A past examination paper had the following question that I found somewhat difficult. I tried having a go at it but haven't come around with any solutions. How would one go about tackling it?

The question is split into three, managed to do the first part;

a) The points A and B have position vectors 2i+j-3k and 4i-j+k, respectively. Find the equation of the line $l_1$ that passes A and B.

Basically, using the equation of a vector line $$r=a+λ(b-a)$$ Which results to; $$2i+j-3k+λ(2i-2j+4k)$$ Correct? or should I put my final answer as $$2i+j-3k+μ(i-j+2k)$$

However, my main problem is this;

b) The points C and D have position vectors ai+j+4k and -3i+3j+bk, respectively. Let $l_2$ be the line that passes through C and D. Find a and b given that $l_1$ and $l_2$ intersect at the midpoint of the line segment $\vec{AB}$.

In case you're wondering what the third question is, it's asking you to find the acute angle, but I'll work it out by dividing the dot product by the length of vector.

I do not wish to ask for a final answer, but more of a guideline on how to achieve it, also, if instead of a and b there was α and β, same method would apply right?

  • $\begingroup$ Hint: at what value of $\lambda$ does the vector $\vec{r}$ point to the midpoint of $AB$? Can you repeat what you did for $AB$ and come up with the vector $\vec{r'}$ that sweeps over CD? What can you say in addition if the line $CD$ passes through the midpoint of $AB$? $\endgroup$ – NickD Apr 22 at 22:25
  • $\begingroup$ To your last question, it doesn’t matter what you call the unknowns as long as those names aren’tt already in use for something else. The answers don’t change just because you’ve switched from the Roman to the Greek alphabet. $\endgroup$ – amd Apr 22 at 22:33

A solution following the presentation I remember from college. The midpoint is $\mathbf M=(\mathbf A+\mathbf B)/2$. With this, the only other information that you need to continue in $(b)$ is the fact that $\mathbf M$ lies also on the line $l_2$. If $l_2$ is the equation $r=\mathbf v + \lambda \mathbf w$, then writing down the 3 components of $$\mathbf M =\mathbf v + \lambda \mathbf w $$ gives you 3 simultaneous equations in the 3 unknowns are $\lambda,a,b$. So if you don't make a mistake and the question doesn't have an error, you can solve for these values.

Explicitly, writing vectors as columns, you might end up with $$ \begin{bmatrix} 3\\0 \\ -1\end{bmatrix} =\begin{bmatrix} a\\1 \\ 4\end{bmatrix} +\lambda \begin{bmatrix} -3-a\\3-1 \\ b-4\end{bmatrix} $$ From this, its clear that you get $\lambda = -1/2$(not so important), $a=1$, and $b=14$.

A 3D plotter (different camera angle) lets you visually see that this is indeed a solution:

enter image description here

  • $\begingroup$ This is equivalent to my solution. The widely-used trick of using a cross product to generate the system of equations avoids introducing the extraneous variable $\lambda$. $\endgroup$ – amd Apr 24 at 21:00
  • $\begingroup$ @amd Yes, I thought also of getting a plane equation out and forcing $M$ to lie in it but didn't attempt to work that out. I posted the answer mainly because (a) OP asked this on reddit and I had it essentially typed out already, (b) I wanted an excuse to try out math3d... $\endgroup$ – Calvin Khor Apr 24 at 21:59

For $CD$ to pass through some point $M$, $M-C$ must be parallel to $D-C$ (i.e., $CD$ and $CM$ must be the same line.) Since you’re working in $\mathbb R^3$, you can take advantage of the fact that two nonzero vectors are parallel iff their cross product vanishes. In this case, then, that means $(M-C)\times(D-C)=0$. This gives you a system of three equations in the two unknowns to solve (although only two of those three equations are independent). The midpoint of $\overline{AB}$ can of course be found via the usual formula $M=\frac12(A+B)$.

Another approach: If $AB$ and $CD$ intersect, then they are coplanar, so in particular all four of the given points are. This constraint is equivalent to $$\det\pmatrix{A&1\\B&1\\C&1\\D&1}=0,$$ which is a linear equation in $a$ and $b$. We want $C$, $D$ and $M$ to be colinear as well, which can also be expressed via a determinant that generates another linear equation in $a$ and $b$: $$\det\pmatrix{C\\D\\\frac12(A+B)}=0.$$ Expanding these determinants involves more work than the cross product solution I described earlier, but this technique of using determinants to express linear dependence can be useful in other problems, and it generalizes more easily than the cross product to higher-dimensional spaces.


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