Q. 29 - Calculus

Question

In Exercises 27–30, use the result from Example 4 to find the distance from the point P to the given plane. $P=(4,0,-3), 12y-5z=7$

Step-by-Step Solution

Answer

$\mathrm{The} \mathrm{distance} \mathrm{from} \mathrm{the} \mathrm{given} \mathrm{point} \mathrm{to} \mathrm{the} \mathrm{given} \mathrm{plane} \mathrm{is}:D=\frac{8 }{13}.$

Step 1. Given information

$\mathrm{A} \mathrm{point}, P=(4,0,-3); \mathrm{a} \mathrm{plane}, 12y-5z=7$

Step 2. Finding the distance from the given point to the given plane

$$\begin{gathered} \mathrm{The} \mathrm{distance} \mathrm{from} \mathrm{the} \mathrm{point} P(x_0,y_0,z_0) \mathrm{to} \mathrm{the} \mathrm{plane} \mathrm{with} \mathrm{equation} ax+by+cz=d \\ \mathrm{is} \mathrm{given} \mathrm{by}, D=\frac{|a{x}_0+b{y}_0+c{z}_0-d| }{\sqrt{a^2+b^2+c^2}}. \\ \mathrm{Substituting} \mathrm{the} \mathrm{given} \mathrm{values}, \mathrm{we} \mathrm{get}, \\ D=\frac{|\left(4\right)\times \left(0\right)+\left(12\right)\times \left(0\right)+\left(-5\right)\times \left(-3\right)-\left(7\right)| }{\sqrt{{\left(0\right)}^2+{\left(12\right)}^2+{\left(-5\right)}^2}} \\ \Rightarrow D=\frac{|0+0+15-7| }{\sqrt{0+144+25}} \\ \Rightarrow D=\frac{\left|8\right| }{\sqrt{169}} \\ \Rightarrow D=\frac{8}{13} \end{gathered}$$