Question

Find the equation of the planes that passes through three points.

(a) (1,1,-1),(6,4,-5),(-4,-2,-3)

(b) (1,1,0),(1,2,1),(-2,2,-1).

Answer 1

(a)The given points are A (1,1,-1), B (6,4,-5), and C (-4,-2,3).

\(\begin{vmatrix}1&1&-1\\6&4&-5\\-4&-2&3\end{vmatrix}\)=(12-10)-(18-20)-(-12+16)

=2+2-4 =0

Since, A, B, C are collinear points, there will be infinite number of planes passing through the given points.

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(b)The given points are A (1,1,0), B (1,2,1), and C (-2,2,-1).

\(\begin{vmatrix}1&1&0\\1&2&1\\-2&-2&1\end{vmatrix}\)=(-2-2)-(2+2)
=-8≠0

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points, (x₁,y₁,z₁), (x₂,y₂,z₂), and (x₃,y₃,z₃), is

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

\(\Rightarrow \begin{vmatrix}x-1&y-1&z\\0&1&1\\-3&1&-1\end{vmatrix}\)=0

\(\Rightarrow\) (-2)(x-1)-3(y-1)+3z=0

\(\Rightarrow\)-2x-3y+3z+2+3=0

\(\Rightarrow\)- 2x-3y+3z=-5
\(\Rightarrow\) 2x+3y-3z=5

This is the cartesian equation of the required plane.