The vector product M → × N → .

Question

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Answer 1

Question

Chapter 11, Problem 1P

To determine

The vector product M→×N→.

Expert Solution & Answer

Answer to Problem 1P

The vector product M→×N→ is 1i^+8j^+22k^.

Explanation of Solution

The vector M→ is 2i^−3j^+k^ and the vector N→ is 4i^+5j^−2k^.

The cross product of the two vectors is written as,

M→×N→=(2i^−3j^+k^)×(4i^+5j^−2k^)=[(2i^×4i^)+(2i^×5j^)+(2i^×(−2k^))+(−3j^×4i^)+(−3j^×5j^)+(−3j^×(−2k^))+(k^×4i^)+(k^×5j^)+(k^×(−2k^))]=[(0)+(10k^)+(4j^)+(12k^)+(0)+(6i^)+(4j^)+(−5i^)+(0)]=1i^+8j^+22k^

Conclusion:

Therefore, the vector product M→×N→ is 1i^+8j^+22k^.

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Answer 2

Question

Chapter 11, Problem 1P

To determine

The vector product M→×N→.

Expert Solution & Answer

Answer to Problem 1P

The vector product M→×N→ is 1i^+8j^+22k^.

Explanation of Solution

The vector M→ is 2i^−3j^+k^ and the vector N→ is 4i^+5j^−2k^.

The cross product of the two vectors is written as,

M→×N→=(2i^−3j^+k^)×(4i^+5j^−2k^)=[(2i^×4i^)+(2i^×5j^)+(2i^×(−2k^))+(−3j^×4i^)+(−3j^×5j^)+(−3j^×(−2k^))+(k^×4i^)+(k^×5j^)+(k^×(−2k^))]=[(0)+(10k^)+(4j^)+(12k^)+(0)+(6i^)+(4j^)+(−5i^)+(0)]=1i^+8j^+22k^

Conclusion:

Therefore, the vector product M→×N→ is 1i^+8j^+22k^.

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Answer 3

Question

Chapter 11, Problem 1P

To determine

The vector product M→×N→.

Expert Solution & Answer

Answer to Problem 1P

The vector product M→×N→ is 1i^+8j^+22k^.

Explanation of Solution

The vector M→ is 2i^−3j^+k^ and the vector N→ is 4i^+5j^−2k^.

The cross product of the two vectors is written as,

M→×N→=(2i^−3j^+k^)×(4i^+5j^−2k^)=[(2i^×4i^)+(2i^×5j^)+(2i^×(−2k^))+(−3j^×4i^)+(−3j^×5j^)+(−3j^×(−2k^))+(k^×4i^)+(k^×5j^)+(k^×(−2k^))]=[(0)+(10k^)+(4j^)+(12k^)+(0)+(6i^)+(4j^)+(−5i^)+(0)]=1i^+8j^+22k^

Conclusion:

Therefore, the vector product M→×N→ is 1i^+8j^+22k^.

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Answer 4

Question

Chapter 11, Problem 1P

To determine

The vector product M→×N→.

Expert Solution & Answer

Answer to Problem 1P

The vector product M→×N→ is 1i^+8j^+22k^.

Explanation of Solution

The vector M→ is 2i^−3j^+k^ and the vector N→ is 4i^+5j^−2k^.

The cross product of the two vectors is written as,

M→×N→=(2i^−3j^+k^)×(4i^+5j^−2k^)=[(2i^×4i^)+(2i^×5j^)+(2i^×(−2k^))+(−3j^×4i^)+(−3j^×5j^)+(−3j^×(−2k^))+(k^×4i^)+(k^×5j^)+(k^×(−2k^))]=[(0)+(10k^)+(4j^)+(12k^)+(0)+(6i^)+(4j^)+(−5i^)+(0)]=1i^+8j^+22k^

Conclusion:

Therefore, the vector product M→×N→ is 1i^+8j^+22k^.

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Answer 5

Chapter 11, Problem 1P

To determine

The vector product M→×N→.

Expert Solution & Answer

Answer to Problem 1P

The vector product M→×N→ is 1i^+8j^+22k^.

Explanation of Solution

The vector M→ is 2i^−3j^+k^ and the vector N→ is 4i^+5j^−2k^.

The cross product of the two vectors is written as,

M→×N→=(2i^−3j^+k^)×(4i^+5j^−2k^)=[(2i^×4i^)+(2i^×5j^)+(2i^×(−2k^))+(−3j^×4i^)+(−3j^×5j^)+(−3j^×(−2k^))+(k^×4i^)+(k^×5j^)+(k^×(−2k^))]=[(0)+(10k^)+(4j^)+(12k^)+(0)+(6i^)+(4j^)+(−5i^)+(0)]=1i^+8j^+22k^

Conclusion:

Therefore, the vector product M→×N→ is 1i^+8j^+22k^.

Answer 6

Chapter 11, Problem 1P

To determine

The vector product M→×N→.

Expert Solution & Answer

Answer to Problem 1P

The vector product M→×N→ is 1i^+8j^+22k^.

Explanation of Solution

The vector M→ is 2i^−3j^+k^ and the vector N→ is 4i^+5j^−2k^.

The cross product of the two vectors is written as,

M→×N→=(2i^−3j^+k^)×(4i^+5j^−2k^)=[(2i^×4i^)+(2i^×5j^)+(2i^×(−2k^))+(−3j^×4i^)+(−3j^×5j^)+(−3j^×(−2k^))+(k^×4i^)+(k^×5j^)+(k^×(−2k^))]=[(0)+(10k^)+(4j^)+(12k^)+(0)+(6i^)+(4j^)+(−5i^)+(0)]=1i^+8j^+22k^

Conclusion:

Therefore, the vector product M→×N→ is 1i^+8j^+22k^.

Answer 7

Chapter 11, Problem 1P

To determine

The vector product M→×N→.

Answer 8

Expert Solution & Answer

Answer to Problem 1P

The vector product M→×N→ is 1i^+8j^+22k^.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The vector M→ is 2i^−3j^+k^ and the vector N→ is 4i^+5j^−2k^.

The cross product of the two vectors is written as,

M→×N→=(2i^−3j^+k^)×(4i^+5j^−2k^)=[(2i^×4i^)+(2i^×5j^)+(2i^×(−2k^))+(−3j^×4i^)+(−3j^×5j^)+(−3j^×(−2k^))+(k^×4i^)+(k^×5j^)+(k^×(−2k^))]=[(0)+(10k^)+(4j^)+(12k^)+(0)+(6i^)+(4j^)+(−5i^)+(0)]=1i^+8j^+22k^