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

બિંદુ \((0, 1, 2)\) માંથી પસાર થતી અને રેખા \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}\)ને લંબ રેખાનું સમીકરણ ............. છે.

  1. A \(\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{3}\)
  2. B \(\frac{x}{3}=\frac{y-1}{-4}=\frac{z-2}{3}\)
  3. C \(\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{-3}\)
  4. D \(\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}\)
Verified Solution

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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