# Two planes pi_1 and pi_2 are orthogonal and intersect in the line L: (x,y,z) = (-4,-4,2) + t<3,-5,0>. If pi_1 contains the point Q(-3,-3,-4), find the equations of the plane pi_2.

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The line

`(x,y,z) =(-4,-4,2) +t<3,-5,0>`

belongs to both planes. It means one common vector of the two planes is `<3,-5,0>` and one common point is `P(-4,-4,2)`

Because plane `pi_1` contains the point `Q(-3,-3,4)` it means the vector `QP =<-3-(-4),-3-(-4),4-2> =<1,1,2>` is contained in plane `pi_1` .

The vector that is perpendicular both to `<3,-5,0>` and `<1,1,2>` will lie in plane `pi_2` :

`<3,-5,0>xx<1,1,2> =|[i,j,k],[3,-5,0],[1,1,2]| =-10*i-6*j+8*k =<-10,-6,8> =2*<-5,-3,4>`

Therefore the plane `pi_2` will be defined by the point `P(-4,-4,2)` and the vectors `<3,-5,0>` and `<-5,-3,4>` . The equation of the plane `pi_2` is

`(x,y,z) = (-4,-4,2) +t<3,-5,0> +s<-5,-3,4>`

**Answer: The equation of plane `pi_2` is` ` **

**(x,y,z) =(-4,-4,2)+t<3,-5,0>+s<-5,-3,4>**