Three Dimensional Geometry Question 350
Question: The equation of the line which passes through the point (1, 1, 1) and intersect the lines $ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} $ and $ \frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4} $ is
Options:
A) $ \frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17} $
B) $ \frac{x-1}{3}=\frac{y-1}{3}=\frac{z-1}{-5} $
C) $ \frac{x-1}{-2}=\frac{y-1}{1}=\frac{z-1}{-4} $
D) $ \frac{x-1}{8}=\frac{y-1}{-2}=\frac{z-1}{3} $
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Answer:
Correct Answer: A
Solution:
[a] Any line passing through the point $ (1,1,1) $ is $ \frac{x-a}{a}=\frac{y-a}{b}=\frac{z-a}{c} $ (i) This line intersects the line $ \frac{x-1}{2}=\frac{y-2}{3} $ $ =\frac{z-3}{4} $ . If $ a:b:c\ne 2:3:4 $ and $ \begin{vmatrix} 1-1 & 2-1 & 3-1 \\ a & b & c \\ 2 & 3 & 4 \\ \end{vmatrix} =0 $
$ \Rightarrow a=2b+c=0 $ ???(ii) Again, line (i) intersects line $ \frac{x-(-2)}{1}=\frac{y-3}{2} $ $ =\frac{z-(-1)}{4} $ . If $ a:b:c\ne 2:3:4 $ and $ \begin{vmatrix} -2-1 & 3-1 & 3-1 \\ a & b & c \\ 1 & 2 & 4 \\ \end{vmatrix} =0 $
$ \Rightarrow 6a+5b-4c=0 $ ???.(iii) From (ii) and (iii) by cross multiplication, we have $ \frac{a}{8-5}=\frac{b}{6+4}=\frac{c}{5+12} $ or $ \frac{3}{a}=\frac{b}{10}=\frac{c}{17} $ So, the required line is $ \frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}. $
BETA
