Find an equation of the plane through the points(0,9,9),(9,0,9) and(9,9,0)

Solution:

Even though 5 points are given a plane can be determined using 3 non-collinear points (0,9,9),(9,0,9) and(9,9,0)

Let (x\(_1\), y\(_1\), z\(_1\)) = (0, 9, 9)

(x\(_2\), y\(_2\), z\(_2\)) = (9, 0, 9)

(x\(_3\), y\(_3\), z\(_3\)) = (9, 9, 0)

We have equation of plane passing through three points as ,

\(\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2} -x_{1} & y_{2}-y_{1} & z_{2}-z_{1}\\ x_{3} - x_{1}& y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix} = 0\)

\(\begin{vmatrix} x-0 & y-9 & z-9 \\ 9-0 & 0-9 & 9-9\\ 9 - 0& 9-9 & 0-9 \end{vmatrix}=0\)

\(\begin{vmatrix} x-0 & y-9 & z-9 \\ 9 & -9 & 0\\ 9 & 0 & -9 \end{vmatrix}=0\)

x(81 - 0) - (y - 9) (-81 - 0) + (z - 9)(0 + 81) = 0

81x - (y - 9)(-81) + (z - 9)(81) = 0

Dividing throughout by 81 we get,

x + (y - 9) + (z - 9) = 0

x + y + z - 18 = 0

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Find an equation of the plane through the points(0, 9, 9), (9, 0, 9), (0, 9, 9), (9, 0, 9), and (9, 9, 0)

Summary:

The equation of the plane through the points (0, 9, 9), (9, 0, 9), and (9, 9, 0) is x + y + z - 18 = 0.