Question

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C (5, 3, −3). 

Answer 1

The given points are A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).
The equation of the plane ABC is given by

\[\begin{vmatrix}x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}x - 2 & y - 5 & z - \left( - 3 \right) \\ - 2 - 2 & - 3 - 5 & 5 - \left( - 3 \right) \\ 5 - 2 & 3 - 5 & - 3 - \left( - 3 \right)\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}x - 2 & y - 5 & z + 3 \\ - 4 & - 8 & 8 \\ 3 & - 2 & 0\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}x - 2 & y - 5 & z + 3 \\ 1 & 2 & - 2 \\ 3 & - 2 & 0\end{vmatrix} = 0\]
\[ \Rightarrow - 4\left( x - 2 \right) - 6\left( y - 5 \right) - 8\left( z + 3 \right) = 0\]
\[ \Rightarrow 2\left( x - 2 \right) + 3\left( y - 5 \right) + 4\left( z + 3 \right) = 0\]
\[ \Rightarrow 2x + 3y + 4z - 7 = 0\]

∴ Distance between the point (7, 2, 4) and the plane

\[2x + 3y + 4z - 7 = 0\]

= Length of perpendicular from (7, 2, 4) to the plane

\[2x + 3y + 4z - 7 = 0\]

\[= \left| \frac{2 \times 7 + 3 \times 2 + 4 \times 4 - 7}{\sqrt{2^2 + 3^2 + 4^2}} \right|\]
\[ = \left| \frac{14 + 6 + 16 - 7}{\sqrt{4 + 9 + 16}} \right|\]
\[ = \left| \frac{29}{\sqrt{29}} \right|\]
\[ = \sqrt{29} \text{ units} \]

Thus, the required distance between the given point and the plane is  \[\sqrt{29}\] units .