SOLUTION: I need help. #1 find the parametric equations for line passing through the points P(5,-2,4) Q(7,2,-4). #2 show that if v is a nonzero vector, then v/ llvll is a unit vector

Question 148541: I need help.
#1
find the parametric equations for line passing through the points P(5,-2,4) Q(7,2,-4).
#2
show that if v is a nonzero vector, then v/ llvll is a unit vector

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
#1
find the parametric equations for line passing through the points P(5,-2,4) Q(7,2,-4).
#2
show that if v is a nonzero vector, then v/ llvll is a unit vector
-------------------
#1
Using determinants:
x y z 1
5 -2 4 1
7 2 -4 1
gives:
2y + z = 0
This can be expressed in parametric terms. Use t for the parameter (time, eg)
y = t
z = -2t
Other coefficients are usable, as long as the ratio is maintained.
-----------------
#2 show that if v is a nonzero vector, then v/ llvll is a unit vector
---
I'll edit later and add this.
Vector v is general, connecting 2 points in space. It can be moved (translated) so that one end is at the origin (0,0,0) and the other end is the point (x,y,z).
The length of the vector is sqrt(x^2+y^2+z^2). This is also the absolute value of the vector.
That's proof that dividing the vector by its abs value gives a unit vector, but in math terms:
v/(ABS(v))= x/ABS + y/ABS + z/ABS
= x/sqrt() + y/sqrt() + z/sqrt()
Now calculate the length
= sqrt{ [x/sqrt()]^2 + [y/sqrt()]^2 + [z/sqrt()]^2}
= sqrt{ (x^2+y^2+z^2)/(x^2+y^2+z^2) } (DEN = ABS value from above)
= 1. So it's a unit vector parallel to the original.
This is similar to proving that 2 = 2, imo.

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