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6  Find the equations of the planes that passes through three points.

  (b)      (1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

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The equation of the plane which passes through the three points A(1,1,0),\ B(1,2,1),\ and\ C(-2,2,-1)  is given by;

Determinant method,

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

As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

Finding the equation of the plane through the points, (x_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2})\ and\ (x_{3},y_{3},z_{3})

\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

After substituting the values in the determinant we get,

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

\Rightarrow(x-1)(-1-1)-(y-1)(0+3)+z(0+3) = 0

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

2x+3y-3z = 5

So, this is the required Cartesian equation of the plane.

Posted by

Divya Prakash Singh

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