How to find equation of line intersecting with two given lines and parallel to the given plane

Question

We have given two lines $a$ and $b$ and a plane $\sum$. We are asked to find equation of a line which is intersecting with the lines $a$ and $b$ and is parallel to the given plane $\sum$.

$$a: \frac{x}{2} = \frac{y-1}{1} = \frac{z-1}{2} \\b: \frac{x}{1} = \frac{y+1}{2} = \frac{z}{3} \\ \sum: x+y+z=0$$

I started from the fact if two lines are intersecting they must belong to the same plane, but I'm not sure if this means that the lines $a$ and $b$ will always belong to the same plane, so I'm not sure how to start solving the task.

Answer 1

We can write the line a as x= 2t, y= t+ 1, z= 2t+ 1 and the second line as x= s, y= 2s- 1, z= 3s. If those two lines were to intersect at some point (x, y, z) we would have to have x= 2t= s, y= t+ 1= 2s- 1, and z= 2t+ 1= 3s. From 2t= s, s= 2t so t+ 1= 2(2t)- 1= 4t 1. 3t= 2 so t= 2/3 and s= 4/3. Then z= 2(2/3)+ = 7/3 and z= 3(4/3)= 4. Those are not the same so, no they do not intersect. Nor are they parallel. They are skew and do not lie in a single plane.

Now, as for the original problem, to find a line that intersects both given lines and is parallel to the given plane. Write the desired line as x= au+ b, y= cu+ d, z= eu+ f. To be parallel to the given plane, a vector in the direction of the line, such as , must be perpendicular to the planes normal vector, <1, 1, 1>. That is, .<1, 1, 1>= a+ c+ e= 0 so we must have e= -a- c. We can write the line as x= au+ b, y= cu+ d, z= (-a- c)u+ f. At the point of intersection of that line with the first given line, we must have 2t= au+ b, t+ 1= cu+ d, and 2t+ 1= (-a- c)u+ f. At the point of intersection of that line with the second given line we must have s= au+ b, 2s- 1= cu+ d, and 3s= (-a- c)u+ f. That gives us 6 equations to solve for the 7 unknowns, a, b, c, d, f, s, and t. There is one more unknown than equations so this is "under determined". There will be an infinite number of possible solutions.

Answer 2

Here's another way of looking at it. Choose an arbitrary point on line a: Line a is given by parametric equations x= 2t, y= t+ 1, z= 2t+ 1. Taking t= 0 gives the point (0, 1, 1). Write the equation of the plane parallel to the given plane through that point: a plane parallel to the given plane is of the form x+ y+ z= A for non-zero A. 0+ 1+ 1= 2 so x+ y+ z= 2 is parallel to x+ y+ z= 0 and contains the point (0, 1 1).

Now, where does line b, given by x= s, y= 2s- 1, z= 3s, intersect the plane x+ y+ z= 2? x+ y+ z= s+ 2s- 1+ 3s= 6s- 1= 2 so 6s= 3 and s= 1/2. The point (1/2, 0, 3/2) lies on line b and on the plane x+ y+ z= 2. The line through (0, 1, 1) and (1/2, 0, 3/2), x= (1/2)t, y= -t+ 1, z= -(1/2)t+ 1, which can also be written [tex]2x= -y+ 1= -2z+ 2[/tex], intersects both line and is parallel to the given plane.

Of course, that original choice of (0, 1, 1) on line a was arbitrary. A different choice of a point on that line would give another of the infinite number of solutions.

Answer 3

​Write the two lines in vector form by setting them equal to some constant $\lambda$ and $\mu$:

$$ a = \begin{cases}\dfrac{x}{2}=\lambda \\ y-1=\lambda \\ \dfrac{z-1}{2}=\lambda \end{cases} \implies \begin{cases}x = 2\lambda \\ y= \lambda + 1\\ z=2\lambda + 1\end{cases} \implies \left(\begin{array}{c}2\lambda \\ \lambda + 1\\ 2\lambda + 1\end{array}\right)$$

$$ b = \begin{cases}x=\mu \\ \dfrac{y+1}{2}=\mu \\ \dfrac{z}{3}=\mu \end{cases} \implies \begin{cases}x = \mu \\ y= 2\mu - 1\\ z=3\mu\end{cases} \implies \left(\begin{array}{c}\mu \\ 2\mu - 1\\ 3\mu\end{array}\right)$$

Since the line is parallel to the plane, the line is simply the connection between the two points where $a$ and $b$ intersect with $\Sigma$ (We'll call them $A$ and $B$). First, find the intersection of $a$ with $\Sigma$:

$$ 2\lambda+\lambda +1 + 2\lambda + 1 = 0 \implies \lambda=-\frac{2}{5} \\ A = \left(\begin{array}{c}2(-\frac{2}{5}) \\ (-\frac{2}{5}) + 1\\ 2(-\frac{2}{5}) + 1\end{array}\right) = \left(\begin{array}{c}-\frac{4}{5} \\ \frac{3}{5}\\ \frac{1}{5}\end{array}\right) $$

Find the intersection of $b$ with $\Sigma$:

$$ \mu + 2\mu - 1 + 3\mu = 0 \implies \mu = \frac{1}{6}\\ B= \left(\begin{array}{c}\frac{1}{6} \\ 2(\frac{1}{6}) + 1\\ 3(\frac{1}{6})\end{array}\right) = \left(\begin{array}{c}\frac{1}{6} \\ -\frac{2}{3}\\ \frac{1}{2}\end{array}\right) $$

Then the line going through the two points is (choosing $A$ as the position vector)

$$ \left(\begin{array}{c}-\frac{4}{5} \\ \frac{3}{5}\\ \frac{1}{5}\end{array}\right) + \gamma \left(\begin{array}{c}\frac{1}{6}+\frac{4}{5} \\ -\frac{2}{3}-\frac{3}{5}\\ \frac{1}{2}-\frac{1}{5}\end{array}\right) = \left(\begin{array}{c}-\frac{4}{5}+\frac{29}{30}\gamma\\ \frac{3}{5}-\frac{19}{15}\gamma \\ \frac{1}{5}+\frac{3}{10}\gamma\end{array}\right) $$

Finally, converting into cartesian form:

$$ \begin{cases}\frac{30}{29}(x+\frac{4}{5})=\gamma \\ \frac{15}{19}(\frac{3}{5}-y)=\gamma \\ \frac{10}{3}(z-\frac{1}{5})=\gamma\end{cases} \implies \frac{30x+24}{29}=\frac{9-15y}{19}=\frac{10z-2}{3} $$