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The solutions to homework 1 of calculus iii, math 233. It includes the calculations for finding the angles at each vertex of a triangle, the area of the triangle, the intersection point of two lines, and the equation and distance of a plane. Essential for students enrolled in calculus iii courses to check their understanding of vector calculations and geometry.
Typology: Assignments
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(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|
cos β =
→ QP ·
→ QR → |QP |
→ |QR|
cos γ =
→ RP ·
→ RQ → |RP |
→ |RQ|
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
The area of the triangle is
1
2
→ P Q ×
→ P R | =
(3) Determine whether the lines r 1 = 〈 2 , − 1 , 3 〉+s〈 1 , 2 , 2 〉 and r 2 = 〈− 3 , − 1 , 9 〉+
t〈 3 , 1 , − 2 〉 intersect. 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 − 2 t
We multiply the first equation by −2 and add to the second. This gives
−5 = 5 − 5 t so 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).
Solution:
n =
→ P Q ×
→ P R=
i j k 2 2 3 2 − 2 − 1
The equation of the plane is
4(x − 1) + 8(y − 1) − 8(z + 1) = 0
which can be simplified to
x + 2y − 2 z − 5 = 0.
The distance to the origin is
| − 5 | √ 1 + 4 + 4