How to prove the following lines are intersecting on 3d?

Question

Question:

Show that the lines $\frac{x-5}{4}=\frac{y-7}{-4}=\frac{z-3}{-5}$ and $\frac{x-8}{4}=\frac{y-4}{-4}=\frac{z-5}{4}$ intersect.

I am confused in seeing the fractions, I could think of separating the variables to the RHS and LHS respectively. But since there are three sides I am not sure what could I do.

If I could prove they are not parallel, then that probably tells that they will meet at some point, Right? But it is for 2d lines,no? But here there are 3 axis.

Answer 1

Setting

$$a=\frac{x-5}{4}=\frac{y-7}{-4}=\frac{z-3}{-5}\tag{1}$$

$$b=\frac{x-8}{4}=\frac{y-4}{-4}=\frac{z-5}{4}\tag{2}$$

allows to obtain an equivalent description of these 3D lines under a parametric representation :

$$(i) \ \begin{cases}x&=& \ \ \ 4a+5\\ y&=&-4a+7\\ z&=&-5a+3\end{cases} \ \ \text{and} \ \ (ii) \ \begin{cases}x&=& \ \ \ 4b+8\\ y&=&-4b+4\\ z&=& \ \ \ 4b+5\end{cases}$$

The fact that these 3D lines have a common point $(x,y,z)$ is equivalent to the fact that there exists a solution to the following system of 3 equations in 2 unknowns $a$ and $b$ :

$$\begin{cases} \ \ \ 4a+5&=& \ \ \ 4b+8\\ -4a+7&=&-4b+4\\ -5a+3&=& \ \ \ 4b+5\end{cases}$$

Indeed, this system has a (unique) solution:

$$a=\frac19 \ \ \text{and} \ \ b=-\frac{23}{36}$$

The coordinates $(x,y,z)$ of the intersection point can be obtained for example by plugging the value $a=\frac19$ in equations (1).

Edit : an alternative solution :

(i) implies that

$$x+y=12\tag{4}$$

the same for (ii).

Therefore the two lines are includes into (remember that the verb "to imply" used between equations means "to be included" for the corresponding sets) the same plane with equation (4). It remains to verify that they are aren't parallel. This is the case because their directing vectors, given by the denominators of (1) and (2), aren't proportional.

Answer 2

Hint: Set $\;t=\dfrac{x-8}{4}=\dfrac{y-4}{-4}=\dfrac{z-5}{4},$ whence $$x=8+4t,\quad y=4-4t,\quad z=5+4t$$ and plug these relations in the first series of equations: $$\frac{3+4t}{4}=\frac{-3-4t}{-4}=\frac{2+4t}{-5}.$$ As you can see there is a redundant equation, and you only have yo solve: $$\frac{3+4t}{4}=\frac{2+4t}{-5}\iff -5(3+4t)=4(2+4t)$$ Solve for $t$, then deduce $x,y,z$.