## Task Condition

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 $a$?

$A(-1 ; 2 ; 3)$

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

$k=2.5$
## 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 $a$:

$a^{\prime}: x-3 y+z+5=0$

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

$-1-3 \cdot 2+3+5=0$

$-1-6+3+5=0$

$1=0$

Since $1 \neq 0$, point $A$ does not belong to the image of the plane $a$.

## Problem Kuznetsov Analytical Geometry 12-20