The complete question is below because the world limit doesn't allow me to post full question in title.

Find the shortest distance between the straight line joining the points A(3,2,-4) and B(1,6,-6) and the straight line joining the points C(-1,1,-2) and D(-3,1,-6). Also find equation of the line of shortest distance and coordinates of the feet of the common perpendicular

I have found the shortest distance by two methods

  1. by determining that lines are skew and proceeding further on e.g. taking dot product with direction vectors and then by finding s and t. Then I found the direction vector and thus by finding its magnitude of that direction vector I got the shortest distance

  2. by finding equation of parallel planes and by solving them using distance formula between two planes

In both cases, I got shortest distance equals to sqrt(21). [is it right]

But my real query is how to find equation of the line of shortest distance and coordinates of the feet of the common perpendicular. Please help.

  • $\begingroup$ You need to show your work so we can help. Did you find the equations of lines $AB$ and $CD$? If yes, please edit your question to add that. Also when you do, please use mathjax. $\endgroup$
    – Math Lover
    Jul 6 at 4:13
  • $\begingroup$ What you did in your first method contains the information to answer your question. If you show what you worked out, it will make it easier to describe what you still need to do. $\endgroup$
    – boojum
    Jul 6 at 5:18
  • $\begingroup$ Please can you help me guide this website as I got flagged in notification by posting this question. Please can someone guide me what I did to deserve flag (I might be unaware as I am new to this community). Please guide me to avoid flags for future posts. $\endgroup$ Jul 6 at 15:06
  • $\begingroup$ Also please guide me how to type your working here like complex type. Integrations, graphs, fractions. Various mathematical signs. Please guide me.. $\endgroup$ Jul 6 at 15:08

Let $PQ$ be the line of shortest distance whose equation is required such that $P$ lies on $AB$ and $Q$ lies on $CD$ and its direction ratios be $(a,b,c)$.

Now, direction ratios of $AB$ and $CD$ are $(2,-4,2)$ and $(2,0,4)$ respectively.

$\therefore$ $2a-4b+2c=0 \tag{i}\label{i}$ since $PQ \perp AB$

and $2a+0+4c=0 \tag{ii}\label{ii}$ since $PQ \perp CD$

In Eqs. $(i)$ and $(ii)$, by cross multiplication, we get $$\frac{a}{4}=\frac{b}{1}=\frac{c}{-2}$$

Therefore $(4,1,-2)$ are direction ratios of PQ $\tag{iii}$

Clearly any point on $AB$ is of the form $X_{AB}(2\mu+3,2-4\mu,2\mu-4)$ and any point on $CD$ is of the form $X_{CD}(2\lambda-1,1,4\lambda-2)$.

So, let $P\equiv(2\mu+3,2-4\mu,2\mu-4)$

and $Q\equiv(2\lambda-1,1,4\lambda-2)$

$\therefore$ Direction ratios of $PQ$ can be given by $(2\mu+3-2\lambda+1,2-4\mu-1,2\mu-4-4\lambda+2) \tag{iv}$

From $(iii)$ and $(iv)$, $$\frac{2\mu+3-2\lambda+1}{4}=\frac{2-4\mu-1}{1}=\frac{2\mu-4-4\lambda+2}{-2}=k (say) \tag{v}$$ Since length of $PQ$ is $\sqrt{21}$ units, $${(2\mu+3-2\lambda+1)}^2+{(2-4\mu-1)}^2+{(2\mu-4-4\lambda+2)}^2=21$$ $\implies {(4k)}^2+k^2+{(-2k)}^2=21$ using $(v)$ $$\implies k=\pm1$$ ($k=-1$ is rejected since it can not satisfy $(v)$ for any real value of $\mu$ and $\lambda$)

Putting $k=1$ in Eq. $(v)$, we get

$\mu=0$ and $\lambda=0$

$\therefore$ Feet of common perpendicular PQ are nothing but the points $A(3,2,-4)$ and $C(-1,1,-2)$

and the required equation of $PQ$ is $$\frac{x-3}{4}=\frac{y-2}{1}=\frac{z+4}{-2}$$ which is same as line $AC$.

  • $\begingroup$ Thank you so so so much Aman. Really Thank you $\endgroup$ Jul 6 at 15:04
  • $\begingroup$ @Haroon Tahir You're Welcome. You can upvote my answer and accept it by clicking on the tick button if you're satisfied by my answer and don't want any other answers now. And for formatting, you can use MathJax (to display mathematical notation written in LaTeX). For example, instead of writing sqrt(21) as text you can write \$\sqrt{21}\$ to make it appear like $\sqrt{21}$. This page can guide you- math.meta.stackexchange.com/questions/5020/… $\endgroup$ Jul 6 at 17:01

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