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
- A \(1\)
- B \(-1\)
- C \(2\)
- D \(-2\)
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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