A) 3
B) 5
C) 7
D) 9
Correct Answer: C
Solution :
The perpendicular distance of (2, 4, ? 1) from the line \[\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}\] is \[\{{{(2+5)}^{2}}+{{(4+3)}^{2}}+(-1-6)\] \[{{\left. -{{\left[ \frac{1(2+5)+4\,(4+3)-9(-1-6)}{\sqrt{1+16+81}} \right]}^{2}} \right\}}^{1/2}}\] \[=\sqrt{147-{{\left( \frac{98}{\sqrt{98}} \right)}^{2}}}\]\[=\sqrt{147-98}=\sqrt{49}=7\].You need to login to perform this action.
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