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) \[\frac{x-1}{8}=\frac{y-1}{-2}=\frac{z-1}{3}\]
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 \[\left| \begin{matrix} 1-1 & 2-1 & 3-1 \\ a & b & c \\ 2 & 3 & 4 \\ \end{matrix} \right|=0\] |
| \[\Rightarrow a=2b+c=0\] ???(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 \[\left| \begin{matrix} -2-1 & 3-1 & 3-1 \\ a & b & c \\ 1 & 2 & 4 \\ \end{matrix} \right|=0\] |
| \[\Rightarrow 6a+5b-4c=0\] ???.(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 \[\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}.\] |
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