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)

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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