Linear Algebra Problem

Note: For t being any real number,

(vector) = (position vector of a point) + t(direction vector)

which is similar to y = c + mx

Similarity
c = y-intercept ~ position vector of a point
m = slope ~ direction vector

If the direction vectors are the same, then the two lines are parallel.


(a)
p = (0, 7) + s(14, -7) = (0, 7) + 7s(2, -1) with direction vector (2, -1)
q = (0,3) + t(3, 6) = (0,3) + 3t(1, 2) with direction vector (1, 2)

p and q (both are 2-dim. lines) have different direction vectors, the two lines intersect.

Note 1:
This is exactly like
　Line 1: y = m₁x + c₁
　Line 2: y = m₂x + c₂
If m₁ ≠ m₂ (not parallel), then the two lines intersects.

Note 2:
If they are 3-dim. lines, then even if they are not parallel, we cannot say they must intersect. There is a situation of a pair of "skew lines" (you can search it easily) that two non-parallel lines do not intersect.


(b)
x = (3, 7, 2) + s(3, 5, -4)
y = (4, 2, 1) + t(-9, -15, 12) = (4, 2, 1) - 3t(3, 5, -4)

The two lines have the same direction vectors as (3, 5, -4), so they are parallel.

x = (3 + 3s, 7 + 5s, 2 - 4s)
y = (4 - 9t, 2 - 15t, 1 + 12t)

Consider x = y
{ 3 + 3s = 4 - 9t　⇒ 3s + 9t = 1　⇒ s + 3t = 1/3
{ 7 + 5s = 2 - 15t　⇒ 5s + 15t = -5　⇒ s + 3t = -1　(inconsistent)
{ 2 - 4s = 1 + 12t

x and y cannot be the same, so the two lines do not intersect.
That is, the two lines are parallel and distinct.


(c)
u = (2, 1) + s(3, 3) = (2, 1) + 3s(1, 1)
v = (8, 4) + t(2, 2) = (8, 4) + 2t(1, 1)

The two lines have the same direction vectors as (1, 1), so they are parallel.

u = (2 + 3s, 1 + 3s)
v = (8 + 2t, 4 + 2t)

Consider u = v
{ 2 + 3s = 8 + 2t　⇒ 3s - 2t = 6
{ 1 + 3s = 4 + 2t　⇒ 3s - 2t = 3　(inconsistent)
{ 2 - 4s = 1 + 12t

u and v cannot be the same, so the two lines do not intersect.
That is, the two lines are parallel and distinct.


總結：
２Ｄ情況

不同 slopes ⇒ 必定 intersect
　同 slopes ⇒ (parallel and distinct) 或 (same line)

３Ｄ情況

不同 slopes (正確說法是 direction vector) 未必 intersect (請查看 skew line)
　同 slopes (正確說法是 direction vector) ⇒ (parallel and distinct) 或 (same line)


2015-07-30 23:58:06 補充：
For (a), you can also solve the intersection:

p = (0, 7) + s(14, -7) = (14s, 7 - 7s)
q = (0, 3) + t(3, 6) = (3t, 3 + 6t)

Consider p = q
{ 14s = 3t
{ 7 - 7s = 3 + 6t

{ 6t = 28s
{ 7s + 6t = 4

7s + 28s = 4
35s = 4
s = 4/35

t = 8/15

p = q = (8/5, 31/5) is the point of intersection.

2015-07-30 23:58:46 補充：
For your reference:

http://math.stackexchange.com/questions/302598/how-to-prove-that-two-lines-in-3d-are-not-parallel-and-do-not-intersect-also-h

https://en.wikipedia.org/wiki/Skew_lines

https://zh.wikipedia.org/wiki/%E6%AD%AA%E6%96%9C