Okay, let's break this down step-by-step:
1) z = 7x - This is an equation describing a plane.
2) 2 − 12x − 5y - This is an equation describing another plane.
3) xyz2 = 2 - This describes a surface. We need to find the tangent planes to this surface at the point P(2, 1, -1).
To find the tangent plane at a point (x0, y0, z0), we use the formula:
(x - x0) + m(y - y0) + n(z - z0) = 0
Where m and n are the slopes of the plane.
For the point P(2, 1, -1):
x0 = 2
y0 = 1
z0 = -1
Now we need to calculate m and n:
m = -∂z/∂y|P = -5 (from the equation xyz2 = 2)
n = -∂z/∂x|P = -14 (from the equation xyz2 = 2)
So the equation of the tangent plane is:
(x - 2) - 5(y - 1) - 14(z + 1) = 0
Does this make sense? Let me know if you have any other questions!
To show the planes are perpendicular, we need to show their normal vectors are perpendicular (dot product is 0). The normal vector of the plane above is <-5, -14, 1>. The normal vectors of the other two planes are <1, 0, 7> and <-12, -5, 0>. Calculating the dot products:
<-5, -14, 1> • <1, 0, 7> = 0
<-5, -14, 1> • <-12, -5, 0> = 0
Therefore, the tangent planes are perpendicular to each other as required. Let me know if you have any other questions!