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MHT CET · Maths · Three Dimensional Geometry

The symmetric equation of lines \(3 x+2 y+z-5=0\) and \(x+y-2 z-3=0\), is

  1. A \(\frac{x-1}{5}=\frac{y-4}{7}=\frac{z-0}{1}\)
  2. B \(\frac{x+1}{5}=\frac{y+4}{7}=\frac{z-0}{1}\)
  3. C \(\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}\)
  4. D \(\frac{x-1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}\)

Step-by-step Solution

Detailed explanation

Let \(a, b, c\) be the direction ratios of required line. \(\therefore 3 a+2 b+c=0\) and \(a+b-2 c=0\)
\(\Rightarrow \quad \frac{a}{-4-1}=\frac{b}{1+6}=\frac{c}{3-2}\)
\(\Rightarrow\)
\(\frac{a}{-5}=\frac{b}{7}=\frac{c}{1}\)
In order to find a point on the required line we put \(z=0\) in the two given equations to obtain, \(3 x+2 y=5\) and \(x+y=3\). Solving these two equations, we get \(x=-1, y=4\). \(\therefore\) Coordinates of point on required line are \((-1,4,0)\)
Hence, required line is \(\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}\)