Question

Show that the line of intersection of the planes x + 2y + 3z = 8 and 2x + 3y + 4z = 11 is coplanar with the line x 1 y 1 z 1 1 2 3      . Also find the equation of the plane containing them.

Answer 1

Answer:

here is the solution

The given line

x 1 y 1 z 1

1 2 3

  

 

…(1) is coplanar with the line determined by the planes

x 2y 3z 8 0    

…(2) and

2x 3y 4z 11 0    

…(3), if we are able to show there exists a

plane passing through intersection of planes (2) and (3) containing the line (1).

Equation of the plane passing through the intersection of planes (2) and (3) is

(x 2y 3z 8) k(2x 3y 4z 11) 0        

…(4) [2]

We find, value of k for which the plane given by (4) passes through the point (–1, –1, –1) lying

on line (1).

Substituting the coordinates of the point (–1, –1, –1) in (4), we get

(–1 –2 –3 –8) + k(–2 – 3 – 4 – 11) = 0

–14 – 20k = 0

k =

7

10



Putting,

7

k

10

 

in (4) we get

7

(x 2y 3z 8) (2x 3y 4z 11) 0

10

       



4x y 2z 3 0    

…(5) [2]

Now we find value of

1 2 1 2 1 2 a a b b c c  

, where

1 1 1 a , b , c and D.

Ratios of the line (1) and

2 2 2 a , b , c and D.

Ratios of the normal to the plane (5)

1 2 1 2 1 2 a a b b c c 1(4) 2(1) 3( 2)      

i.e.,

1 2 1 2 1 2 a a b b c c 0   

[1]

which implies line (1) lies in plane (5)

Hence the two lines are coplanar and the equation of the plane containing them is

4x y – 2z 3 0

Step-by-step explanation: