What is the distance from the point A, to the line through the points B and C where A, B, and C...

Question:

What is the distance from the point A, to the line through the points B and C where A, B, and C have the following coordinates (3, 0, 4), (-1, 4, 1) and (2, 1, 0)?

Distance From a Point to the given Line in 3D:


In three dimensional coordinate system, to find the distance from point to the given line is found as given below:

(i) Find the coordinates of a general point on the line and take it as the foot of perpendicular.

(ii) Find the direction ratios of the line joining the foot of perpendicular and the given point.

(iii) Apply the condition that the sum of the product of the direction ratios of two perpendicular lines is zero to get the coordinates of the general point.

(iv) Use distance formula between the two points to find the required distance.

Answer and Explanation:


The given points are:

{eq}A\left ( 3, 0, 4 \right ),\ B\left ( -1, 4, 1 \right ), \ C \left ( 2, 1, 0 \right ) {/eq}

The equation of the line passing through the points {eq}B {/eq} and {eq}C {/eq} is:

{eq}\\\\ \begin{align*} \frac{x+1}{2+1} & = \frac{y-4}{1-4}=\frac{z-1}{0-1} \\\\\Rightarrow \frac{x+1}{3} & = \frac{y-4}{-3}=\frac{z-1}{-1}=\lambda \texttt{ (say)} \ \ \ ...(1) \\\\ \end{align*} {/eq}

Thus, the coordinates of any point {eq}P {/eq} on the line can be taken as:

{eq}P \left ( 3\lambda -1, -3\lambda +4, -\lambda +1 \right ) {/eq}

If {eq}P {/eq} be the foot of perpendicular from the point {eq}A {/eq} on the line {eq}(1) {/eq} then direction ratios of {eq}AP {/eq} are:

{eq}3\lambda -1-3, -3\lambda +4-0, -\lambda +1-4 = 3\lambda -4, -3\lambda +4, -\lambda -3 {/eq}

Since, {eq}AP {/eq} is perpendicular to the line {eq}(1) {/eq} therefore, the sum of the product of direction ratios would be zero.

Thus, we have:

{eq}\\\\ \begin{align*} 3(3\lambda -4)-3(-3\lambda +4)-1(-\lambda -3)& =0 \\\\\Rightarrow 9\lambda -12 + 9\lambda -12 +\lambda + 3 & = 0 \\\\\Rightarrow 19\lambda & = 21 \\\\\Rightarrow \lambda & = \frac{21}{19} \\\\ \end{align*} {/eq}

Thus, the point {eq}P {/eq} is:

{eq}\left ( \frac{63}{19} -1,-\frac{63}{19}+4,-\frac{21}{19}+1 \right ) =\left ( \frac{44}{19},\frac{13}{19},-\frac{2}{19} \right ) {/eq}

Hence, the required distance {eq}d {/eq} from the the point {eq}A {/eq} to the line {eq}(1) {/eq} is:

{eq}\\\\ \begin{align*} AP & = \sqrt{\left (\frac{44}{19}-3\right)^{2}+\left (\frac{13}{19}-0\right)^{2}+\left (\frac{2}{19}-4\right)^{2}} \\\\& = \sqrt{\left (-\frac{13}{19}\right)^{2}+\left (\frac{13}{19}\right)^{2}+\left (-\frac{74}{19}\right)^{2}} \\\\& = \frac{1}{19}\sqrt{169+169+5476} \\\\& = \frac{1}{19}\sqrt{5814} \\\\& \approx 4.01 \texttt{ units} \\\\ \end{align*} {/eq}


Learn more about this topic:

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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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