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.