JEE Mains · Maths · STD 12 - 11. three dimension geometry
The perpendicular distance from the origin to the plane containing the two lines,\(\frac{{x + 2}}{3} = \frac{{y - 2}}{5} = \frac{{z + 5}}{7}\) and \(\frac{{x - 1}}{1} = \frac{{y - 4}}{4} = \frac{{z + 4}}{7}\), is
- A \(11\sqrt 6 \)
- B \(11/\sqrt 6\)
- C \(11\)
- D \(6\sqrt {11} \)
Answer & Solution
Correct Answer
(B) \(11/\sqrt 6\)
Step-by-step Solution
Detailed explanation
Equation of plane is \(\left| {\begin{array}{*{20}{c}} {x + 2}&{y - 2}&{z + 5}\\ 3&5&7\\ 1&4&7 \end{array}} \right| = 0\) \(\Rightarrow(x+2) 7-(y+2) 14+(z+5) 7=0\) \(\Rightarrow x-2 y+z+11=0\) The perpendicular distance from the origin to the plane is…
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