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A line in the 3-dimensional space makes an angle $\theta \left ( 0< \theta \leq \frac{\pi }{2} \right )$ with both the x and y axes. Then the set of all values of $\theta$ is the interval :

Option 1)
$\left ( 0,\frac{\pi }{4} \right )$

Option 2)
$\left ( \frac{\pi }{6},\frac{\pi }{3} \right )$

Option 3)
$\left ( \frac{\pi }{4},\frac{\pi }{2} \right )$

Option 4)
$\left ( \frac{\pi }{3},\frac{\pi }{2} \right )$

Answers (1)

best_answer

As we learnt in

Direction Cosines -

If $\alpha, \beta, \gamma$ are the angles which a vector makes with positive X-axis,Y-axis and Z-axis respectively then $\cos \alpha, \cos \beta, \cos \gamma$ are known as diresction cosines, generally denoted by (l,m,n).

$l=\cos \alpha, m=\cos \beta, n=\cos \gamma$

$l^{2}+m^{2}+n^{2}= 1$

$\cos^{2} \alpha+ \cos^{2} \beta+\cos^{2} \gamma= 1$

- wherein

Let the direction angles be ( $\left ( \alpha, \beta, \gamma \right )$ $\left ( \alpha, \beta, \gamma \right )\alpha =\theta\;and\; \beta= \theta$

We get

$\cos ^{2}\theta +\cos ^{2}\theta+\cos ^{2}\gamma =1$

$\\2\cos ^{2}\theta+\cos ^{2}\gamma =1$

$cos ^{2}\theta =\frac{1}{2}-\cos ^{2}\gamma$

So $\frac{1}{2}$

Thus

$\cos ^{2}\theta \epsilon \left ( 0,\frac{1}{2} \right )$

$cos\ \theta \epsilon \left ( 0,\frac{1}{\sqrt{2}} \right )\\$

$\theta \epsilon \left ( \frac{\pi}{4},\frac{\pi}{2} \right )$

Option 1)

$\left ( 0,\frac{\pi }{4} \right )$

This option is incorrect

Option 2)

$\left ( \frac{\pi }{6},\frac{\pi }{3} \right )$

This option is incorrect

Option 3)

$\left ( \frac{\pi }{4},\frac{\pi }{2} \right )$

This option is correct

Option 4)

$\left ( \frac{\pi }{3},\frac{\pi }{2} \right )$

This option is incorrect

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divya.saini

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