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Calculus III, Math 233: Homework 1 Solutions - Triangle Angles, Area, Intersection, Assignments of Analytical Geometry and Calculus

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

Pre 2010

Uploaded on 09/02/2009

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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)
116=2
116
cos β=
QP ·
QR
|QP |
|QR|
=3·2+1·0+(1)(3)
1113 =9
1113
cos γ=
RP ·
RQ
|RP |
|RQ|
=1(2) + 1 ·0+2·3
613 =4
613
Using a calculator we find the angles
α75.7, β 41.2, γ 63.1.
(2) Find the area of the triangle in problem (1).
Solution:
P Q ×
P R=
i j k
31 1
112
=h3,7,2i.
The area of the triangle is
1
2|
P Q ×
P R |=1
29 + 49 + 4 = 1
262 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).
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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|

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