## 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 $a$?

$A(-3; -2; 4)$

$a: 2x - 3y + z - 5 = 0$

$k = -\frac{4}{5}$
## 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 $a$:

$a^{\prime}: 2 x-3 y+z+4=0$

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

$2 \cdot(-3)-3 \cdot(-2)+4+4=0$

$-6+6+4+4=0$

$8=0$

Since $8 \neq 0$, point $A$ does not belong to the image of the plane $a$.

## Problem Kuznetsov Analytical Geometry 12-26