# Consider the points A(-5,1,3), B(4,1,2), C(1,-2,1) and D(9,0,1). Find the standard equation of the plane which contains all four points.

*print*Print*list*Cite

### 2 Answers

You need to find the equations of the lines AC and BD, such that:

`(AC): (x - x_A)/(x_C - x_A) = (y - y_A)/(y_C - y_A) = (z - z_A)/(z_C - z_A)`

`(AC): (x + 5)/(1 + 5) = (y - 1)/(-2 - 1y_B - y_A) = (z - 3)/(1 - 3)`

`(AC): (x + 5)/6 = (y - 1)/(-3) = (z - 3)/(-2)`

You need to identify the direction al vector of the line AB such that:

`bar(AB) = <6,-3,-2>`

`(BD): (x - x_B)/(x_D - x_B) = (y - y_B)/(y_D - y_B) = (z - z_B)/(z_D - z_B)`

`(BD): (x - 4)/(9 - 4) = (y - 1)/(0 - 1) = (z - 2)/(1 - 2)`

`(BD): (x - 4)/5 = (y - 1)/(- 1) = (z - 2)/(-1)`

You need to identify the direction al vector of the ine AB such that:

`bar(BD) = <5,-1,-1> `

You need to evaluate the cross product of the directional vectors, such that:

`bar(AB) X bar(BD) = [(bar i, bar j, bar k),(6 , -3 , -2),(5 , -1 , -1)]`

`bar(AB) X bar(BD) = 3bar i - 6 bar k - 10 bar j + 15 bar k - 2bar i + 6 bar j`

`bar(AB) X bar(BD) = bar i - 4 bar j + 9 bar k`

The vector `bar n = bar(AB) X bar(BD)` represents the normal vector to the plane whose equation you need to find.

You may write the equation of the plane using the normal vector such that:

`P: x - 4y + 9z = ` 0

**Hence, evaluating the equation of the plane P, under the given conditions, yields **`P: x - 4y + 9z = 0.`

The above answer is mostly correct, however, the vector denoted as AB should read AC (just a typo), and you must plug any of the four points into the final equation to find the constant term, for example, plugging-in point A(-5, 1, 3) yields (-5)-4(1)+9(3)=18, so the final answer is x-4y+9z=18.