Can Vector Cross Product Determine Intersection Point and Angle Between Lines?

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Homework Statement



4. Consider the following two lines:

L1: [1 2 1] + s [2 -1 1] , L2 = [3 0 1] + t [1 1 2](by the way, all of those are column vectors. I just don't know how to format them correctly.)

(a) These two lines intersect at a point P . Find the co-ordinates of P .

(b) What is the cosine of the acute angle θ between these two lines ?

The Attempt at a Solution



Well for a), I don't know if we are supposed to get the normal vector or not. If so, then I know to do the cross product. If that's not the proper solution then can I merely do P1P2 = P2 - P1, or is that completely irrelevant? I'm just really confused about the wording!
 
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If they intersect at point P, then the x, y and z coordinates of each line are the same at that point right?
 
I don't understand what you mean. Each line has separate points and different direction vectors as well. :S
 
They have a lot of separate points, but one special point P they have in common. If this point is given by P = [p1 p2 p3] then you would agree that 1+2s = p1 for some s and 3+t = p1 for some t right? It looks like you've got the makings of a set of linear equations.
 
For example, the x coordinate of L1 is given by 1+2s while the x coordinate is given by 3+ t. Where the lines intersect, those must be the same: 1+ 2s= 3+ t. The same is true for the y and z coordinates.
 
But it says to find the coordinates of P, not the related scalar equations...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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