Find an equation of the plane passing through (0 comma negative 4 comma 1 )(0,−4,1) that is orthogonal to the planes 55xplus+33ynegative 5−5zequals=00 and negative 5−5xplus+44yplus+22zequals=99.

26x + 15y + 35*z = -25

Step-by-step explanation:

Given the planes:

5x + 3y -5z = 0

-5x + 4y + 2z = 9

First we have to find the cross product of the normal vectors of these planes:

<5, 3, -5> x <-5, 4, 2> = <3*2 - 4*(-5), -[5*2 - (-5)*(-5)], 5*4 - (-5)*3> =

<26, 15, 35>

Then substitute these and the point (0, -4, 1) into the standard plane equation:

ax + by + cz = d

26*0 + 15*(-4) + 35*1 = d

-25 = d

triangle and cube

step-by-step explanation: