JEE Mains · Maths · STD 12 - 11. three dimension geometry
The equation of the line through the point \((0,1,2)\) and perpendicular to the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}\) is
- A \(\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{3}\)
- B \(\frac{x}{3}=\frac{y-1}{-4}=\frac{z-2}{3}\)
- C \(\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{-3}\)
- D \(\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r\) \(\Rightarrow P(x, y, z)=(2 r+1,3 r-1,-2 r+1)\) Since, \(\overline{Q P} \perp(2 \hat{i}+3 \hat{j}-2 \hat{k})\) \(\Rightarrow 4 r+2+9 r-6+4 r+2=0\) \(\Rightarrow r =\frac{2}{17}\)…
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