# What is the implicit equation for the plane passing through the points (-2,5,5), (2,8,5) and (-5,5,3)

*print*Print*list*Cite

Expert Answers

justaguide | Certified Educator

The equation of the plane through the points A(-2,5,5), B(2,8,5) and C(-5,5,3) has to be determined.

The vector AB is (2+2)i + (8-5)j + (5-5)k = 0 => 4i + 3j = 0

The vector AC is (-5+2)i +(5-5)j + (3 - 5)k = 0 => -3i - 2k = 0

The cross product of AB and AC gives the normal of the plane and is equal to `[[i,j,k],[4,3,0],[-3,0,-2]]` = i(-6 - 0) - j(-8 - 0) + k(0 + 9) = -6i + 8j + 9k

Using the normal and the point A(-2,5,5) the equation of the plane is -6(x +2) + 8(y - 5) +9(z - 5) = 0

=> -6x - 12 + 8y - 40 + 9z - 45 = 0

=> -6x + 8y + 9z - 97 = 0

**The equation of the plane passing through A(-2,5,5), B(2,8,5) and C(-5,5,3) is -6x + 8y + 9z - 97 = 0**