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: The equation of two planes are 5x+3y-z =4; 5x-4z =4.

To find: The value of a, b and c

If the given three planes are co-planar then scalar triple product of theirs normal is 0

First plane: $5x+3y-z=4$

$\therefore n_1=<5,3,-1>$

Second plane: $5x-4z=4$

$\therefore n_2=<5,0,-4>$

Third plane: $ax+by+cz=d$

$\therefore n_3=<a,b,c>$

Now, find the scalar triple product of $n_1,n_2\text{ and } n_3$

$\Rightarrow <5,3,-1>\cdot <5,0,-4>\times <a,b,c>=0$

$\Rightarrow <5,3,-1>\cdot <4b,-5c-4a,5b>=0$

$\Rightarrow 20b-15c-12a-5b=0$

$\Rightarrow -12a+15b-15c=0$

$\Rightarrow 4a-5b+5c=0$

One equation and three variable.

Therefore, the value of a, b and c are infinite.

Using hit-trial we will get the value of a, b and c.

So, $a=10,b=3,c=-5$

These values make three planes a common line.