## 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 $a$?

$A(1; -2; 1)$

$a: 5x + y - z + 6 = 0$

$k = \frac{2}{3}$
## Solution

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

$a: A x+B y+C z+D=0$ and the coefficient $k$, it transitions to the plane

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

$a^{\prime}: 5 x+y-z+4=0$

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

$5 \cdot 1-2-1+4=0$

$5-2-1+4=0$

$6=0$

Since $6 \neq 0$, point $A$ does not belong to the image of the plane $a$.

## Problem Kuznetsov Analytical Geometry 12-8