Answer:
The equation of plane is
[tex]22x -3y+18z = 21[/tex]
Step-by-step explanation:
We have to find the equation of plane passing through the point (0,-1,1) and orthogonal to the planes
[tex]n_1:3x + 4y-3z = 0\\n_2:-3x + 2y + 4z = 5[/tex]
Thus, we can write:
[tex]n_1:<3,4,-3>\\n_2:<-3,2,4>[/tex]
We will evaluate:
[tex]n = n_1\times n_2\\\\n = \left[\begin{array}{ccc}i&j&k\\3&4&-3\\-3&2&4\end{array}\right] \\\\n = i(16 + 6)-j(12-9) +k(6+12)\\n = 22i-3j+18k\\n = <22,-3,18>[/tex]
The required plane passes through the point (0,-1,1)
Thus, the equation of plane is
[tex]n.<x-0,y+1,z-1>=0\\\Rightarrow <22,-3,18>.<x,y+1,z-1>=0\\\Rightarrow 22(x) -3(y+1) + 18(z-1) =0\\\Rightarrow 22x - 3y - 3 + 18z - 18 = 0\\\Rightarrow 22x -3y+18z = 21[/tex]
is the required equation of the plane.
