Find the equation of the plane passes through the point `(2,3,-4)` and `(1,-1,3)` and parallel to x-axis.

Explanation

The generarl equation of the plane passing throught the point
A(2,3,-4) is given by
a(x-2)+b(u-3)+c (z+4)=0
Since it passes throiugh the point B (1,0-1,3) we have
a(x-2) +b(y-3)+c(z+4)=0
since it passes through the point B (1,-1,3) we have
If this plane is parallel to x axis then the normal to the plane is perpendicular tihe x axis
Dr 's of normal to plane (i) are a,b c and r's of hte x axis are 1 0,0
`therefore axx1+bx0+cxx0=0 rarr a=0`
putting a=0 in (ii) we get 7c - 4b =0
Let `b=lambda` Then 7c `=4 lambda rarr c=(4 lamdba)/(7)` in (i) we get
`0xx(x-2)+lambda(y-3)+(4 lambda)/(7)(z+4)=0`
`rarr (y-3)+4/7(z+4)=0 rarr 7y-21 + 4z+16 =0`
`rarr 7y + 4z-5=0`
Hence the required equation of hte plane is 7y + 4z -5 =0