ivory oracle
(b).
We can take that v just as the difference of vectors of points they go through:
v = r1 - r0 = {0, 1, 0} - {2, 1, 0} = {-2, 0, 0}
(c).
First, we find v = r1 - r0 again.
Then we need to find its component that is perpendicular to the direction vector s of both lines:
perp(s, v) = v - proj(s, v) = v - (v·s/(s·s))s
Then the distance between the lines is just the length of that vector:
d(L0, L1) = |perp(s, v)| = |v - (v·s/(s·s))s|
We can stop there, but I will bring it to a more common form first.
Let's consider the squared distance.
d(L0, L1)^2 = (v - (v·s/(s·s))s)^2 = v·v + (v·s/(s·s))^2 s·s - 2(v·s/(s·s))v·s = |v|^2 + (v·s)^2/|s|^2 - 2(v·s)^2/|s|^2 = |v|^2 - (v·s)^2/|s|^2 = (|v|^2 |s|^2 - (v·s)^2)/|s|^2
And now we can apply the following identity: (v·s)^2 + (v⨯s)^2 = |v|^2 |s|^2.
d(L0, L1)^2 = (v⨯s)^2)/|s|^2
Thus:
d(L0, L1) = |(r1 - r0)⨯s|/|s|
(d).
The normal vector of the plane must be perpendicular to both s and v. So, we can take it as their cross product:
n = s⨯v
Now we can pick a point that the plane passes through: either r0 or r1, doesn't matter. I'll pick r0.
Then the equation of the plane is, as usual:
n·(r - r0) = 0
Q6.
Just substitute these coordinates into the resulting equation of the plane and see that you don't get a valid equality.
fluid inlet
