Multivariable calculus and vector analysis

Distance from point to plane example

Example

Find the distance from the point $P=\left(4,-4,3\right)$ to the plane $2x-2y+5z+8=0$. (This is pictured in the distance to plane CVT in its original view, which you can restore by pressing Home).

Solution: We use formula (2) from the distance to plane reading. From the equation for the plane we substitute $A=2$, $B=-2$, $C=5$, $D=8$. From the point $P$, we substitute $x_1=4$, $y_1=-4$, and $z_1=3$. The distance from $P$ to the plane is

$$\begin{array}{lll}\hfill d=\frac{|2\cdot 4+\left(-2\right)\cdot \left(-4\right)+5\cdot 3+8|}{\sqrt{2^2+{\left(-2\right)}^2+5^2}}=\frac{39}{\sqrt{33}}\approx 6.8& & \hfill \end{array}$$

Contents by topic

1. Geometry in higher dimensions

1a. Preliminaries

1b. Lines and planes

• Parametrization of a line
• Forming planes
• A line or a plane or a point?
• Parametrization of a plane
• Distance from point to plane
• Parametrization of a line examples
• Forming plane examples
• Plane parametrization example
• Distance from point to plane example
• Intersecting planes example

1c. Matrices

1d. Visualizing functions in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material