of numbers and e. Next, complete the table by calculating LB when 0 = 120°. Hint: To do this efficiently, use only unit vectors û, v, ŵ. Hint: Use the distributive property of the vector dot-product as shown in Section 2.9.1 and Homework 2.4. Note: Synthesis problems are difficult. Think, talk, draw, sleep, walk, get help,... (if necessary, read Section 3.3). A-20-B ------- 15 LA 8 LB ·10 Lc û Distance between A and B Distance between B and C Distance between N, and B Distance along back wall (see picture) Q's height above No Distance along side wall (see picture) LA: Length of cable joining A and Q LB: Length of cable joining B and Q Le: Length of cable joining C and Q No 7û+5v + 8ŵ 20 m 15 m 8 m Using a dot-product, show the angle 3 between lines B No and BQ is 3 Optional: Verify the calculation of 3 using the law of cosines. 7m 5 m 8 m 16.9 m 8.1 m 14.2 m e- 7 Note: The floor is horizontal, the walls are vertical. Lc=√ 128 Result: LA = √202 Cos (0) LB=√122 + 112 cos(0) Vocabulary: This is inverse kinematics. The position of "end-effector" Q is known and you determine the cable lengths. 68.33°

of numbers and e. Next, complete the table by calculating LB when 0 = 120°. Hint: To do this efficiently, use only unit vectors û, v, ŵ. Hint: Use the distributive property of the vector dot-product as shown in Section 2.9.1 and Homework 2.4. Note: Synthesis problems are difficult. Think, talk, draw, sleep, walk, get help,... (if necessary, read Section 3.3). A-20-B ------- 15 LA 8 LB ·10 Lc û Distance between A and B Distance between B and C Distance between N, and B Distance along back wall (see picture) Q's height above No Distance along side wall (see picture) LA: Length of cable joining A and Q LB: Length of cable joining B and Q Le: Length of cable joining C and Q No 7û+5v + 8ŵ 20 m 15 m 8 m Using a dot-product, show the angle 3 between lines B No and BQ is 3 Optional: Verify the calculation of 3 using the law of cosines. 7m 5 m 8 m 16.9 m 8.1 m 14.2 m e- 7 Note: The floor is horizontal, the walls are vertical. Lc=√ 128 Result: LA = √202 Cos (0) LB=√122 + 112 cos(0) Vocabulary: This is inverse kinematics. The position of "end-effector" Q is known and you determine the cable lengths. 68.33°

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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of numbers and 0. Next, complete the table by calculating LB when 0 = 120°.
Hint: To do this efficiently, use only unit vectors û, v, ŵ.
Hint: Use the distributive property of the vector dot-product as shown in Section 2.9.1 and Homework 2.4.
Note: Synthesis problems are difficult. Think, talk, draw, sleep, walk, get help,... (if necessary, read Section 3.3).
15
LA
LB
5
Lc.
Note: The floor is horizontal, the walls are vertical.
û
8
Distance between A and B
Distance between B and C
Distance between N, and B
Distance along back wall (see picture)
Q's height above No
Distance along side wall (see picture)
LA: Length of cable joining A and Q 16.9 m
LB: Length of cable joining B and Q8.1 m
Lc: Length of cable joining C and Q 14.2 m
No 7ũ + 5v + 8w
20 m
15 m
8 m
• Using a dot-product, show the angle 3 between lines B N. and BQ is ß ≈ 68.33°
Optional: Verify the calculation of 3 using the law of cosines.
7m
5 m
8 m
- 128
Result: LA = √/202-
cos(0) LB=√122 + 112 cos(0) Lc =
Vocabulary: This is inverse kinematics. The position of "end-effector" Q is known and you determine the cable lengths.

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