JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the distance of the point \((1,-2,3)\) from the plane \(x+2 y-3 z+10=0\) measured parallel to the line, \(\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}\) is \(\sqrt{\frac{7}{2}},\) then the value of \(\mid m \mid\) is equal to ....... .
- A \(1\)
- B \(4\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(DC\) of line \(\equiv\left(\frac{3}{\sqrt{ m ^{2}+10}}, \frac{- m }{\sqrt{ m ^{2}+10}}, \frac{1}{\sqrt{ m ^{2}+10}}\right)\) \(Q \equiv\left(1+\frac{3 r }{\sqrt{ m ^{2}+10}},-2+\frac{- mr }{\sqrt{ m ^{2}+10}}, 3+\frac{ r }{\sqrt{ m ^{2}+10}}\right)\) \(Q\) lies on…
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