A) \[\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5}\]
B) \[\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+5}{3}\]
C) \[\frac{x+1}{1}=\frac{y+2}{2}=\frac{z+3}{-5}\]
D) \[\frac{x+1}{1}=\frac{y+2}{2}=\frac{z-5}{3}\]
Correct Answer: A
Solution :
The line passes through point (1, 2, 3) is \[\frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c}\] and it is perpendicular to the plane \[x+2y-5z+9=0,\] therefore the line must be \[\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5}\] because \[\sin \theta =\frac{1\,\,.\,\,1+2\,\,.\,\,2+(-\,5)\,\,(-\,5)}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{5}^{2}}}\,.\,\,\sqrt{{{1}^{2}}+{{2}^{2}}+{{5}^{2}}}}=1\] \[\,\Rightarrow \,\,\theta ={{90}^{o}}\].
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