Question

Find the equation for the plane through the points Po(3,-2,5), Qo(-3,-1,-5), and Ro(0,-4,4) The equation of the plane is Type an equation.)

Answer 1

Answer:

- 21 x + 24 y + 15 z =120

Step-by-step explanation:

Given that

Po(3,-2,5), Qo (-3,-1,-5), and Ro (0,-4,4) ,These are the point in the space.

We know that equation of a plane is given as

$\begin{vmatrix}x-x_1 & y-y_1 &z-z_1 \\ x_2-x_1 & y_2-y_1 &z_2-z_1 \\ x_3-x_1 &y_3-y_1 & z_3-z_1\end{vmatrix}=0$

$\begin{vmatrix}x-0 & y+4 &z-4 \\ 3-0 & -2+4 &5-4 \\ -3-0 &-1+4 & -5-4\end{vmatrix}=0.$

$\begin{vmatrix}x & y+4 &z-4 \\ 3 & 2 &1 \\ -3 &3 & -9\end{vmatrix}=0.$

Now by solving above determinate we get

x( -18 -3 ) -(y+4 ) ( -27 +3 ) + ( z- 4) (9+6) = 0

-21 x +24 y -24 x 4 + 15 z - 24 = 0

- 21 x + 24 y + 15 z -120 = 0

- 21 x + 24 y + 15 z =120

Therefore the equation of the plane will be

- 21 x + 24 y + 15 z =120