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A line makes angles \alpha ,\beta ,\gamma with the coordinate axis. If \alpha + \beta =90^{\circ},then \gamma =?

  • Option 1)

    0

  • Option 2)

    90^{\circ}

  • Option 3)

    180^{\circ}

  • Option 4)

    None of these

 

Answers (1)

As 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

 

 we have,         \lambda +\beta =90^{\circ}

\Rightarrow \cos \left ( \lambda +\beta \right )= 0

\Rightarrow \cos \lambda \cos \beta -\sin \lambda \sin \beta =0

\Rightarrow \cos ^{2}\lambda \cos ^{2}\beta =\sin^{2} \lambda \sin^{2} \beta

\Rightarrow l^{2}m^{2}= \left ( 1-l^{2} \right )\left ( 1-m^{2} \right )

\Rightarrow l^{2}m^{2}= 1-l^{2}-m^{2}+l^{2}m^{2}

\Rightarrow l^{2}+m^{2}= 1

Now,

\Rightarrow l^{2}+m^{2}+n^{2}= 1

\Rightarrow 1 + n^{2} = 1 \ \ \ \Rightarrow n^{2}=0\Rightarrow n=0

So, \cos \gamma =0 \Rightarrow \gamma =90^{\circ}

 


Option 1)

0

This option is incorrect

Option 2)

90^{\circ}

This option is correct

Option 3)

180^{\circ}

This option is incorrect

Option 4)

None of these

This option is incorrect

Posted by

Vakul

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