Find the distance from (2, 2, 5) to the line L: x(t) = 3 + 2t, y(t) = -1-t, z(t) = t . Please...

Question:

Find the distance from {eq}(2,\ 2,\ 5) {/eq} to the line {eq}L {/eq} : {eq}x(t) = 3 + 2t, \ \ y(t) = -1-t, \ \ z(t) = t . {/eq} Please use the orthogonal projection method to solve the problem.

Distance From a Point to a Line:

Suppose a point {eq}P {/eq} and a line {eq}L {/eq} are given. One way to find the distance of {eq}P {/eq} from {eq}L {/eq} is to first identify two points {eq}Q {/eq} and {eq}R {/eq} on the line {eq}L {/eq}. Next, we find the vectors {eq}\overrightarrow {QR} {/eq} and {eq}\overrightarrow {QP} {/eq}. The distance from {eq}P {/eq} to the line {eq}L {/eq} equals the orthogonal projection {eq}\displaystyle \frac{|\overrightarrow {QP}\times\overrightarrow {QR}|}{|\overrightarrow {QR}|} {/eq} of the vector {eq}\overrightarrow {QP} {/eq} on the vector {eq}\overrightarrow {QR} {/eq}.

Answer and Explanation:

Given: a point, say {eq}P:\,(2,\,2,\,5) {/eq}, and a line {eq}L:\,\langle x,\,y,\,z \rangle = \langle 3 + 2t,\, -1-t,\,t \rangle {/eq}


Two points on the line {eq}L {/eq} are found as follows:

Substituting {eq}t=0 {/eq}, we obtain the point {eq}(3 + 2*0,\, -1-0,\,0 )=(3,\,-1,\,0) {/eq} on the line. Let's denote this point as {eq}Q {/eq}.

Substituting {eq}t=\,-1 {/eq}, we obtain the point {eq}(3 + 2*(-1),\, -1-(-1),\,-1 )=(1,\,0,\,-1) {/eq} on the line. Call this point {eq}R {/eq}.


Let

{eq}\begin{align*} \vec a=\overrightarrow {QR}&=\vec{R}-\vec{Q}\\[2ex] &=\langle 1,\,0,\,-1 \rangle-\langle 3,\,-1,\,0 \rangle\\[2ex] &=\langle -2,\,1,\,-1 \rangle \end{align*} {/eq}

and

{eq}\begin{align*} \vec b=\overrightarrow {QP}&=\vec{P}-\vec{Q}\\[2ex] &=\langle 2,\,2,\,5 \rangle-\langle 3,\,-1,\,0 \rangle\\[2ex] &=\langle -1,\,3,\,5 \rangle \end{align*} {/eq}


The distance from the point {eq}P:\,(2,\,2,\,5) {/eq} to the line {eq}L {/eq} is given by the orthogonal projection of {eq}\vec b {/eq} on {eq}\vec a {/eq}:

{eq}\begin{align*} \hspace{0.4cm}&\displaystyle \frac {|\mathbf{a}\times\mathbf{b}|}{|\mathbf{a}|}\\[2ex] &= \frac {|\begin{vmatrix} \mathbf{\vec i}& \mathbf{\vec j} & \mathbf{\vec k}\\ -2&1&-1\\ -1&3&5 \end{vmatrix} |}{|\langle -2,\,1,\,-1 \rangle|}\\[2ex] &=\displaystyle \frac{|(1)(5)-(3)(-1))\mathbf{\vec i}-((-2)*5-(-1)(-1))\mathbf{\vec j}+(-2*3-(-1)*1)\mathbf{k}|}{\sqrt{(-2)^2+1^2+(-1)^2}}\\[2ex] &=\displaystyle\frac{|8\mathbf{\vec i}+8\mathbf{\vec j}-5\mathbf{\vec k}|}{\sqrt 4}\\[2ex] &=\frac{\sqrt{8^2+8^2+(-5)^2}}2\\[2ex] &=\frac {\sqrt{153}}2\\[2ex] &=\color{Blue}{\frac {3\sqrt{17}}2\hspace{0.8cm}\mathrm{(Answer)}} \end{align*} {/eq}


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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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