JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \((1,-2,3)\) from the plane \(x-y+z=5\) measured parallel to the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}\) is
- A \(7\)
- B \(1\)
- C \(\frac{1}{7}\)
- D \(\frac{7}{5}\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
equation of line parallel to \(\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}\) passes \(\begin{array}{c}\text { through }(1,-2,3) \text { is } \\\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r \\x=2 r+1 \\y=3 r-2 \\z=-6 r+3\end{array}\) So \(\quad 2 r+1-3 r+2-6 r+3=5\)…
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