Show that the lines $\frac{\mathrm{x}-5}{7}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{z}}{1}$ and $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}$ are perpendicular to each other.

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Show that the lines $\frac{\mathrm{x}-5}{7}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{z}}{1}$ and $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}$ are perpendicular to each other.

Solution:

The equations of the given lines are
$\frac{\mathrm{x}-5}{7}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{z}}{1}$ and $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}$
Two lines with direction ratios is given as
$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
The direction ratios of the given lines are $7,-5,1$ and $1,2,3$
i.e., $a_{1}=7, b_{1}=-5, c_{1}=1$ and
$a_{2}=1, b_{2}=2, c_{2}=3$
Considering
$\begin{array}{l} a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=7 \times 1+(-5) \times 2+1 \times 3 \\ =7-10+3 \\ =-3+3 \\ =0 \end{array}$
As a result, the two lines are perpendicular to each other.