JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance ofthe point \((1,3,-7)\) from the plane passing through the point \(\left( {1, - 1, - 1} \right)\) having normal perpendicular to both the lines \(\frac{{x - 1}}{1} = \frac{{y + 2}}{{ - 2}} = \frac{{z - 4}}{3}\) and \(\frac{{x - 2}}{2} = \frac{{y + 1}}{{ - 1}} = \frac{{z + 7}}{{ - 1}}\) is . . . .
- A \(\frac{{10}}{{\sqrt {74} }}\;\)
- B \(\frac{{20}}{{\sqrt {74} }}\)
- C \(\frac{{10}}{{\sqrt {83} }}\)
- D \(\frac{5}{{\sqrt {83} }}\)
Answer & Solution
Correct Answer
(C) \(\frac{{10}}{{\sqrt {83} }}\)
Step-by-step Solution
Detailed explanation
Let the plane be \(a(x-1)+b(y+1)+c(z+1)=0\) Normal vector \(\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 1&{ - 2}&3\\ 2&{ - 1}&{ - 1} \end{array}} \right| = 5\hat i + 7\hat j + 3\hat k\) So plane is \(5(x-1)+7(y+1)+3(z+1)=0\) \(\Rightarrow 5 x+7 y+3 z+5=0\)…
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