Why Are There No Solutions to This Vector Equation Problem?

[LCSphysicist]

Homework Statement

Get a vector equation of the line that pass by the point P and intersect with the straight line R and S

Relevant Equations

All below.

I think that we can say that PPR = α*PRPS
where PR and PS are the points where occurs the intersection on the line R and S.
Obs: line r and s are found by knowing that the straight line intersection of two planes are
n1 X n2 [cross product]

Lr = (0,1,-2) + y(-1,1,1)
Ls = (0,1,-1) + u(1,2,1)

This leave us to three equation and three incognits

Three real solution, but the answer is that is impossible, why?

 

[LCKurtz]

I'm not quite sure what you actually want, but I don't think your equation for $L_r$ satisfies your definition for line $r$.

 

[andrewkirk]

I think that we can say that PPR = α*PRPS
where PR and PS are the points where occurs the intersection on the line R and S.

I agree

Obs: line r and s are found by knowing that the straight line intersection of two planes are
n1 X n2 [cross product]

I presume you are referring to (1) the plane containing point P and line R and (2) the plane containing point P and line S. Yes it's correct, and a good insight, that the line we seek lies in the intersection of those two planes.

Lr = (0,1,-2) + y(-1,1,1)

Parametrising on y, I get Lr = (-1, 0, -1) + y(-1, 1, -1), which differs from yours in 4 out of 6 places.
Why did you not parametrise on x, which is easier? :
Lr = (0, 1, -2) + x(1, -1, 1)
If we replace your y by x, what you wrote is closer to this, but still differs in two places.

Ls = (0,1,-1) + u(1,2,1)

I agree with this one, and we could replace u by x so as to parametrise in the same way as we did for Lr, to write
Ls = (0,1,-1) + x(1,2,1)

This leave us to three equation and three incognits

View attachment 265641

Three real solution, but the answer is that is impossible, why?

I couldn't quite follow this. I expected to see the equation written as follows:
$$y = a_yx + b_y$$
$$z = a_zx + b_z$$
(assuming the line is not perpendicular to the x axis)

I haven't done the calcs but perhaps if you correct your parametrisations as per above, your solution will match that in the book.

 

[haruspex]

Obs: line r and s are found by knowing that the straight line intersection of two planes are
n1 X n2 [cross product]

I presume you are referring to (1) the plane containing point P and line R and (2) the plane containing point P and line S.

I interpreted @LCSphysicist's remark as referring to the fact line R (and likewise line S) is given in terms of two planes, z=z(x) and y=y(x). Taking the cross product of their normals gives a vector parallel to their line of intersection.

 

[haruspex]

I think that we can say that PPR = α*PRPS

Lr = (0,1,-2)+ y(-1,1,1)

Please explain your PPR, PRPS notation.
Did you mean Lr = (0,1,-2)+ y(1,-1,1)?
Reusing y is confusing. How about u and v as the parameters?

No idea how you got your last trio of equations, but I would rebase the origin to be at P. Then all you need is a point on each of R and S (i.e. u and v values) such that their vector representations are collinear (cross product is zero). Add back in P at the end.

The attachment appears to be a different question.

 

[haruspex]

Update: my relatively simple method still gave crazy answers. I believe the problem is that the line S is parallel to the plane containing P and R (and likewise mutatis mutandis). So there are no solutions.

Fwiw, here's my method in general:
Normalise so that P is the origin. Represent the points on lines R and S parametrically as $\vec R+r\vec R'$, etc.
If r and s are the parameters for the points where the sought line intersects them, the vectors for these points are collinear, so have a zero cross product:
$(\vec R+r\vec R')\times(\vec S+s\vec S')=0$
Expanding, and taking the dot product with $\vec S'$ to eliminate references to s:
$r=-\frac{\vec S'.(\vec R\times\vec S)}{\vec S'.(\vec R'\times\vec S)}=-\frac{\vec R.(\vec S'\times\vec S)}{\vec R'.(\vec S'\times\vec S)}$