JEE Mains · Maths · STD 12 - 11. three dimension geometry
A plane \(P\) meets the coordinate axes at \(A, B\) and \(C\) respectively. The centroid of \(\Delta ABC\) is given to be \((1,1,2)\) . Then the equation of the line through this centroid and perpendicular to the plane \(P\) is
- A \(\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\)
- B \(\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}\)
- C \(\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-2}{1}\)
- D \(\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-2}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}\)
Step-by-step Solution
Detailed explanation
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) \(A \equiv(a, 0,0), B \equiv(0, b, 0), C \equiv(0,0, c)\) Centroid \(\equiv\left(\frac{ a }{3}, \frac{ b }{3}, \frac{ c }{3}\right)=(1,1,2)\) \(a=3, b=3, c=6\) Plane \(: \frac{x}{3}+\frac{y}{3}+\frac{z}{6}=1\) \(2 x+2 y+z=6\) line…
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