The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
A. Perpendicular
B. Parallel
C. intersect y–axis
D. passes through
The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
A. Perpendicular
B. Parallel
C. intersect y–axis
D. passes through

Solution:

Given:
First Plane:
$2 x-y+4 z=5$ [On multiply both the sides by $2.5]$
We obtain,
$5 x-2.5 y+10 z=12.5 \ldots$
Second Plane:
$5 x-2.5 y+10 z=6 \ldots$
Therefore,
$\begin{array}{l}
\frac{a_{1}}{a_{2}}=\frac{2}{5} \\
\frac{b_{1}}{b_{2}}=\frac{2}{5} \\
\frac{c_{1}}{c_{2}}=\frac{4}{10}=\frac{2}{5}
\end{array}$
Thus,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
It is clear that the direction ratios of normal of both the plane (1) and (2) are same.
As a result, both the given planes are parallel.
Therefore option (B) is correct.