Distance between a point and a line and between two lines

Question

Let $P = (-5, 3, 4)$, $Q = (-6, 0, 3)$, $R = (-7, 1, 6)$ and $S = (-4, 2, 2)$. Let $A$ be the line passing through $P$ and $Q$, and let $B$ be the line passing through $R$ and $S$.

a) What is the distance between $R$ and $A$?

b) What is the distance between $A$ and $B$?

I am quite confused on how to start with this problem. Firstly, I am not entirely sure how I will find the distance between the point and the line. Would that distance simply be the normal vector multiplied by the projection? If so, how exactly would I calculate the projection here? No equations for the lines are given so I am quite confused.

Also, for the shortest distance between two lines, will it be a similar approach of finding the normal vector and projection?

I am not entirely sure how to proceed here. Any help would be highly appreciated!

Answer 1

Although what @AmateurMathGuy says is VERY useful :P there is a formula you can use called the point to line formula that basically says: $$d = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$ where the line is $ax+by+c=0$ and the point is $(x_0,y_0)$.

The correct way to do it on the other hand, without using this formula is to first find the line through the point perpendicular to the line in question. Then, find the intersection of these two lines. Then, find the distance from the point in question to the intersection point. That is your distance.

Answer 2

Let P = (-5, 3, 4), Q = (-6, 0, 3), R = (-7, 1, 6) and S = (-4, 2, 2). Let A be the line passing through P and Q, and let B be the line passing through R and S.

a)What is the distance between R and A?

parametrise A to get $$x=-5-t,y=3-3t,z=4-t$$

The distance is minimized when $$D^2 =(t-2)^2+(3t-2)^2+(t+2)^2$$ is minimized.

Upon differentiation and solving for t, you find the minimum distance between R and A.

b) What is the distance between A and B?

parametrise B to get $$x=-7+3s,y=1+s,z=6-4s$$

The distance is minimized when

$$ D^2= (-5-t+7-3s)^2 +(3-3t-1-s)^2+(4-t-6+4s)^2 $$ is minimized.

Differentiate with respect to t and s.

Solve for t and s to find the minimum distance.

Answer 3

Even if you're only in Calc I, you can still do this. Write an equation for the distance from the point to an arbitrary point on the line and then differentiate the equation you come up with with respect to $x$. The value of the derivative will be zero when your equation for the distance from the point to the line is at a minimum and this will give you the $x$ value on the line where the point is closest to. You can then get the $y$ and $z$ values from the original equation for the line.

Answer 4

(a) Let $X$ be a point on $A$ such that $RX$ is the minimum. Then $X=t(-5,3,4)+(1-t)(-6,0,3)$ for some $t\in\mathbb{R}$. $RX$ is peprpendicular to $A$. We can find $t$ by dot product.

(b) Let $Y$ be the point on $A $ and $Z$ be the point on $B$ such that $YZ$ is the minimum. Then $Y=t(-5,3,4)+(1-t)(-6,0,3)$ and $Z=s(-7,1,6)+(1-s)(-4,2,2)$ for some $s,t\in\mathbb{R}$. $YZ$ is perpendicular to both $A$ and $B$. Again, we can find $s$ and $t$ by taking dot products.