Straight Lines in Space.

    1. Equations. Let A(a₁,a₂,a₃) be a point, and N(n₁,n₁,n₃) is a given vector. Then the line through A with N as direction, (i.e., parallel to N) is made up of points X(x,y,z) that satisfies the condition that X-A is parallel to N, so that X-A = t N, or X = A + t N for some real number t.
We say X = A + t N . . . . . (i) is the vector equation of the line. t is called a parameter for the line, because different values of t give different point X on the line. We say X is a vector function of t, and we can write X(t) = A + t N as a one-one mapping from the set of all real numbers into 3-dim space.

    In terms of coordinates, this is really 3 equations in 4 variables, x, y, z, t.
x = a₁ + t n₁, y = a₂ + t n₂, z = a₃ + t n₃ . . . . (ii) called parametric form.
Some people eliminates t and write
(x-a₁)/n₁ = (y-a₂)/n₂ = (z-a₃)/n₃ . . . . (iii), which is called symmetric form.

    Given 2 points A(a₁,a₂,a₃) and B(b₁,b₂,b₃). The line through them is from (i)
X-A = t(B-A), or X = (1-t)A + t B . . . . (iv) called the two-points form.

    Example 1. Convert the equations 3 x + 1 = 2- 3 y = 2 z into parametric form.
    Solution. We want each coefficient of x, y, z be 1, so divide each of the 3 quantities top-and-bottom by something, thus getting
(x - (-1/3) )/(1/3) = (y - 2/3)/ (-1/3) = (z- 0)/(1/2) as the symmmetric form.
Let A = (-1/3, 2/3, 0) and N = (1/3, -1/3, 1/2), then A is a point on the line and N is a direction vector, and the line is X = A + t N, or (x,y,z) = (-1/3, 2/3, 0) + t (1/3, -1/3, 1/2).

    Example 2. Find the line of intersection of the planes 2 x - y + z = 3 and x + 2 y - z = 1.
    Solution.By hook or by crook, find some point A on the line. E.g., let y = 0, then the equations become 2 equations in 2 unknowns, which can be solved, and we get x = 4/3, z = 1/3, so that (4/3, 0, 1/3) is a point on the line. The direction vector V is perpendicular to the normal vector (2, -1, 1) of the first plane and the normal vector (1, 2, -1) of the second plane,
hence V = their cross product (2, -1, 1) × (1, 2, -1)
= ( -1×-1 - 2×1, 1×1 - 2×-1, 2×2 - 1×-1)
= (-1, 3, 5). Therefore the line is (x, y, z) = (4/3, 0, 1/3) + t (-1, 3, 5).
    Remark. Different people may come up with different answers, e.g., if we had set y to 1 instead of 0, we would have gotten x = 1 and z = 2, so that your answer may be
(x, y, z) = (1, 1, 2) + t (1, -3, -5). However, the two directional vectors in the two answers are parallel, and the difference of coordinates between the point (4/3, 0, 1/3) and the point (1, 1, 2), namely, (4/3-1, 0-1, 1/3-2) = (1/3, -1, -5/3) is also parallel to the direction vector. So that is OK.

    Example 3. Given a line L: X(t) = (x, y, z) = (1, 2, 3) + t (2, 1, -1) and a point P(3, 2, 1), find (a) the distance of P from L, (b) the equation of the perpendicular line from P to L, (c) the foot of perpendicular from P to L, (d) the plane containing P and L, (e) the plane containing P perpendicular to L, (f) the plane containing P parallel to L and at maximum distance to L, (g) the lines through P crossing L at 60 degrees.
    Solution. Let Q be the point X(t) on L. Then PQ ^ L if 0 = (Q-P)·N, where N is the direction vector (2,1,-1) of L. Thus 0 = (1 + 2 t - 3) 2 + (2 + 1 t - 2) 1 + (3 -t -1) -1, from which we get 6 t -6 = 0, or t = 1. Hence Q is (3, 3, 2). Check: Vector P-Q is (3-3, 2-3, 1-2) = (0,-1,-1), which is perpendicular to N, because their dot product = 0.
    The equation of line PQ is (x, y, z) = P + t (Q -P), i.e., (x, y, z) = (3, 2, 1) + t(0, 1, 1). The distance of P from L = |P-Q| = Ö(0² + 1² + 1²) = Ö2. Thus (a),(b),(c) are done.
    Let P₁ be the plane containing P and L, then its normal N₁ = N × (Q-P) = (2,1,-1)×(0,1,1) = (1- -1, 0-2, 2-0) = (2,-2,2). So we can take N₁ to be (1, -1, 1).
Hence the equation of the plane P₁ is (X-P)·N₁ = 0, i.e.,
(x-3)1 + (y-2)(-1) + (z-1)1 = 0, or x-y+z-2 = 0.
    The plane P₂ through P or Q ^ L is easy, because it has N as normal, so that it is
(x-3)2 + (y-2)1 + (z-1)(-1) = 0, i.e., 2 x + y -z -7 = 0.
    The plane P₃ containing P, parallel to L, and at maximal distance from L is ^ to P₁ and P₂, hence its normal is P-Q, so that its equation is (X-P)·(Q-P) = 0, or
(x-3)0 + (y-2)1 + (z-1)1 = 0, or y + z -3 = 0.
    Check: The normals to the 3 planes are ^ each other.
    Finally, let S be any point X(t) on L. We want PS to be at 60° with L.
So we want (P-S)·N = |N| |P-S| cos 60°,
i.e., [(3-(1+2t))2 + (2-(2+t))1 + (1-(3-t))(-1)]² = [2²+1²+(-1)²] [(2-2t)²+t²+(t-2)²] /4, or
[4-4 t -t + 2-t]² = 3 [8 -12 t +6 t²]/2, or
36 (1-t)² = 3(4 -6 t + 3 t²), or 9 t² -18 t +8=0, or (3t-4)(3t-2)=0
Thus t = 4/3 or 2/3.
Hence S has two solutions, (1 + 2(4/3), 2+4/3, 3+(-1)4/3), or (1 + 2(2/3), 2+2/3, 3+(-1)2/3),
i.e., S₁(11/3, 10/3, 5/3) or S₂(7/3, 8/3, 7/3).
Hence PS₁ is (x,y,z) = (3,2,1) + t(2/3,4/3,2/3), PS₂ is (x,y,z) = (3,2,1) + u(-2/3,2/3,4/3).

    Example 4. (Skew Lines) Given a line L: X(t) = (2,1,4)+ t(1,-1,1), and line
L': X(u) = (1,-2,2) + u(1,3,-1). Find the minimum distance between them and the mutual perpendicular line. Find the equation of the two parallel planes, one containing L, one containing L'.
    Solution. Let A be (2,1,4), N be (1,-1,1), A' be (1,-2,2), and N' be (1,3,-1). Let Q be the point X(t), and Q' the point X(u). We want QQ' to be ^ N and ^ N', hence QQ' must be parallel to
N×N', which is (1,-1,1)×(1,3,-1) = (1-3, 1- -1, 3- -1) = (-2,2,4), or (-1,1,2).
    Now vector QQ' is Q'-Q, = ((1+u)-(2+t), (-2+3 u)-(1-t), (2-u)-(4+t))
= (-1+u-t, -3+3 u+t, -2-u-t). Since Q'-Q is parallel to N×N', we have
(-1+u-t)/(-1) = (-3+3 u+t)/1 = (-2-u-t)/2, or
2 - 2 u + 2 t = -6 + 6 u + 2 t = -2 -u -t, so that
u = 1 and t = -1. Therefore Q is (1,2,3) and Q' is (2,1,1).
The distance between them is Ö((1-2)²+(2-1)²+(3-1)²) = Ö6.
The line through them is (x,y,z) = (1,2,3) + s (2-1,1-2,1-3) or (x,y,z) = (1,2,3)+ s (1,-1,-2).
     The parallel planes P and P' containing L and L' each has Q'-Q as normal,
so P is (x-1)1 + (y-2)(-1) + (z-3)(-2) = 0, or x -y -2 z + 7 = 0.
P' is (x-2)1 + (y-1)(-1) + (z-1)(-2) = 0, or x -y -2 z + 1 = 0.

    Example 5. (Bird's eyeview) Assume L and L' as in Example 4. A bird is perching at a point K(-3, 2, 4). Find the direction at which the two lines appear to cross, from his point of view.
    Solution. Define Q as X(t) on L and Q' as X(u) on L'. We want Q-K to be parallel to A'-K, so that
(2+t- -3, 1-t-2, 4 + t-4) is parallel to (1+u - -3, -2 + 3 u-2, 2-u-4), or
(5+t, -1-t, t) is parallel to (4+u, -4+3 u, -2-u). . . .(i)
\ (5+t)/(4+u)= t/(-2-u) and (1+t)/(4-3 u) = t/(-2-u) . . . (ii), i.e.,
4 t + t u + 10 + 2 t + 5 u + t u = 0 and 4 t - 3 t u + 2 + 2 t + u + t u = 0,
10 + 6 t + 5 u + 2 t u = 0 and 2 + 6 t + u -2 t u = 0.
Adding gives 12 + 12 t + 6 u = 0, or u = -2-2 t. Subtracting gives 8 + 4 u + 4 t u = 0, or 2 + u (1 + t) = 0.
So 2 + -2(1 + t)² = 0, or 1 + t = 1 or -1, so that t = 0 or -2, then u = -2 or 2.
The first set of solution (t = 0, u = -2) is unacceptable, because it makes the RHS of (ii) 0/0.
So the only solution is t = -2 and u = 2, giving the direction from (i) as (3, 1, -2). In fact, the line through K crossing both L and L' is X(w) = (-3,2,4) + w(3, 1, -2).
    2nd Solution (more ingenious). Let P be the plane containing K and L, P' be the plane containing K and L'. Then P intersects P' at the line of sight of the bird.
    Now P-A is (-5,1,0), P-A' is (-4, 4, 2). Hence the normal M of the plane containing K and L is
(P-A)×N = (-5,1,0)×(1,-1,1) = (1, 5, 4), and the normal M' of the plane containing K and L' is
(P-A')×N' = (-2,2,1)×(1,3,-1) = (-5,-1,-8).
Hence the direction of the bird's sight is M×M' = (1,5,4)×(5,1,8) = (36, 12, -24) or (3,1,-2).