Part (a): Finding the area of the parallelogram using the dot product method. The area of a parallelogram in 3D space can be found using the magnitudes of the vectors forming the parallelogram and the angle between them. The formula is: Area = |ABX AC First, let's find vectors AB and AC: AB-B-A-(18-(-10), 12-25,-30-7)=(28,-13,-37) AC-C-A (5-(-10), 35-25, 13-7)-(15,10,6) Now, calculate the cross product of AB and AC: k AB AC 28 -13 -37 15 10 6 =((-13)(6)-(-37)(10)) - (28(6)-(-37)(15)) + (28(10) − (−13) (15)) -(-78 (-370))-3(168 +555)+(280+195) =(292) -(-723) + (475) (292,-723, 475) Now, calculate the magnitude of the cross product: x |AB × AC|-√√292² + (−723)² + 475² √85304 +523329 + 225625 833258 = 913.06 So, the area of the parallelogram using the dot product method is approximately 913.06. Part (b): Determining the unit vector perpendicular to the parallelogram's plane. The normal vector to the parallelogram's plane is the cross product we calculated earlier, AB × AC, divided by its magnitude to obtain a unit vector: ABAC (22,-723,475) 913.06 (0.319,-0.792, 0.521) So, the unit vector perpendicular to the parallelogram's plane is approximately (0.319,-0.792,0.521). Part (c): Verifying the results using the cross product method. We already did the cross product method in part (a), which gave us the area of the parallelogram. We'll use the same result here. The magnitude of the cross product we obtained in part (a) was 913.06, which is the area of the parallelogram. So, both methods confirm each other. A(10,25,7) (-251151-18) D (18,171-30) (5135,13)
Part (a): Finding the area of the parallelogram using the dot product method. The area of a parallelogram in 3D space can be found using the magnitudes of the vectors forming the parallelogram and the angle between them. The formula is: Area = |ABX AC First, let's find vectors AB and AC: AB-B-A-(18-(-10), 12-25,-30-7)=(28,-13,-37) AC-C-A (5-(-10), 35-25, 13-7)-(15,10,6) Now, calculate the cross product of AB and AC: k AB AC 28 -13 -37 15 10 6 =((-13)(6)-(-37)(10)) - (28(6)-(-37)(15)) + (28(10) − (−13) (15)) -(-78 (-370))-3(168 +555)+(280+195) =(292) -(-723) + (475) (292,-723, 475) Now, calculate the magnitude of the cross product: x |AB × AC|-√√292² + (−723)² + 475² √85304 +523329 + 225625 833258 = 913.06 So, the area of the parallelogram using the dot product method is approximately 913.06. Part (b): Determining the unit vector perpendicular to the parallelogram's plane. The normal vector to the parallelogram's plane is the cross product we calculated earlier, AB × AC, divided by its magnitude to obtain a unit vector: ABAC (22,-723,475) 913.06 (0.319,-0.792, 0.521) So, the unit vector perpendicular to the parallelogram's plane is approximately (0.319,-0.792,0.521). Part (c): Verifying the results using the cross product method. We already did the cross product method in part (a), which gave us the area of the parallelogram. We'll use the same result here. The magnitude of the cross product we obtained in part (a) was 913.06, which is the area of the parallelogram. So, both methods confirm each other. A(10,25,7) (-251151-18) D (18,171-30) (5135,13)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter8: Applications Of Trigonometry
Section8.4: The Dot Product
Problem 2E
Related questions
Question
Use the dot product method to find the area of the parallelogram attached below. Provide detailed steps in the solution.
Transcribed Image Text:Part (a): Finding the area of the parallelogram using the dot product method.
The area of a parallelogram in 3D space can be found using the magnitudes of the vectors forming the
parallelogram and the angle between them. The formula is:
Area = |ABX AC
First, let's find vectors AB and AC:
AB-B-A-(18-(-10), 12-25,-30-7)=(28,-13,-37)
AC-C-A (5-(-10), 35-25, 13-7)-(15,10,6)
Now, calculate the cross product of AB and AC:
k
AB AC 28 -13 -37
15 10
6
=((-13)(6)-(-37)(10)) - (28(6)-(-37)(15)) + (28(10) − (−13) (15))
-(-78 (-370))-3(168 +555)+(280+195)
=(292) -(-723) + (475)
(292,-723, 475)
Now, calculate the magnitude of the cross product:
x
|AB × AC|-√√292² + (−723)² + 475²
√85304 +523329 + 225625
833258
= 913.06
So, the area of the parallelogram using the dot product method is approximately 913.06.
Part (b): Determining the unit vector perpendicular to the parallelogram's plane.
The normal vector to the parallelogram's plane is the cross product we calculated earlier, AB × AC,
divided by its magnitude to obtain a unit vector:
ABAC
(22,-723,475)
913.06
(0.319,-0.792, 0.521)
So, the unit vector perpendicular to the parallelogram's plane is approximately
(0.319,-0.792,0.521).
Part (c): Verifying the results using the cross product method.
We already did the cross product method in part (a), which gave us the area of the parallelogram. We'll
use the same result here.
The magnitude of the cross product we obtained in part (a) was 913.06, which is the area of the
parallelogram.
So, both methods confirm each other.
Transcribed Image Text:A(10,25,7)
(-251151-18)
D
(18,171-30)
(5135,13)
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