Question:
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 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 :
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If the plane cuts intercepts \(\alpha, \beta, \gamma\), the centroid is \((\alpha/3, \beta/3, \gamma/3)\); the plane is \(\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1\) and its normal gives the line's direction.
Updated On: Nov 5, 2026
- \(\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}\)
- \(\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-2}{1}\)
- \(\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}\)
- \(\frac{x-1}{1} = \frac{y-1}{2} = \frac{z-2}{2}\)
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The Correct Option is C
Solution and Explanation
Let the plane be \(\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1\), so \(A(\alpha,0,0)\), \(B(0,\beta,0)\), \(C(0,0,\gamma)\).
Centroid \(= \left(\frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3}\right) = (1, 1, 2)\), so \(\alpha = 3\), \(\beta = 3\), \(\gamma = 6\).
Plane: \(\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1\), i.e. \(2x + 2y + z = 6\). Its normal has direction ratios \((2, 2, 1)\).
The required line passes through \((1, 1, 2)\) and is parallel to the normal: \(\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}\).
Answer: (3) \(\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}\)
Centroid \(= \left(\frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3}\right) = (1, 1, 2)\), so \(\alpha = 3\), \(\beta = 3\), \(\gamma = 6\).
Plane: \(\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1\), i.e. \(2x + 2y + z = 6\). Its normal has direction ratios \((2, 2, 1)\).
The required line passes through \((1, 1, 2)\) and is parallel to the normal: \(\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}\).
Answer: (3) \(\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}\)
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