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Question

If the direction cosines of line are $$\displaystyle\frac{1}{c}, \frac{1}{c}, \frac{1}{c}$$, then.


A
0<c<1
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B
c>2
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C
c>0
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D
c=±3
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Solution

The correct option is D $$c =\pm \sqrt{3}$$
Given that the direction cosines of a line are $$\dfrac {1}{c}, \dfrac {1}{c}, \dfrac {1}{c}$$.
We know the sum of the squares of the direction cosines is one.
i.e. $$\left(\dfrac {1}{c}\right)^2+\left (\dfrac {1}{c}\right)^2+\left (\dfrac {1}{c}\right)^2=1$$
$$\Rightarrow \dfrac {3}{c^2}=1$$
$$\Rightarrow c^2=3$$
$$\Rightarrow c=\pm \sqrt 3$$

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