## 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 $a$?

$A(4 ; 3 ; 1)$

$a: 3x - 4y + 5z - 6 = 0$

$k = \frac{5}{6}$
## 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 $k$, the plane transitions to

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

$a^{\prime}: 3 x-4 y+5 z-5=0$

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

$3 \cdot 4-4 \cdot 3+5 \cdot 1-5=0$

$12-12+5-5=0$

$0=0$

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

## Problem Kuznetsov Analytical Geometry $12-21$