Find the values of p so that the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

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Find the values of p so that the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

Solution:

The standard form of a pair of Cartesian lines is:
$\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{a}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{c}_{1}} \text { and } \frac{\mathrm{x}-\mathrm{x}_{2}}{\mathrm{a}_{2}}=\frac{\mathrm{y}-\mathrm{y}_{2}}{\mathrm{~b}_{2}}=\frac{\mathrm{z}-\mathrm{z}_{2}}{\mathrm{c}_{2}} \ldots \text { (1) }$
So according to the standard form the given equations can be written, i.e.
$\begin{array}{l} \frac{-(x-1)}{3}=\frac{7(y-2)}{2 p}=\frac{z-3}{2} \text { and } \frac{-7(x-1)}{3 p}=\frac{y-5}{1}=\frac{-(z-6)}{5} \\ \frac{x-1}{-3}=\frac{y-2}{2 p / 7}=\frac{z-3}{2} \quad \frac{x-1}{-3 p / 7}=\frac{y-5}{1}=\frac{z-6}{-5} \end{array}$
Now, on comparing the eq. (1) and eq.(2), we get
$\mathrm{a}_{1}=-3, \mathrm{~b}_{1}=\frac{2 \mathrm{p}}{7}, \mathrm{c}_{1}=2 \quad \mathrm{a}_{2}=\frac{-3 \mathrm{p}}{7}, \mathrm{~b}_{2}=1, \mathrm{c}_{2}=-5$
The direction ratios of the lines are
$-3,2 p / 7,2$ and $-3 p / 7,1,-5$
Now, as both the lines are at right angles,
Therefore, $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
$\begin{array}{l} (-3)(-3 p / 7)+(2 p / 7)(1)+2(-5)=0 \\ 9 p / 7+2 p / 7-10=0 \\ (9 p+2 p) / 7=10 \\ 11 p / 7=10 \\ 11 p=70 \\ p=70 / 11 \end{array}$
$\therefore$ The value of $p$ is $70 / 11$