A) \[ax-by\pm z\sqrt{{{a}^{2}}+{{b}^{2}}}\cdot \cot \alpha =0\]
B) \[ax-by\pm z\sqrt{{{a}^{2}}+{{b}^{2}}}\cdot \tan \alpha =0\]
C) \[ax+by\pm z\sqrt{{{a}^{2}}+{{b}^{2}}}\cdot \cot \alpha =0\]
D) \[ax+by\pm z\sqrt{{{a}^{2}}+{{b}^{2}}}\cdot \tan \alpha =0\]
Correct Answer: C
Solution :
Given planes are \[ax+by=0\] ... (i) and \[z=0\] ... (ii) \[\therefore \]Equation of any plane passing through the line of intersection of planes (i) and (ii) may be taken as \[ax+by+Kz=0\] ... (iii) The \[DC's\] of a normal to the plane (iii) are \[\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{K}^{2}}}},\,\,\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{K}^{2}}}},\,\,\frac{K}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{K}^{2}}}}\]The \[DC's\] of a normal to the plane (i) are \[\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}},\,\,\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=0\] Since, the angle between the planes (i) and (ii) is\[\alpha \]. \[\therefore \] \[\cos \alpha =\frac{a\cdot a+b\cdot b+c\cdot c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{K}^{2}}}\sqrt{{{a}^{2}}+{{b}^{2}}}}\] On putting the value of K in Eq. (iii), we get the equation of plane as \[ax+by\pm z\sqrt{{{a}^{2}}+{{b}^{2}}}\tan \alpha =0\]
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