Question

The value of (a,b,c) for which the given three planes have a line is common are 5x+3y-z =4; 5x-4z =4 ; ax +by+cz=d​

Answer 1

Given: Two equations: 5x + 3y - z = 4 and 5x - 4z = 4

To find: The value of a, b and c?

Solution:

• Now we have given that three planes have a line is common, so if three planes are co-planar then the box product or scalar triple product of their normal will be 0.

• So the first plane given is: 5x + 3y - z = 4

            n = 5 , 3 , -1

• The second plane is: 5x - 4z = 4

           m = 5 , 0 , -4

• and the third plane is in the form of a, b, c and d, that is: ax +by+cz=d​

           l = a , b , c

• Scalar triple product: [ a b c ] = a . ( b x c )

• So now we can find the triple product of these three terms, we get:

          (5 , 3 , -1) . { (5 , 0 , -4) x (a , b , c) } = 0

          (5 , 3 , -1) . ( 4b, -5c-4a , 5b ) = 0

          20b - 15c - 12a - 5b = 0

          4a - 5b + 5c = 0

• But here we can see that there are three variables but only one equation. So from this we can conclude that the value of a, b and c can be infinite.

• Taking hit and try, we get:

           a = 10, b = 3 and c = -5

Answer:

        So the value of a, b, c are infinite and one of them are: 10. 3 and -5