Question

A variable plane moves so that the sum of reciprocal 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 is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ ...$\left(1\right)$

The intercept on the coordinate axes are $a,b,c$.

The sum of reciprocals of intercepts is constant $\lambda$, therefore

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\lambda \Rightarrow \frac{\left(1/\lambda \right)}{a}+\frac{\left(1/\lambda \right)}{b}+\frac{\left(1/\lambda \right)}{c}=1$

$\therefore (\frac{1}{\lambda },\frac{1}{\lambda },\frac{1}{\lambda })$ lies on the plane $\left(1\right)$

Hence, the vatiable plane $\left(1\right)$ always passes through the fixed point $(\frac{1}{\lambda },\frac{1}{\lambda },\frac{1}{\lambda })$