## 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 $a$?

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

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

$k=1.5$
## 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 $a$:

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

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

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

$1-1+3-3=0$

$0=0$

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

Problem Kuznetsov Analytical Geometry 12-6