Question

Q 10 Find the value of 'p'; if the lines given by the following equations are at right angles,
(4 -x)/6 = (9y - 18)/2p = (z -3)/2
(7- 7x)/3p = (y - 4) = (4 - Z)/ 5
Ops: A. О 148/630
B. O 630/148
C. O 315/18
D. O 18/315​

Answer 1

The value of p if the lines are at right angles is $\frac{315}{88}$

Given : Equations of line

$\frac{4-x}{6}$ = $\frac{9 y-18}{2p}$ = $\frac{z-3}{2}$  and $\frac{7-7x}{3p}$ = ${ y-4}$ = $\frac{4-z}{5}$

To Find : The vale of p if the lines are at right angles

Solution : The value of p if the lines are at right angles is $\frac{315}{88}$

The equation of the lines is given as

$\frac{4-x}{6}$ = $\frac{9 y-18}{2p}$ = $\frac{z-3}{2}$  and $\frac{7-7x}{3p}$ = ${ y-4}$ = $\frac{4-z}{5}$

Writing the equation of lines in standard form

L1 is   $\frac{x-4}{-6}$ = $\frac{ y-2}{2p/9}$ = $\frac{z-3}{2}$

and L2 is  $\frac{x-1}{-3p/7}$ = ${ y-4}$ = $\frac{z-4}{-5}$

Two lines are at right angles if the dot product of their direction cosines is zero.

Direction cosines of line L1

x1  = -6 , y1 = 2p/9 and z1 = 2

Direction cosines of line L2

x2  = -3p/7  , y2 = 1 and z1 = -5

Taking dot product of the direction cosines of both the lines we have

$\frac{18 p}{7} + \frac{2p}{9} +( -10) = 0$

$\frac{18 p}{7} + \frac{2p}{9} = 10$

$\frac{176p}{63} = 10$

$p = \frac{630}{176}$

$p =\frac{315}{88}$

So the value of p if the lines are at right angles is $\frac{315}{88}$

#SPJ3