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Question

If the direction cosines of a line are $$\left(\dfrac {1}{c},\dfrac {1}{c},\dfrac {1}{c}\right)$$ then show that $$c=\pm \sqrt {3}$$.


Solution

Direction cosines 

$$\alpha =\dfrac{v_x}{\sqrt{v_x^2+v_y^2+v_z^2}}=\dfrac{1}{c}$$

$$\beta  =\dfrac{v_x}{\sqrt{v_x^2+v_y^2+v_z^2}}=\dfrac{1}{c}$$

$$\gamma  =\dfrac{v_x}{\sqrt{v_x^2+v_y^2+v_z^2}}=\dfrac{1}{c}$$

$$\Rightarrow v_x=v_y=v_z=1$$

and

$$\sqrt{v_x^2+v_y^2+v_z^2} =c$$

$$\sqrt{1^2+1^2+1^2} =c$$

$$\pm \sqrt{3}=c$$

Maths

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