MHT CET · Maths · Three Dimensional Geometry
The cartesian equation of a line passing through \((1,2,3)\) and parallel to planes \(x-y+2 z=5\) and \(3 x+y+z=6\) is
- A \(\frac{x-1}{-3}=\frac{y-2}{-5}=\frac{z-3}{4}\)
- B \(\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}\)
- C \(\frac{x-1}{13}=\frac{y-2}{-1}=\frac{z-3}{1}\)
- D \(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{1}\)
Answer & Solution
Correct Answer
(B) \(\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}\)
Step-by-step Solution
Detailed explanation
Required equation
\(\frac{x-1}{(-1) \times 1-1 \times 2}=\frac{y-2}{3 \times 2-1 \times 1}=\) \(\frac{z-3}{1 \times 1-3 \times(-1)}\)
\(\Rightarrow \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}\)
\(\frac{x-1}{(-1) \times 1-1 \times 2}=\frac{y-2}{3 \times 2-1 \times 1}=\) \(\frac{z-3}{1 \times 1-3 \times(-1)}\)
\(\Rightarrow \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}\)
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