Question

A variable plane moves such that the sum of reciprocals of its intercepts on the three coordinate axes is constant $\lambda$. It passes through a fixed point, which has coordinates

• A. $\lambda ,\lambda ,\lambda$
• B. $(\frac{1}{\lambda },\frac{1}{\lambda },\frac{1}{\lambda })$
• C. $-\lambda ,-\lambda ,-\lambda$
• D. $(-\frac{1}{\lambda },-\frac{1}{\lambda },-\frac{1}{\lambda })$

Answer 1

The correct option is B $(\frac{1}{\lambda },\frac{1}{\lambda },\frac{1}{\lambda })$

Let the equation of the variable plane be
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\cdots \left(i\right)$
The intercepts on the coordinate axes are a,b,c.
The sum of reciprocals of intercept is constant $\lambda$
$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\lambda \Rightarrow \frac{\left(\frac{1}{\lambda }\right)}{a}+\frac{\left(\frac{1}{\lambda }\right)}{b}+\frac{\left(\frac{1}{\lambda }\right)}{c}=1$
$\therefore (\frac{1}{\lambda },\frac{1}{\lambda },\frac{1}{\lambda })$ lies on the plane $\left(i\right)$
Hence the variable plane $\left(i\right)$ always passes through the fixed point $(\frac{1}{\lambda },\frac{1}{\lambda },\frac{1}{\lambda })$