Calculus III, Math 233, Homework 1
(1) Consider the triangle with vertices P(1,2,3), Q(−2,1,4), R(0,1,1).
Find the angle at each vertex of the triangle.
Solution: Let the angles at P, Q, R be denoted by α, β, γ , respectively.
Then
cos α=
→
P Q ·
→
P R
→
|P Q|
→
|P R|
=(−3)(−1) + (−1)(−1) + 1(−2)
√11√6=2
√11√6
cos β=
→
QP ·
→
QR
→
|QP |
→
|QR|
=3·2+1·0+(−1)(−3)
√11√13 =9
√11√13
cos γ=
→
RP ·
→
RQ
→
|RP |
→
|RQ|
=1(−2) + 1 ·0+2·3
√6√13 =4
√6√13
Using a calculator we find the angles
α≈75.7◦, β ≈41.2◦, γ ≈63.1◦.
(2) Find the area of the triangle in problem (1).
Solution:
→
P Q ×
→
P R=
i j k
−3−1 1
−1−1−2
=h3,−7,2i.
The area of the triangle is
1
2|
→
P Q ×
→
P R |=1
2√9 + 49 + 4 = 1
2√62 ≈3.94.
(3) Determine whether the lines r1=h2,−1,3i+sh1,2,2iand r2=h−3,−1,9i+
th3,1,−2iintersect. If they do find the intersection point.
Solution: We try to solve the equations
2 + s=−3+3t
−1+2s=−1 + t
3+2s= 9 −2t
We multiply the first equation by −2 and add to the second. This gives
−5 = 5 −5tso t= 2. Then from the first equation s= 1. We check that
s= 1, t= 2 is a solution of all three equations. Therefore, the lines intersect
at (3,1,5).
(4) Find an equation for the plane containing the points P(1,1,−1), Q(3,3,2),
R(3,−1,−2) and determine the distance of this plane to the origin (0,0,0).
