Distance between $(-3, 0, 1)$ and the line $(2t, -t, -4t)$

Question

Could you please help me with this problem?

How do I calculate the distance between $(-3, 0, 1)$ and the line with the following parametric equation: $(x, y, z) = (2t, -t, -4t)$?

I'm really lost with this one.

Thanks in advance!

Answer 1

To calculate the distance between a line $\ell: x = x_0 + tr$, $x_0 \in \mathbf R^3$, $r \in \mathbf R^3 - \{0\}$ (in your case $x_0 = 0$ and $r = (2,-1,-4)$) and a point $p \in \mathbf R^3$, you can do the following:

Denote by $E$ the plane perpendicular to $\ell$ containing $p$. The point of intersection between $E$ and $\ell$ is the point on $\ell$ which is nearest to $p$. Note that $E$ has $r$ (the direction of $\ell$) as a normal vector, that is, $E$ is given by $$ E = \{x \in \mathbf R^3 \colon \def\<#1>{\left<#1\right>}\<x-p, r> = 0 \}. $$ In your case: $$ E : 2x_1 - x_2 - 4x_3 = \<p,r> = -10 $$ To calculate the point of intersection, plug $\ell$ into the equation of $E$, we have \begin{align*} 4t + t + 16t &= -10\\ \iff t &= -\frac{10}{21} \end{align*} So $E$ and $\ell$ intersect in $$ q = \left(-\frac{20}{21}, \frac{10}{21}, -\frac{40}{21}\right) $$ Now calculate the distance between $p$ and $q$.

Answer 2

Hint: Find the expression for the distance between your point and an arbitrary point on the line for a parameter value of $t$ and then square it. This defines a geometric curve in $t$. Finding the minimum distance is now easily accomplished by differentiating.

Answer 3

Hint: To determine the distance from a point $\vec p$ to the straight line given by $t\vec v$ just calculate the length of the cross product of $\vec p$ and the unit vector of $\vec v$. Do you see why this is the desired distance?

Answer 4

Let $A=(-3,0,1)$. Note the nearest point on the line is the point $H$ such that $\overrightarrow{AH}$ is orthogonal to the directing vector $\vec u=(2,-1,-4)$, and the distance is equal to $\;\lVert\overrightarrow{AH}\rVert$.

Hence we have to solve first: $$\overrightarrow{AH}\cdot\vec u=0\iff 2(2t+3)-(-t)-4(-4t-1)=21t+10=0\iff t=-\frac{10}{21}.$$ The distance is $$\sqrt{\Bigl(\frac{43}{21}\Bigr)^2+\frac{10}{21}\Bigr)^2+\frac{61}{21}\Bigr)^2}=3\sqrt{\frac{10}7}.$$