## 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 $a$?

$A(-2 ; 3 ;-3)$

$a: 3 x+2 y-z-2=0$

$k=\frac{3}{2}$
## 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 $a$:

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

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

$3 \cdot(-2)+2 \cdot 3-(-3)-3=0$

$-6+6+3-3=0$

$0=0$

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

## Problem Kuznetsov Analytical Geometry 12-11