The question is: Find a vector equation of the line that is parallel to the line of intersection o f 3x - y + 2z = 6 and 2x + y - z = 4 and passes through P(5, -1, -3).

Solution: 3x - y + 2z = 6 Let this be #1

2x + y - z = 4 Let this be #2
Now we subtract #2 from #1 to elimate 'y' and we get 5x + z = 10

Then we say x = t (HOW can we say this?)

The we say that because x = t, we can say that using 5x + z = 10 that this becomes
5t + z = 10 and this becomes z = -5t + 10

Then we substitute x = t and z = -5 + 10 into 2x + y + 5t - 10 = 4
which gives us 2t + y + 5t - 10 = 4
y = -7t + 14

Then we say the vector equation of the line of intersection is
l(1) = (0, 14, 10) + t(1, -7.-5) BUT HOW DO YOU FIGURE THIS TO BE SO??

and the vector equation of the line through P(5,-1,-3) is l(2)- (5, -1, -3) + s(1, -7,-5)
HOW DO YOU GET THIS??

The line of intersection of the two planes is parallel to the cross product of their normal vectors.



Cross product:



Passes through (5,-1,-3):

Quote:

---

Originally Posted by ankara55t

Then we say x = t (HOW can we say this?)

---

Both x and t are variables, you can say that, or even x = a, or x = b etc. However, you cannot say x = y, or x = z since these variables are already used and already have a particular significance.

Quote:

---

Then we say the vector equation of the line of intersection is
l(1) = (0, 14, 10) + t(1, -7.-5) BUT HOW DO YOU FIGURE THIS TO BE SO??

---

That is :



And this means:



And when you expand, you get:
x= t
y = 14 - 7t or y = -7t + 14
z = 10 - 5t or z = -5t + 10

which you obtained before.

Quote:

---

and the vector equation of the line through P(5,-1,-3) is l(2)- (5, -1, -3) + s(1, -7,-5)
HOW DO YOU GET THIS??

---

You take only the 'gradient' part of the equation of the parallel line, that is you keep: and you pout the point (5,-1,-3) as .

Then, you put it as

where s is a variable.

Hope it helped! :)