Point of intersection when only direction ratios are given

Question

I am starting out with 3D Geometry. In one of the test booklets, I found a question for which I have no idea where and how to start from.

If a line with direction ratio $2:2:1$ intersects the line $\frac{x-7}{3}$ = $\frac{y-5}{2}$ = $\frac{z-3}{2}$ and $\frac{x-1}{2}$ = $\frac{y+1}{4}$ = $\frac{z+1}{3}$ at A and B, find AB.

I know the algorithm for finding the points of intersection when a line is in the symmetrical form. I tried with an approach wherein I tried to find A by equating:

$\frac{x}{2}$ = $\frac{y}{2}$ = $\frac{z}{1}$ and $\frac{x-7}{3}$ = $\frac{y-5}{2}$ = $\frac{z-3}{2}$

However, on equating, I found out that these lines never even intersect.

How do I approach this sum? How to start off with this sum?

Answer 1

Let $a$ be the common value of $$\frac{x-7}{3}=\frac{y-5}{2}=\frac{z-3}{2}=a$$

We can re-write these equations under the following equivalent parametric form:

$$\tag{1}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}7+3a\\5+2a\\3+2a\end{pmatrix}$$

In the same way, the generic point of the second straight line is : $$\tag{2}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\ \ 1+2b\\-1+4b\\-1+3b\end{pmatrix}$$

Then you just have to express the proportionality of $\vec{AB}$ with the given vector, yielding the following system of 2 equations and 2 unknowns:

$$\dfrac{(1+2b)-(7+3a)}{2}=\dfrac{(-1+4b)-(5+2a)}{2}=\dfrac{(-1+3b)-(3+2a)}{1}$$

giving $a=-2/3$ and $b=1/3$.

It remains to plug these values in (1) and (2) to get the coordinates of

$$A=\begin{pmatrix}5\\11/3\\5/3\end{pmatrix} \ \ \ \text{and} \ \ \ B=\begin{pmatrix}5/3\\1/3\\0\end{pmatrix}.$$

Answer 2

Parametric equation of the two given line 1 and line 2 are:

$$\frac{x-7}{3} = \frac{y-5}{2} = \frac{z-3}{2}\implies(7,5,3)+t(3,2,2)$$

and

$$\frac{x-1}{2} = \frac{y+1}{4} = \frac{z+1}{3}\implies(1,-1,-1)+s(2,4,3)$$

The parametric equation for the unknown line3 is

$$(a,b,c)+h(2,2,1)$$

The intersection between line 1 and line 3 require that both lines are in the same plane, we can find its normal by cross product

$$n_{13}=\begin{vmatrix} i&j&k\\3&2&2\\2&2&1 \end{vmatrix}=-2i+j+2k$$

thus the plane equation containing both line 1 and line 3 is

$$\pi_{13}:-2x+y+2z+d=0$$

we find "d" by $(7,5,3)\in \pi_{13}$

$$-14+5+6+d=0\implies d=3$$

thus

$$\pi_{13}:2x-y-2z-3=0$$

Similarly we find the plane equation containing both line 2 and line 3

$$n_{23}=\begin{vmatrix} i&j&k\\2&4&3\\2&2&1 \end{vmatrix}=-2i+4j-4k$$

$$\pi_{23}:-2x+4y-4z+d=0$$

we find "d" by $(1,-1,-1)\in \pi_{23}$

$$-2-4+4+d=0\implies d=2$$

thus

$$\pi_{23}:2x-4y+4z-2=0$$

Therefore the equation for line 3 is given by

$$\pi_{13}\cap \pi_{23} \begin{cases} 2x-y-2z-3=0\\ 2x-4y+4z-2=0 \end{cases}$$

The intersection point A between line 1 and line 3 is

$$\begin{cases} 2x-y-2z-3=0\\ 2x-4y+4z-2=0\\(7,5,3)+t(3,2,2) \end{cases}$$

$$2(7+3t)-4(5+2t)+4(3+2t)-2=0\implies 6t+4=0\implies t=-\frac{2}{3}$$

$$A=(7,5,3)-\frac{2}{3}(3,2,2)=\left(5,\frac{11}{3},\frac{5}{3}\right)$$

The intersection point B between line 2 and line 3 is

$$\begin{cases} 2x-y-2z-3=0\\ 2x-4y+4z-2=0\\(1,-1,-1)+s(2,4,3) \end{cases}$$

$$2(1+2s)-(-1+4s)-2(-1+3s)-3=0\implies -6s+2=0\implies s=\frac{1}{3}$$

$$B=(1,-1,-1)+\frac{1}{3}(2,4,3)=\left(\frac{5}{3},-\frac{1}{3},0\right)$$