Q. 27 - Calculus

Question

In Exercises 27–30, use the result from Example 4 to find the distance from the point P to the given plane.

$P=\left(3, 5, -2\right)$, $x-3y+4z=8$

Step-by-Step Solution

Answer

The distance is $\frac{14\sqrt{26}}{13}$.

Step 1. Given Information.

The point is $P=\left(3, 5, -2\right)$ and the equation of the plane is $x-3y+4z=8$. 

Step 2. Explanation.

The distance between point $P\left(x_0, y_{0,} z_0\right)$ to the plane with equation $ax+by+cz=d$ is $D=\frac{\left|a{x}_0+b{y}_0+c{z}_0-d\right|}{\sqrt{a^2+b^2+c^2}}$.

So, the distance between point $P=\left(3, 5, -2\right)$and the line is calculated as,

  $$\begin{gathered} D=\frac{\left|3\times 1+5\times \left(-3\right)+\left(-2\right)\times 4-8\right|}{\sqrt{1^2+{\left(-3\right)}^2+4^2}} \\ =\frac{14\sqrt{26}}{13} \end{gathered}$$