AP EAMCET · Maths · Three Dimensional Geometry
The Cartesian equation of the line passing through the point \((-1,3,-2)\) and perpendicular to the lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and
\(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\) is
- A \(\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}\)
- B \(\frac{x-1}{-2}=\frac{y+3}{-7}=\frac{z-2}{-4}\)
- C \(\frac{x+1}{2}=\frac{y+3}{7}=\frac{z+2}{4}\)
- D \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\)
Step-by-step Solution
Detailed explanation
\(P=(-1,3,-2)\) Let \(\mathrm{dr}\) 's or required line be \((a, b, c)\) Dr's of line \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are \((1,2,3)\) Given that required line perpendicular to above line \(\Rightarrow \quad a_1 a_2+b_1 b_2+c_1 c_2=0\) \(\Rightarrow \quad a+2 b+3 c=0\)…
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