Question:
If a variable plane in 3-dimensional space moves in such a way that the sum of the reciprocals of its intercepts on the \(x\) and \(y\)-axes exceeds the reciprocal of its intercept on the \(z\)-axis by 2, then all such planes will pass through the point :
If a variable plane in 3-dimensional space moves in such a way that the sum of the reciprocals of its intercepts on the \(x\) and \(y\)-axes exceeds the reciprocal of its intercept on the \(z\)-axis by 2, then all such planes will pass through the point :
Show Hint
Write the plane in intercept form and convert the given condition into a linear relation among \(\frac1a,\frac1b,\frac1c\).
Updated On: Nov 12, 2026
- \(\left(\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)\)
- \(\left(\frac{1}{2},-\frac{1}{2}, \frac{1}{2}\right)\)
- \(\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\)
- \(\left(\frac{1}{2}, \frac{1}{2},-\frac{1}{2}\right)\)
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The Correct Option is D
Solution and Explanation
Let the intercepts on the axes be \(a, b, c\). The plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\).
Given: \(\frac1a+\frac1b=\frac1c+2\), that is \(\frac1a+\frac1b-\frac1c=2\).
Divide by 2: \(\frac12\cdot\frac1a+\frac12\cdot\frac1b+\left(-\frac12\right)\cdot\frac1c=1\).
This says the point \(\left(\frac12,\frac12,-\frac12\right)\) satisfies \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) for every such plane.
Answer: (4) \(\left(\frac{1}{2}, \frac{1}{2},-\frac{1}{2}\right)\)
Given: \(\frac1a+\frac1b=\frac1c+2\), that is \(\frac1a+\frac1b-\frac1c=2\).
Divide by 2: \(\frac12\cdot\frac1a+\frac12\cdot\frac1b+\left(-\frac12\right)\cdot\frac1c=1\).
This says the point \(\left(\frac12,\frac12,-\frac12\right)\) satisfies \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) for every such plane.
Answer: (4) \(\left(\frac{1}{2}, \frac{1}{2},-\frac{1}{2}\right)\)
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