Mathematics II. Homework 1
All answers have to be reasonably explained and justified.
Contents: Days 1, 2 and 3.
1. In R3, find the equation of the line that goes through points (1,2,3) and (3,−2,−1). Find a
point on the line different from the given ones.
SOLUTION:
(x, y, z) = (1,2,3) + t(2,−4,−4) or x−1
2=y−2
−4=z−3
−4or 2x+y−4 = 0
y−z+ 1 = 0 A point
on the line could be the one we get for t= 1/2 : (2,0,1).
2. Find the equation of the plane that goes through points (1,2,3), (4,0,5), (−1,1,0). Find a
point on the plane different from the given ones.
SOLUTION:
(x, y, z) = (1,2,3) + t·(3,−2,2) + s·(−2,−1,−3) (vector equation) or 8x+ 5y−7z+ 3 = 0
( general equation). A different point could be the one we get for x= 2, y =−1. Replacing
in the equation provides z= 2: (2,−1,2).
3. Find the equation of a plane parallel to 8x+ 5y−7z+ 3 = 0 and containing the point
(−1,−1,−1).
Find also the parallel plane through the origin of coordinates.
SOLUTION:
Any parallel plane has 8x+ 5y−7z+D= 0 as general equation. The one demanded is
8x+ 5y−7z+ 6 = 0. The one through the origin clearly has D= 0: 8x+ 5y−7z= 0.
4. (a) Prove that point (0,−1,1) lies on the plane −x+ 2y+ 3z= 1.
(b) Find the equation of the line through (−2,1,−1) which is perpendicular to the plane
found in the previous item.
SOLUTION:
(a) One checks that the point satisfies the equation.
(b) (x, y, z) = (−2,1,−1) + t·(−1,2,3).
5. Is there any value of ksuch that the line kx +y−z= 3
2x−y+ 2z= 5 is perpendicular to the plane
x=z?
SOLUTION:
k=−1 is the only solution.
6. Consider f(x, y) = x2−2xy + 3y3. Compute:
(a) f(2,−1).
(b) f(−1/x, 1/y).
(c) f(x, y +k)−f(x, y)
k.
