Find the equation of line `x+y-z-3=0=2x+3y+z+4` in symmetric form. Find the direction of the line.

Explanation

In the section of planes we will see the equation of the form `ax+by+cz+d=0` is the equation of the plane in the space.
Now equation of line in the form `x+y-z-3=0=0=2x+3y+z+4` means set of those points in space which are common to the planes `x+y-z-3=0 and 2x+3y+z+4=0`, which lie on the line of intersection of planes.
For example, equation of x-axis is y=z=0 where xy-plane (z=0) and xz-plane (y=0) intersect.
Now to get the equation of line in symmetric form, in above equations, first we eliminate any one of the variables, say z. Then adding `x+y-z-3=0 and 2x+3y+z+4=0`, we get
`" "3x+4y+1=0 or 3x=-4y-1=lamda("say")`
or `" "x=(lamda)/(3), y=(lamda+1)/(-4)`
Putting these values in `x+y-z-3=0,` we have `(lamda)/(3)+(lamda+1)/(-4)-z-3=0`
or `" "lamda=39+12z`
Comparing values of `lamda`, we have equation of line as
`" "3x=-4y-1=12z+39`
or `" "(3x)/(12)=(-4y-1)/(12)=(12z+39)/(12) or (x)/(4)=(y+(1)/(4))/(-3)=(z+(13)/(4))/(1)`
Hence, the line is passing through point `(0, -(1)/(4), -(13)/(4))` and have direction ratios 4, -3, 1.
If we eliminate x or y first we will get equation of line having same direction ratio but with different point on the line.