Can someone explain how this is done:

Let L1 be the line through the points P(4, -1,1) and Q(1,0,-3) and

L2: (x-3)/4=(y+2)/-2=(z+5)/3

1. Find the vector equations of the lines L1 and L2.

2. Find the point of intersection between the lines L1 and L2.

Can you please make the explanation very detailed as I have the answer but don't know how it was derived. Thanks

The vector equation of a line is generally of the form

r = a + tb,

where a is a vector pointing from the origin to some point on the line, b is any vector which points from the origin in the same direction as the line, t is some parametric variable which can be varied infinitely to give you all the points on the line, and r is the resultant vector which actually aims at each point on the line for the various values of t.

For part 1, we can easily find a good candidate for a by just arbitrarily picking the point P as one that we know lies on the line. Thus,

a = 4i -j +k,

where i, j, and k are the unit vectors in the x, y, and z directions, respectively.

To find a vector pointing in the direction of the line, we need simply find the vector between the points P and Q:

b = P - Q = 4i -j +k - (i - 3k) = 3i -j +4k

Thus one vector equation of the line L1 is:

r = 4i -j +k + t(3i -j +4k)

Of course, there are plenty of other equations describing the same line (for example, if we had chosen to point the vector a at the point Q instead of P)

For part 2, the easiest first step is to write the vector equation for that line as well. We can do that by arbitrarily setting some variable s equal to the rest of the elements of the equation. (We'd usually use t, but I don't want to confuse it with the t from above, so lets use s instead):

s = (x-3)/4
s = (y+2)/-2
s = (z+5)/3

No we can rearrange to solve for x, y, and z:

x = 3 + 4s
y = -2 - 2s
z = -5 + 3s

Now lets multiply each equation by unit vectors so we can form a vector equation:

xi = 3i + s(4i)
yj = -2j + s(-2j)
zk = -5k + s(3k)

Now our vector r can be written as

r = xi + yj + zk = 3i - 2j - 5k + s(4i -2j +3k),

and we have our line in vector equation form.

Now, if the two lines intersect each other, we should be able to find some vector r which satisfies both vector equations simultaneously and find the values of the parametric variables s and t which make it happen.

But it seems like there is only one equation with two unknowns - what do we do? The answer is that since i, j, and k are all othogonal to each other, the i, j, and k components of each vector equation must be independently equal to each other. This turns our one equation into three:

4i -j +k + t(3i -j +4k) = 3i - 2j - 5k + s(4i -2j +3k)

So

4i + t(3i) = 3i + s(4i)
-j + t(-j) = -2j + s(-2j)
k + t(4k) = -5k + s(3k)

or

4 + 3t = 3 + 4s
-1 - t = -2 - 2s
1 + 4t = -5+ 3s

Since there are only two variables, we only actually need two equations to solve for them. We can solve the second equation for t:

t = 2s + 1

Plugging back into the first yields:

4 + 3(2s + 1) = 4s + 3
s = -2

which means

t = 2x + 1 = -3

If the two lines truly intersect in three dimensions, the third equation from above must be true as well for the values of s and t we calculated. Let's try:

1 + 4t = -5+ 3s
1 + 4(-3) = -5 + 3(-2)
-11 = -11

It works! The lines do indeed intersect and they do so at a point described by either of our vector equations, as long as we plug in the appropriate value of s or t. Just to check our work, let's make sure that both equations give us the same answer:

r = 4i -j +k + t(3i -j +4k) = 3i - 2j - 5k + s(4i -2j +3k)
r = 4i -j +k - 3(3i -j +4k) = 3i - 2j - 5k - 2(4i -2j +3k)
r = 4i -j +k - 9i + 3j - 12k = 3i - 2j - 5k - 8i + 4j - 6k
r = -5i + 2j - 11k = -5i + 2j - 11k

Yup! Both equations yield the same point of intersection:

P(-5, 2, -11)

Hopefully that agrees with the answer you have.

Josh