A) \[-\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\]
B) \[\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\]
C) \[-\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\]
D) \[\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}\]
Correct Answer: A
Solution :
The direction cosines of the normal to the plane are \[\frac{1}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}},\frac{2}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}},\frac{-\,3}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}}\] i.e., \[\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{-\,3}{\sqrt{14}}\]. But \[x+2y-3z+4=0\] can be written as \[-x-2y+3z-4=0\]. Thus the direction cosines are \[\frac{-1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{3}{\sqrt{14}}\].
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