by $\mathrm{L} = \mathrm{Ax} / |\mathrm{A}|$ , $\mathrm{M} = \mathrm{Ay} / |\mathrm{A}|$ , $\mathrm{N} = \mathrm{Az} / |\mathrm{A}|$ and hence $\mathrm{L}^{\wedge}2 + \mathrm{M}^{\wedge}2 + \mathrm{N}^{\wedge}2 = 1$ .

Solution

Fig.1

The cosines of the angles $\alpha, \beta$ , and $\gamma$ in Fig. 1 are called the direction cosines and are designated by $l, m$ , and $n$ , respectively. Thus, in terms of $A, A_x, A_y$ , and $A_z$

$$\left\{ \begin{array}{l} l = \cos \alpha = A _ {x} / A \\ m = \cos \beta = A _ {y} / A \\ n = \cos \gamma = A _ {z} / A \end{array} \right.$$

According to the Pythagorean theorem

$$\left(A _ {x}\right) ^ {2} + \left(A _ {y}\right) ^ {2} + \left(A _ {z}\right) ^ {2} = A ^ {2}$$

Then

$$\left(A _ {x} / A\right) ^ {2} + \left(A _ {y} / A\right) ^ {2} + \left(A _ {z} / A\right) ^ {2} = 1$$

So, from Eq.(1)

$$(1)^2 + (m)^2 + (n)^2 = 1$$

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