A) \[180{}^\circ \]
B) \[165{}^\circ \]
C) \[150{}^\circ \]
D) \[135{}^\circ \]
Correct Answer: B
Solution :
| [b] We know that sum of square of direction cosines = 1 |
| i.e. \[{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1\] |
| \[\Rightarrow {{\cos }^{2}}45{}^\circ +{{\cos }^{2}}\beta +{{\cos }^{2}}\beta =1\] |
| (As given\[\alpha =45{}^\circ and\beta =\gamma \]) |
| \[\Rightarrow \frac{1}{2}+2{{\cos }^{2}}\beta =1\] |
| \[\Rightarrow {{\cos }^{2}}\beta =\frac{1}{4}\] |
| \[\Rightarrow \cos \beta =\pm \frac{1}{2},\] Negative value is discarded, since the line makes angle with positive axes. |
| Hence, \[\cos \beta =\frac{1}{2}\] |
| \[\Rightarrow \cos \beta =\cos 60{}^\circ \] |
| \[\beta =60{}^\circ \] |
| \[\therefore \] Required sum \[=\alpha +\beta +\gamma =45{}^\circ +60{}^\circ +60{}^\circ \] |
| \[=165{}^\circ \] |
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