1. ## Vectors

Find the point of intersection of the line through the points (2, 0, 1) and (-1, 3, 4) and the line through the points (-1, 3, 0) and (4, -2, 5).

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So far I did:

Let the line l₁ has points c = (2, 0, 1) and d = (-1, 3, 4).

So the vector equation of a straight line is

r = c + t(d - c)

d - c = (-3, 3, 3)

So l₁: r = (2, 0, 1) + t(-3, 3, 3)

Similarly I found

l₂: r = (-1, 3, 0) + s(5, -5, 5)

But it seems not right. When I equate the x & y components & solving simultaneously I found no solution. Actually what to do in this case?

2. Originally Posted by geton
Find the point of intersection of the line through the points (2, 0, 1) and (-1, 3, 4) and the line through the points (-1, 3, 0) and (4, -2, 5).

-------------

So far I did:

Let the line l₁ has points c = (2, 0, 1) and d = (-1, 3, 4).

So the vector equation of a straight line is

r = c + t(d - c)

d - c = (-3, 3, 3)

So l₁: r = (2, 0, 1) + t(-3, 3, 3)

Similarly I found

l₂: r = (-1, 3, 0) + s(5, -5, 5)

But it seems not right. When I equate the x & y components & solving simultaneously I found no solution. Actually what to do in this case?
That's strange because when I equate the components I get:

i-component: 2 - 3t = -1 + 5s => 5s + 3t = 3.

j-component: 3t = 3 - 5s => 3t + 5s = 3.

These two equations are the same.

k-component: 1 + 3t = 5s => 5s - 3t = 1 => 3t - 5s = -1.

There's no trouble here ..... Just solve 5s + 3t = 3 and 3t - 5s = -1 simultaneously for t (or s, or both if you want to check the answer).

3. Originally Posted by mr fantastic
That's strange because when I equate the components I get:

i-component: 2 - 3t = -1 + 5s => 5s + 3t = 3.

j-component: 3t = 3 - 5s => 3t + 5s = 3.

These two equations are the same.

k-component: 1 + 3t = 5s => 5s - 3t = 1 => 3t - 5s = -1.

There's no trouble here ..... Just solve 5s + 3t = 3 and 3t - 5s = -1 simultaneously for t (or s, or both if you want to check the answer).
Yes there is no problem. I solved it There's silly mistake.

Thank you.