Question

4. What is the distance between the point (1,2,4) and the plane 3x+ 2y+ 6z= 5?

5. What is the distance between the parallel planes ax+by+cz=d1 and ax+by+cz=d2?
You may wish to try picking a point on one plane that you can specify exactly and
working out the distance from that point to the other plane.

Answer 1

$(4) \textbf{\textit{Find the distance between}}\hspace{0.1cm} \textit{(1, 2, 4)} \\\hspace{0.1cm} \textbf{\textit{and the plane}} \hspace{0.1cm} \textit{3x + 2y + 6z = 5}.\\ \textrm{The distance between} \hspace{0.1cm} (x_1, y_1, z_1) \hspace{0.1cm}\\ \textrm{and the plane} \hspace{0.1cm} ax + by + cz + d = 0\\\textrm{is given by}\hspace{0.1cm} D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\\ \therefore \textsf{The distance between} \hspace{0.1cm} (1, 2, 4)\hspace{0.1cm} \textsf{and}\hspace{0.1cm}\\3x + 2y + 6z = 5 \hspace{0.1cm}\textsf{is}\hspace{0.1cm} \\ D = \frac{|3(1) + 2(2) + 6(4) - 5|}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{|3 + 4 + 24 - 5|}{\sqrt{9 + 4 + 36}} =\frac{26}{7} \approx 3.714 \hspace{0.1cm}\textit{(3 d.p)} \\ (5) \textbf{\textit{What is the distance between}}\\\textbf{\textit{the parallel planes}}\\ ax + by + cz = d_1 \hspace{0.1cm} \textbf{\textit{and}}\\ax + by + cz = d_2?\\ \textrm{Let} \hspace{0.1cm}X(x_1, y_1, z_1) \hspace{0.1cm} \textrm{be an}\\ \textrm{arbitrary point away from} \\\textrm{the two planes}. \\\textrm{If the distance between}\\\textrm{ the point}\hspace{0.1cm} X \hspace{0.1cm} \textrm{and the first} \\\textrm{plane is given by} \hspace{0.1cm}D_1\\ \textrm{and that between the} \\\textrm{point}\hspace{0.1cm}X\hspace{0.1cm} \textrm{and the second}\\ \textrm{plane is given by}\hspace{0.1cm} D_2. \\ \textrm{The distance between the}\\\textrm{two plates is given by } \hspace{0.1cm} D = D_2 - D_1 \\ D_1 = \frac{|ax_1 + by_1 + cz_1 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \hspace{0.1cm} \textrm{and}\\ \hspace{0.1cm} D_2 = \frac{|ax_1 + by_1 + cz_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}\\ D = D_2 - D_1 = \frac{|ax_1 + by_1 + cz_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} - \\\frac{|ax_1 + by_1 + cz_1 - d_1|}{\sqrt{a^2 + b^2 + c^2}} = \\\frac{|(ax_1 + by_1 + cz_1) - (ax_1 + by_1 + cz_1) + d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} \\=\frac{| d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}\\ \therefore \hspace{0.1cm} \textrm{If} \hspace{0.1cm}ax + by + cz = d_1 \hspace{0.1cm} \textrm{and} \\\hspace{0.1cm} ax + by + cz = d_2 \\\hspace{0.1cm} \textrm{are plane equations, then the} \\\hspace{0.1cm} \textrm{distance} \hspace{0.1cm}D\hspace{0.1cm} \textrm{between the planes}\\ \textrm{can be found using} \\D = \frac{| d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$