Three Dimensional Geometry Question 228
Question: The perpendicular distance of the point (2, 4, ?1) from the line $ \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} $ is
[Kurukshetra CEE 1996]
Options:
A) 3
B) 5
C) 7
D) 9
Show Answer
Answer:
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) $ $ {{. -{{[ \frac{1(2+5)+4,(4+3)-9(-1-6)}{\sqrt{1+16+81}} ]}^{2}} }}^{1/2}} $ $ =\sqrt{147-{{( \frac{98}{\sqrt{98}} )}^{2}}} $ $ =\sqrt{147-98}=\sqrt{49}=7 $ .
BETA
