A) \[\frac{x-4}{1}=\frac{y-5}{1}=\frac{z-5}{-1}\]
B) \[\frac{x-2}{2}=\frac{y-3}{2}=\frac{z+3}{3}\]
C) \[\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}\]
D) \[\frac{x+3}{3}=\frac{4-y}{3}=\frac{z+1}{-\,2}\]
Correct Answer: C
Solution :
\[\frac{x-4}{2}=\,\,\frac{y-5}{2}\,\,=\,\,\frac{z-3}{1}\,\,=\,\,\lambda \,\,\,(any\,\,point)\,\] \[P\left( 2\lambda \,+\text{ }4,\text{ }2\lambda +5,\text{ }\lambda +3 \right)\]\[P\left( 2\lambda ,+\text{ }4,\text{ }2\lambda +5,\text{ }\lambda +3 \right)\] lies on plane \[x+y+z=2\] \[\therefore \,\,\,\,2\lambda +4+2\lambda +5+\lambda +3=2\] \[\lambda \,\,=\,-2\,\,\Rightarrow \,\,P\,(0,\,\,1,\,\,1)\] Hence option is correctYou need to login to perform this action.
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