Dtjytyk

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.

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.

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.

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

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

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

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 $Sigma$:

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

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

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

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

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

Finally, converting into cartesian form:

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