- #1

PsychonautQQ

- 784

- 11

**Thx for help tonight yall :D 3 more questions X_x**

## Homework Statement

(1 pt)

First of all i'm a bit confused.. it wants the equation for a plane in all three questions.. the linear equation of a plane?

(A) If the positive z-axis points upward, an equation for a horizontal plane through the point (-5,4,-4) is

So it's a horizontal plane which means that the z direction number will be zero..

a(x-xi) + b(y-yi) + c(z-zi) = 0

ax - a5 + by - b4 = -

so i'll make a equal 4 and b = -5?

4x - 20 -5y +20 = 0

4x - 5y = 0

program says this is wrong..

.

(B) An equation for the plane perpendicular to the x-axis and passing through the point (-5,4,-4 is:

attempt at solution:

a plane that hits the x axis at a 90 degree angel would mean that it always has the same value for x if i'm picturing this correctly? so the direction number for x would be zero i believe.

a(x-xi) + b(y-yi) + c(z-zi) = 0

b(y-(-5)) + c(z-(-4)) = 0

by + 5b + cz + 4z = 0

so i could make b = 4 and z = -5?

4y + 20 - 5z - 20 = 0

4y - 5z = 0...

program says i'm doing it wrong X_x

.

(C) An equation for the plane parallel to the xz-plane and passing through the point (-5,4,-4) is:

attempt at solution:

If it's parallel to the xz plane that means the direction on it's y component must be zero..

a(x-xi) + b(y-yi) + c(z-zi) = 0

a(x-(-5)) + b(y-4) + c(z-(-4)) = 0

ax + 5a + by - 4b + cz + 4c = 0

well b has to be zero so it's parallel to the xz plane...

so xa + 5a -cz + 4c = 0

c could be -5 and a could be 4... ?

4x + 20 +5z -20 = 0

4x + 5z = 0

something went wrong here...?