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

If the length of the perpendicular from the point \((\beta , 0, \beta )\, (\beta  \neq 0)\) to the line \(\frac{x}{1} = \frac{{y - 1}}{0} = \frac{{z + 1}}{{ - 1}}\) is \(\sqrt {\frac{3}{2}} \), then \(\beta \) is equal to

  1. A \(1\)
  2. B \(-1\)
  3. C \(2\)
  4. D \(-2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-1\)

Step-by-step Solution

Detailed explanation

\(\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}=\lambda\) A point on this line is \(A(0,1,-1)\) \(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}=0\) We get \(\lambda=\frac{-1}{2}\) \(\therefore C=\left(-\frac{1}{2}, 1, \frac{-1}{2}\right)\)…
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