A point on the plane determined by the points | vedclass

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Question

(A) The equation of a plane passing through three points $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by the determinant equa

Vedclass · Accepted Answer

$(1,2,-2)$

Vedclass · Answer

(A) The equation of a plane passing through three points $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by the determinant equation:$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$Substituting the given points $A(1,1,-1)$,$B(2,-1,0)$,and $C(-1,0,2)$:$\begin{vmatrix} x-1 & y-1 & z+1 \ 2-1 & -1-1 & 0+1 \ -1-1 & 0-1 & 2+1 \end{vmatrix} = 0$$\begin{vmatrix} x-1 & y-1 & z+1 \ 1 & -2 & 1 \ -2 & -1 & 3 \end{vmatrix} = 0$Expanding along the first row:$(x-1)(-6+1) - (y-1)(3+2) + (z+1)(-1-4) = 0$$-5(x-1) - 5(y-1) - 5(z+1) = 0$Dividing by $-5$:$(x-1) + (y-1) + (z+1) = 0$$x + y + z - 1 = 0$Now,check the options by substituting the coordinates into the equation $x + y + z - 1 = 0$:For $(1,2,-2)$: $1 + 2 - 2 - 1 = 0$. This satisfies the equation.Thus,the point $(1,2,-2)$ lies on the plane.

$A$ point on the plane determined by the points $A(1,1,-1)$,$B(2,-1,0)$,and $C(-1,0,2)$ among the following is:

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