## Problem Statement

Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?

$A\left(1 ; \frac{1}{3} ;-2\right)$

$a: x-3 y+z+6=0$

$k=\frac{1}{3}$
## Solution

When transforming similarity with the center at the origin of the plane

$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane

$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:

$a^{\prime}: x-3 y+z+2=0$

Substitute the coordinates of point $A$ into the equation $a^{\prime}:$

$1-3 \cdot \frac{1}{3}-2+2=0$

$1-1-2+2=0$

$0=0$

Since $0=0$, point $A$ belongs to the image of the plane $a$.

## Problem Kuznetsov Analytical Geometry $12-5$