Three Dimensional Geometry Question 124
Question: The line which passes through the origin and intersect the two lines $ \frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{3},\frac{x-4}{2}=\frac{y+3}{3}=\frac{z-14}{4}, $ is
Options:
A) $ \frac{x}{1}=\frac{y}{-3}=\frac{z}{5} $
B) $ \frac{x}{-1}=\frac{y}{3}=\frac{z}{5} $
C) $ \frac{x}{1}=\frac{y}{3}=\frac{z}{-5} $
D) $ \frac{x}{1}=\frac{y}{4}=\frac{z}{-5} $
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Answer:
Correct Answer: A
Solution:
[a] Let the line be $ \frac{x}{a}=\frac{y}{b}=\frac{z}{c} $ (i) If line (i) intersects with the line $ \frac{x-1}{2} $ $ =\frac{y+3}{4}=\frac{z-5}{3}, $ Then $ \begin{vmatrix} a & b & c \\ 2 & 4 & 3 \\ 4 & -3 & 14 \\ \end{vmatrix} =0\Rightarrow 9a-7b-10c=0 $ (ii) From (i) and (ii), we have $ \frac{a}{1}=\frac{b}{-3}=\frac{c}{5} $
$ \therefore $ The line is $ \frac{x}{1}=\frac{y}{-3}=\frac{z}{5} $
BETA
