# Find the distance from the point P(0, \ 2, \ 2) to the line x=2t, \ y= 6 2t, \ z=2+t .

## Question:

Find the distance from the point {eq}P(0, \ 2, \ 2) {/eq} to the line {eq}x=2t, \ y= 6-2t, \ z=2+t {/eq}.

## Distance between Two Points:

Any point between the 3-dimensional cartesian space can be expressed as {eq}\displaystyle (x, y, z) {/eq}. The distance between any two points can be solved with the equation, {eq}\displaystyle d = \sqrt{(x_0-x_1)^2+(y_0-y_1)^2+(z_0-z_1)^2} {/eq}, where one point is {eq}\displaystyle (x_0,y_0, z_0) {/eq} and the other is {eq}\displaystyle (x_1,y_1,z_1) {/eq}.

## Answer and Explanation:

For this problem, we need to find the distance of the vector that is perpendicular from the given line with one end as the given point, *P*. Any point on the line, *S*, can be expressed as {eq}\displaystyle S(2t, 6-2t, 2+t) {/eq}. Now, we express the length of any point on the line to *P* as

{eq}\begin{align} \displaystyle \vec{ SL} &= P - S\\ &= (0-2t, 2-6+2t, 2-2-t)\\ &= (-2t, -4+2t, -t)\\ \end{align} {/eq}

The direction of the line is known as {eq}\displaystyle \vec{L} = (2, -2, 1) {/eq}. Now, we know that {eq}\displaystyle \vec{L} {/eq} and {eq}\displaystyle \vec{SL} {/eq} are perpendicular with each other, so it must be that, {eq}\displaystyle \vec{L}\cdot \vec{SL} = 0 {/eq}. We find *t* that satisfies the equation.

{eq}\begin{align} \displaystyle \vec{L}\cdot \vec{SL} &= 0\\ (2, -2, 1)\cdot (-2t, -4+2t, -t)&= 0\\ -4+8-4t-t &= 0\\ -5t &= -4\\ t &= \frac{4}{5} \end{align} {/eq}

Therefore, the distance of *P* to *S* is the distance between *P* and the point on the line *S* at {eq}\displaystyle t = \frac{4}{5} {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle d &= |P -S_{t = \frac{4}{5}}|\\ &= \left| (0, 2, 2) - \left( 2\cdot \frac{4}{5}, 6- 2\cdot \frac{4}{5}, 2 + \frac{4}{5} \right) \right|\\ &= \left|(0,2,2) - \left(\frac{8}{5}, \frac{22}{5}, \frac{14}{5} \right) \right|\\ &= \sqrt{\left(0 - \frac{8}{5} \right)^2-\left(2 - \frac{22}{5} \right)^2-\left(2- \frac{14}{5} \right)^2}\\ &=\boxed{\frac{4\sqrt{14}}{5}} \end{align} {/eq}

#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3