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JEE Mains · Maths · STD 12 - 11. three dimension geometry

ધારો કે રેખા \(\mathrm{L}\) એ, રેખાઓ \(x-2=-y=z-1,2(x+1)=2(y-1)=z+1\) ને છે, તથા રેખા \(\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}\) ને સમાંતર છે. તો નીચેના બિંદૂઓ પૈકી ક્યું \(L\) પર આવેલ છે ?

  1. A  \(\left(-\frac{1}{3}, 1,1\right)\)
  2. B  \(\left(-\frac{1}{3}, 1,-1\right)\)
  3. C  \(\left(-\frac{1}{3},-1,-1\right)\)
  4. D  \(\left(-\frac{1}{3},-1,1\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B)  \(\left(-\frac{1}{3}, 1,-1\right)\)

Step-by-step Solution

Detailed explanation

\(\mathrm{L}_1: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\lambda\) \(\mathrm{L}_2: \frac{\mathrm{x}+1}{\frac{1}{2}}=\frac{\mathrm{y}-1}{\frac{1}{2}}=\frac{\mathrm{z}+1}{1}=\mu\) dr of line \(MN\) will be…
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