Question

Find the equation of the line intersecting the line x-a/1 = y/1 = z-a/1 and x+a/1 = y/1 = z+a/2 and parallel to the line x-a/2 = y-a/1 = z-2a/3​

Answer 1

Hi !

Solution :- x-a/1 = y/1 = z-a/1 = r (Say)___(1)

And , x+a/1 = y/1 = z+a/2 = L ___(2)

any point on the line (1) is p(r+a, r,r+a)

any point on the line (2) is Q (L -a , L ,2L - a(

line (1) and (2) will intersect if P and Q coincide for same value of r and L

r + a = L - a => r - L = -2a ___(3)

r = L => r - L ______(4)

r + a = 2L - a => r - 2L = -2a

solving (3) and (4) we get , r =(2a, a, 2a)

Required equation is z-2a/2 = y-a/1 = z-2a/3 Answer

____________________________

Hope it's helpful

#Answerwithquality&#BAL

Answer 2

Given :

• $\frac{x - a}{1}$
• $\frac{x+a}{1} = \frac{y}{1} = \frac{z + a}{2}$
• $\frac{x - a}{2} = \frac{y - a}{1} = \frac{z - 2a}{3}$

According to the question :

Let AB : $\frac{x - a }{1} = \frac{y}{1} = \frac{z + a }{1}$

CD : $\frac{x + a }{1} = \frac{y}{1} = \frac{z + a}{2}$

Let ,

⟹ P $( x_1 y_1 z_1 )$ lies on AB and

⟹ Q $( x_2 y_2 z_2 )$ lies on CD

Let $\frac{ x_1 - a }{1} = \frac{y_1}{1} = \frac{z_1 + a }{1} = r_1$

Let, $x_1 = a + r_1 , y_1 = r_1 and\:z_1 = r_1 + a$

Let $\frac{x_2 + a }{1} = \frac{y_2}{1} = \frac{z_2 + a }{2} = r_2$

$x_2 = -a + r_2 , y_2 = r_2 , z_3 = -a + 2r_2$

Direction ratio of PQ are,

⟹ $( x_2 - x_1 , y_2 - y_1 , z_2 - z_1 )$

⟹ $( r_2 - r_1 , -2a , r_2 - r_1 , 2r_2 - r_1 - 2a )$

Since, PQ is Parallel to line,

⟹ $\frac{ x - a }{2} = \frac{y - a }{1} = \frac{ z - za}{3}$

∴ $\frac{r_2 - r_1 - 2a }{2} = \frac{r_2 - r_1}{1} = {2r_2 - r_1 - 2a }{3}$

⟹ $\bold r_1 = 0$

Equation of line passing through $( x_1\:y_1\:z_1 )$

⟹ $( a, 0 , a )$ and parallel to $\frac{x - a}{2}$

⟹ $\frac{y - a }{1} = \frac{z - 2a}{3}$ is

⟹ $\frac{x - a }{2} = \frac{y}{1}$

⟹ $\bold {\frac{z - 2a}{3}}$

So, It's Done !!