Math 211 Spring 2008: Solutions: HW #1
Instructor: S. Cautis
1. section 1.1, #14
The line is (0, 2, 1) + t(2, 0, −1) = (2t, 2, 1 − t).
2. section 1.1, #22 We need to solve t + 4 = 2s + 3, 4t + 5 = s + 1 and
t − 2 = 2s − 3. From the first two we get −7t − 6 = 1 so t = −1 and s = 0.
This satisfies the third equation so the two lines intersect at (3, 1, −3).
3. section 1.1, #26
We put one of the vertices at the origin so the two sides are given by vectors
v and w. Then their midpoints are v/2 and w/2 so the line segment joining
the midpoints is v/2 − w/2 = (v − w)/2. This is half the third side v − w
so this segment is parallel to the third side and half its length.
4. section 1.2, #16
(a) (2, b, 0) · (−3, 2, 1) = −6 + 2b which is zero when b = 3.
(b) (2, b, 0) · (0, 0, 1) = 0 so any value of b will do.
5. section 1.2, #18
Let (−1 + t0 , −2 + t0 , −1 + t0 ) be the intersection point. We want ((−1 +
t0 , −2+t0, −1+t0)−(3, 1, −2) to be perpendicular to (1, 1, 1) (the direction
of the line). So we want
−4 + t0 − 3 + t0 + 1 + t0 = 0
or t0 = 2. So the intersection point is (1, 0, 1) and the line is (3, 1, −2) +
s(2, 1, −3).
6. section 1.3, #12
(2, −4, 3) × (−4, 8, −6) = (0, 0, 0) so the unit vectors orthogonal are all the
unit vectors in the plan perpendicular to (2, −4, 3).
7. section 1.3, #18
The intersection of two planes is a line. To find the line we just need to
find two points on it. Two such points are (0, 0, 0) and (1, 2, −5) so the
line is (t, 2t, −5t).
8. section 1.3, #30
The plane is parallel to the vectors (3, 2, −2) and (2, 3, −3) so its perpendicular to their cross product which is (0, 5, 5). Hence the plane is
0(x − 3) + 5(y − 2) + 5(z + 1) = 0 which is y + z − 1 = 0.