The distance of the point (5,3,-1) from the p | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$ units

Vedclass · Answer

(A) The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$. Substituting the points $(2,1,0)$,$(3,-2,4)$,and $(1,-3,3)$: $\begin{vmatrix} x-2 & y-1 & z-0 \ 3-2 & -2-1 & 4-0 \ 1-2 & -3-1 & 3-0 \end{vmatrix} = 0 \implies \begin{vmatrix} x-2 & y-1 & z \ 1 & -3 & 4 \ -1 & -4 & 3 \end{vmatrix} = 0$. Expanding the determinant: $(x-2)(-9 - (-16)) - (y-1)(3 - (-4)) + z(-4 - 3) = 0$ $(x-2)(7) - (y-1)(7) + z(-7) = 0$ $7x - 14 - 7y + 7 - 7z = 0 \implies 7x - 7y - 7z - 7 = 0 \implies x - y - z - 1 = 0$. The distance $d$ of a point $(x_0, y_0, z_0)$ from the plane $ax + by + cz + d = 0$ is given by $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$. For the point $(5,3,-1)$ and plane $x - y - z - 1 = 0$: $d = \frac{|1(5) - 1(3) - 1(-1) - 1|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|5 - 3 + 1 - 1|}{\sqrt{3}} = \frac{2}{\sqrt{3}}$ units.

The distance of the point $(5,3,-1)$ from the plane passing through the points $A(2,1,0)$,$B(3,-2,4)$,and $C(1,-3,3)$ is:

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