A 3d line intersecting 2 other 3d lines

Question

The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$

My solution is as follow

$\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4}$ represent the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the line and intersecting the line ${L_1}:9x + y + z + 4 = 0 = 5x + y + 3z$ & ${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$

Let $\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4} = {\lambda _i}$ where $\left( {x,y,z} \right) = \left( {a + 2{\lambda _i},b + 3{\lambda _i},c + 4{\lambda _i}} \right)$

$ \Rightarrow 9a + 18{\lambda _1} + b + 3{\lambda _1} + c + 4{\lambda _1} + 4 = 0 \Rightarrow \frac{{9a + b + c + 4}}{{ - 25}} = {\lambda _1}$ after putting the values in $L_1$

$5a + 10{\lambda _1} + b + 3{\lambda _1} + 3c + 12{\lambda _1} = 0 \Rightarrow \frac{{5a + b + 3c}}{{ - 25}} = {\lambda _1}$

$9a + b + c + 4 = 5a + b + 3c \Rightarrow 4a - 2c + 4 = 0$

${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$

$\left( {x,y,z} \right) = \left( {a + 2{\lambda _2},b + 3{\lambda _2},c + 4{\lambda _2}} \right)$

$a + 2{\lambda _2} + 2b + 6{\lambda _2} - 3c - 12{\lambda _2} - 3 = 0 = 2a + 4{\lambda _2} - 5b - 15{\lambda _2} + 3c + 12{\lambda _2} + 3$

$ \Rightarrow a + 2b - 3c - 3 - 4{\lambda _2} = 0 = 2a - 5b + 3c + 3 + {\lambda _2} \Rightarrow \frac{{a + 2b - 3c - 3}}{4} = {\lambda _2} = \frac{{2a - 5b + 3c + 3}}{{ - 1}}$

$ \Rightarrow a + 2b - 3c - 3 = - 8a + 20b - 12c - 12 \Rightarrow 9a - 18b + 9c + 9 = 0 \Rightarrow a - 2b + c + 1 = 0$

$ \Rightarrow a - 2b + c + 1 = 0\& 2a - c + 2 = 0$

Let $b = - 1 \Rightarrow a + c + 3 = 0\& 2a - c + 2 = 0 \Rightarrow c = - \frac{4}{3}\& a = - \frac{5}{3}$

$\frac{{x + \frac{5}{3}}}{2} = \frac{{y + 1}}{3} = \frac{{z + \frac{4}{3}}}{4} = \lambda \Rightarrow \left( {x,y,z} \right) = \left( { - \frac{5}{3}\hat i - \hat j - \frac{4}{3}\hat k} \right) + \lambda \left( {2\hat i + 3\hat j + 4\hat k} \right)$

But my answer is not matching. Can you tell me the error that I have made. Each steps is elaborated

Answer 1

To find a line parametric equation, we have to find a point on the line, and then use the cross product of the normal vectors of the the two planes given.

The two planes for the first line are

$9x + y + z = -4 $

$5 x + y + 3 z = 0 $

To find a point on both these planes, set $x = 0$ and solve the system

$ y + z = -4 $

$ y + 3 z = 0 $

The solution is $y = -6, z = 2 $ so our point is $(0, -6, 2) $

Next the direction vector of the line is along the cross product of the first normal and second normal

$d = (9, 1, 1) \times (5, 1, 3) = (2, -22, 4) = 2 (1, -11, 2) $

So the parametric equation of the first line is

$ P_1(t) = (0, -6, 2) + t (1, -11, 2) $

Similarly, for the second line, the planes are

$ x + 2 y - 3 z = 3 $

$ 2 x - 5 y + 3 z = -3 $

To find a point of the line, it must satisfy both equations, so set $y = 0$ and solve the system

$ x - 3 z = 3 $ $ 2 x + 3 z = -3 $

The solution is $x = 0 , z = -1 $, so our point is $(0, 0, -1) $

And the direction vector is

$ (1, 2, -3) \times (2, -5, 3) = -9 (1, 1, 1) $

So the parametric equation of the second line is

$ P_2 (s)= (0, 0, -1) + s (1, 1, 1) $

Now, since the line whose equation is requested is parallel to $(2,3,4)$ and it intersects both the first and second lines obtained above, then the setup is as follows:

$ P_1(t) - P_2(s) = \lambda (2, 3, 4) $

For some $t, s, \lambda$. Writing the coordinates as column vectors, this last equation becomes the linear system

$ \begin{bmatrix} 1 && -1 && -2 \\ -11 && -1 && -3 \\ 2 && -1 && - 4 \end{bmatrix} \begin{bmatrix} t \\ s\\ \lambda \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ -3 \end{bmatrix} $

Its solution is $ (t, s, \lambda ) = ( - \dfrac{3}{5} , - 3, \dfrac{6}{5} ) $

Thus the first point of intersection on the first line is

$ P_1^* = (0, -6, 2) +\left( -\dfrac{3}{5} \right)(1, -11, 2) = (-\dfrac{3}{5}, \dfrac{3}{5}, \dfrac{4}{5} ) $

And the intersection with the second line is

$P_2^* = (0, 0, -1) +(- 3) (1, 1, 1) = (-3, -3, -4) $

We can use either point for the parametric equation of our line. Using the second point, we get

$ Q(t) = (-3, -3, -4) + t (2, 3, 4) $

Answer 2

Your way of squeezing more than one equation in a line is confusing me and I will do it the usuual way, one line for an equation.

A linear equation represent a plane. A line is represented as the intersection of two planes, that is a systeme of two equations. The coefficients of $x,y,$ and $z$ of a linear equation are equal to a normal vector of the plane. If two planes are parallel then they have parallel normal vectors. so we can get all parallel planes to a plane by changing the constant term of the plane equation.

The equations $$\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$$ represent two planes $$\frac{x}{2} = \frac{y}{3}\\ \frac{y}{3} = \frac{z}{4}$$ All other equations that can deduced from this notation are equations that are dependent from the two equations we already selected and therefore can be ignored. These two equations can be transfomed to

$$3x-2y=0\\ 4y-3z=0$$

The line we are looking for is parallel to these planes and therefor lies on planes parallel to these planes. These parallel planes have the equations

$$3x-2y=a \tag{e1}\\ 4y-3z=b$$

(Maybe your way to establish the equations of the parallel line is wrong)

we want to calculate $a$ and $b$. We have two other lines

$$9x + y + z + 4 = 0 \tag{e2} \\ 5x + y + 3z=0$$

$$x + 2y - 3z - 3 = 0 \tag{e3}\\ 2x - 5y + 3z + 3=0$$

Both lines intersect line($e1)$.

So we intersect line $(e1)$ with $(e2)$, this gives the systems

$$3x-2y=a \tag{e4}\\ 4y-3z=b\\ 9x + y + z + 4 = 0 \\ 5x + y + 3z=0$$

and

$$3x-2y=a \tag{e5}\\ 4y-3z=b\\ x + 2y - 3z - 3 = 0 \\ 2x - 5y + 3z + 3=0$$

$(x,y,z)$ are the coordinates of the intersection point. So these are differents for system $(e4)$ and $(e5)$, because the intersection points are different. $a$ and $b$ are the same in both systems.

Each system has $4$ equations and $5$ variables. We can eliminate $3$ variables $x$,$y$,$z$ and then an equation with two variables $a$,$b$ is left. We get the two equations

$$2a+b+6=0 \tag{e6}\\ a-b+3=0$$

And if we solve this we get $$a=-3\\b=0\tag{e7}$$ And so the line we are looking for is

$$3x-2y=-3 \tag{e8}\\ 4y-3z=0$$

The intersection of $(e2)$ and $(e8)$ is the point $$x=-\frac{3}{5},y=\frac{3}{5},z=\frac{4}{5}$$ the intersection of $e3$ and $e8$ is the point $$x=-3,y=-3,z=-4$$ You can use them to check the solution.