1.
Let L be the line in R^3 passing through the point (2, 2, 5) and parallel to
the vector (2,-1, 1).Let S be the the line segment consisting of the points on L whose
coordinates are all nonnegative (that is, S is the intersection of L with
the first octant). Find the vector equation of S.
Solution
The vector equation of the line will be
r = 2i +2j + 5k + t*(2i -j +k )
r = (2+2t) i + (2-t) j + (5+t) k
Now we require the part which lies in the first octant i.e all co-ordinates are >=0
2+2t >=0 & 2-t >= 0 & 5+t >=0
This gives us that t>=-1 &t<=2 and t>=-5 respectively.
Taking their intersection you'll get that -1<=t<=5
So the equation will be the same but we only impose a condition on t that -1<=t<=5