Three-Dimensional-Geometry Question 415
Question: The direction cosines of the line $ \frac{3x+1}{-3}=\frac{3y+2}{6}=\frac{z}{-1} $ are
Options:
A) $ ( \frac{1}{3},\frac{2}{3},,0 ) $
B) $ ( -1,\frac{2}{3},,1 ) $
C) $ ( -\frac{1}{2},,1,,-\frac{1}{2} ) $
D) $ ( -\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
Change the given equation in standard form, we get, $ \frac{x+\frac{1}{3}}{-1}=\frac{y+\frac{2}{3}}{2}=\frac{z}{-1} $ So direction cosine are, $ ( \frac{-1}{\sqrt{6}},,\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}} ) $ .
BETA
