MHT CET · Maths · Three Dimensional Geometry
If the points \((1,1, \lambda)\) and \((-3,0,1)\) are equidistant from the plane \(3 x+4 y-12 z+13=0\), then integer value of \(\lambda\) is
- A \(2\)
- B \(1\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{A}(1,1, \lambda)\) and \(\mathrm{B}(-3,0,1)\) are equidistant from \(3 \mathrm{x}+4 \mathrm{y}-12 \mathrm{z}+13=0\)
\(\therefore\left|\frac{3(1)+4(1)-12 \lambda+13}{\sqrt{9+16+144}}\right| \)
\( \therefore\left|\frac{20-12 \lambda}{13}\right|=\left|\frac{-8}{13}\right| \)
\( \therefore 20-12 \lambda=\pm 8 \)
\( \therefore 20-12 \lambda=8 \text { or } 20-12 \lambda=-8 \Rightarrow \lambda=1 \text { or } \lambda=\frac{7}{3}\)
\(\therefore\left|\frac{3(1)+4(1)-12 \lambda+13}{\sqrt{9+16+144}}\right| \)
\( \therefore\left|\frac{20-12 \lambda}{13}\right|=\left|\frac{-8}{13}\right| \)
\( \therefore 20-12 \lambda=\pm 8 \)
\( \therefore 20-12 \lambda=8 \text { or } 20-12 \lambda=-8 \Rightarrow \lambda=1 \text { or } \lambda=\frac{7}{3}\)
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