Question:
A plane meets the coordinate axes at points A, B and C, such that the centroid of a triangle ABC is \((2,-\frac{2}{3},\frac{1}{2})\). The perpendicular distance from the origin to this plane is...
A plane meets the coordinate axes at points A, B and C, such that the centroid of a triangle ABC is \((2,-\frac{2}{3},\frac{1}{2})\). The perpendicular distance from the origin to this plane is...
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Centroid coordinates are one third of the intercepts.
Updated On: Oct 1, 2026
- \(\frac{6}{\sqrt{26}}\)
- \(\frac{5}{\sqrt{26}}\)
- \(\frac{4}{\sqrt{26}}\)
- \(\frac{3}{\sqrt{26}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the intercepts
If the plane meets the axes at \((a,0,0)\), \((0,b,0)\), \((0,0,c)\), the centroid is \(\left(\frac a3,\frac b3,\frac c3\right)\).
So \(a=6\), \(b=-2\), \(c=\frac32\).
Step 2: Plane equation
\(\frac x6-\frac y2+\frac{2z}{3}=1\). Multiply by \(6\): \(x-3y+4z=6\).
Step 3: Distance
\[ d=\frac{6}{\sqrt{1+9+16}}=\frac{6}{\sqrt{26}} \]
Option (A).
Final Answer:
The distance is \(\frac{6}{\sqrt{26}}\), option (A). \[ \boxed{\text{(A)}} \]
If the plane meets the axes at \((a,0,0)\), \((0,b,0)\), \((0,0,c)\), the centroid is \(\left(\frac a3,\frac b3,\frac c3\right)\).
So \(a=6\), \(b=-2\), \(c=\frac32\).
Step 2: Plane equation
\(\frac x6-\frac y2+\frac{2z}{3}=1\). Multiply by \(6\): \(x-3y+4z=6\).
Step 3: Distance
\[ d=\frac{6}{\sqrt{1+9+16}}=\frac{6}{\sqrt{26}} \]
Option (A).
Final Answer:
The distance is \(\frac{6}{\sqrt{26}}\), option (A). \[ \boxed{\text{(A)}} \]
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