Find the distance from the point P= (1, 2, 3) to the straight line whose parametric equation is...

Question:

Find the distance from the point P= (1, 2, 3) to the straight line whose parametric equation is given by {eq}x= t + 1,\ y= t - 1,\ z= 2t. {/eq}

Line and Distance from a Point:

The distance of a straight line from a given point is nothing but the minimum distance possible. So, we will find an expression that will give distance as a function of the parameter t. Then, by performing the derivative test, we will find the minimum distance.

The distance between two points is given as:

{eq}\displaystyle d=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2} {/eq}

Answer and Explanation:

{eq}\begin{align} d^2&=(x-1)^2+(y-2)^2+(z-3)^2=D & \left[D \text{ is a linear variable where distance }d^2=D \right]\\ \Rightarrow D&=[(t + 1)-1]^2+[(t - 1)-2]^2+[2t -3]^2\\ &=[t ]^2+[t -3]^2+[2t -3]^2\\ \frac{dD}{dt} &=2t+2[t -3]+2[2t -3](2) & \left[\text{ Find the first derivative equate it to 0 to minimise the function } \right]\\ &= 2t+2t -6+8t -12 \\ \frac{dD}{dt} &= 12t -18 \\ 12t -18 &=0\\ 12t &=18\\ t &=\frac{3}{2}\\ \frac{d^2D}{dt^2} &= 12\\ &\gt 0 & \left[\text{ The second derivative is positive. This confirms that the function has a minimum value } \right]\\ \end{align} {/eq}

So, distance is minimum at {eq}t =\frac{3}{2}\\ {/eq}

{eq}\begin{align} d^2&=(x-1)^2+(y-2)^2+(z-3)^2 \\ &=[t ]^2+[t -3]^2+[2t -3]^2\\ &= \left[ \frac{3} {2} \right]^2+ \left[ \frac{3} {2} -3\right]^2+ \left[2 \frac{3} {2} -3\right]^2 & \left[ \text{ Now, plug in the value of } t=\frac{3}{2} \right]\\ &= \frac{9} {4} + \frac{9} {4}\\ &= \frac{9} {2}\\ d&= 4.5\rm{units} \end{align} {/eq}

Therefore, {eq}\displaystyle \boxed{\color{blue} { d= 4.5\rm{units} }} {/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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