Thx for help tonight yall :D 3 more questions X_x 1. The problem statement, all variables and given/known data (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...?