The value of b + c, 5b + 2 c, b ⋅ c,and b × c .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 1, Problem 1.1P

To determine

The value of b+c, 5b+2c, b⋅c,and b×c.

Expert Solution & Answer

Answer to Problem 1.1P

The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Explanation of Solution

The sum of the given vectors is,

b+c=(x^+y^)+(x^+z^)=2x^+y^+z (I)

The sum of the given two vectors is 2x^+y^+z.

The sum of the given two vectors to find 5b+2c is,

5b+c=5(x^+y^)+2(x^+z^)=x^(5+2)+5y^+2z^=7x^+5y^+2z^ (II)

The value of b⋅c is,

b⋅c=(x^+y^)⋅(x^+z^)=x^⋅x^+x^⋅z^+y^⋅z^=1+0+0+0=1 (III)

The vector or cross product of the vectors is,

b×c=det(x^yz^110101)

Solve the above dot product,

b×c=x^(1−0)−y^(1−0)+z^(0−1)=x^−y^−z^ (IV)

Conclusion:

Therefore, The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

suggest a reason ultrasound cleaning is better than cleaning by hand?

Checkpoint 4 The figure shows four orientations of an electric di- pole in an external electric field. Rank the orienta- tions according to (a) the magnitude of the torque on the dipole and (b) the potential energy of the di- pole, greatest first. (1) (2) E (4)

What is integrated science. What is fractional distillation What is simple distillation

Answer 2

Question

Chapter 1, Problem 1.1P

To determine

The value of b+c, 5b+2c, b⋅c,and b×c.

Expert Solution & Answer

Answer to Problem 1.1P

The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Explanation of Solution

The sum of the given vectors is,

b+c=(x^+y^)+(x^+z^)=2x^+y^+z (I)

The sum of the given two vectors is 2x^+y^+z.

The sum of the given two vectors to find 5b+2c is,

5b+c=5(x^+y^)+2(x^+z^)=x^(5+2)+5y^+2z^=7x^+5y^+2z^ (II)

The value of b⋅c is,

b⋅c=(x^+y^)⋅(x^+z^)=x^⋅x^+x^⋅z^+y^⋅z^=1+0+0+0=1 (III)

The vector or cross product of the vectors is,

b×c=det(x^yz^110101)

Solve the above dot product,

b×c=x^(1−0)−y^(1−0)+z^(0−1)=x^−y^−z^ (IV)

Conclusion:

Therefore, The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 1, Problem 1.1P

To determine

The value of b+c, 5b+2c, b⋅c,and b×c.

Expert Solution & Answer

Answer to Problem 1.1P

The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Explanation of Solution

The sum of the given vectors is,

b+c=(x^+y^)+(x^+z^)=2x^+y^+z (I)

The sum of the given two vectors is 2x^+y^+z.

The sum of the given two vectors to find 5b+2c is,

5b+c=5(x^+y^)+2(x^+z^)=x^(5+2)+5y^+2z^=7x^+5y^+2z^ (II)

The value of b⋅c is,

b⋅c=(x^+y^)⋅(x^+z^)=x^⋅x^+x^⋅z^+y^⋅z^=1+0+0+0=1 (III)

The vector or cross product of the vectors is,

b×c=det(x^yz^110101)

Solve the above dot product,

b×c=x^(1−0)−y^(1−0)+z^(0−1)=x^−y^−z^ (IV)

Conclusion:

Therefore, The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 1, Problem 1.1P

To determine

The value of b+c, 5b+2c, b⋅c,and b×c.

Expert Solution & Answer

Answer to Problem 1.1P

The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Explanation of Solution

The sum of the given vectors is,

b+c=(x^+y^)+(x^+z^)=2x^+y^+z (I)

The sum of the given two vectors is 2x^+y^+z.

The sum of the given two vectors to find 5b+2c is,

5b+c=5(x^+y^)+2(x^+z^)=x^(5+2)+5y^+2z^=7x^+5y^+2z^ (II)

The value of b⋅c is,

b⋅c=(x^+y^)⋅(x^+z^)=x^⋅x^+x^⋅z^+y^⋅z^=1+0+0+0=1 (III)

The vector or cross product of the vectors is,

b×c=det(x^yz^110101)

Solve the above dot product,

b×c=x^(1−0)−y^(1−0)+z^(0−1)=x^−y^−z^ (IV)

Conclusion:

Therefore, The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 1, Problem 1.1P

To determine

The value of b+c, 5b+2c, b⋅c,and b×c.

Expert Solution & Answer

Answer to Problem 1.1P

The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Explanation of Solution

The sum of the given vectors is,

b+c=(x^+y^)+(x^+z^)=2x^+y^+z (I)

The sum of the given two vectors is 2x^+y^+z.

The sum of the given two vectors to find 5b+2c is,

5b+c=5(x^+y^)+2(x^+z^)=x^(5+2)+5y^+2z^=7x^+5y^+2z^ (II)

The value of b⋅c is,

b⋅c=(x^+y^)⋅(x^+z^)=x^⋅x^+x^⋅z^+y^⋅z^=1+0+0+0=1 (III)

The vector or cross product of the vectors is,

b×c=det(x^yz^110101)

Solve the above dot product,

b×c=x^(1−0)−y^(1−0)+z^(0−1)=x^−y^−z^ (IV)

Conclusion:

Therefore, The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Answer 6

Chapter 1, Problem 1.1P

To determine

The value of b+c, 5b+2c, b⋅c,and b×c.

Expert Solution & Answer

Answer to Problem 1.1P

The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Explanation of Solution

The sum of the given vectors is,

b+c=(x^+y^)+(x^+z^)=2x^+y^+z (I)

The sum of the given two vectors is 2x^+y^+z.

The sum of the given two vectors to find 5b+2c is,

5b+c=5(x^+y^)+2(x^+z^)=x^(5+2)+5y^+2z^=7x^+5y^+2z^ (II)

The value of b⋅c is,

b⋅c=(x^+y^)⋅(x^+z^)=x^⋅x^+x^⋅z^+y^⋅z^=1+0+0+0=1 (III)

The vector or cross product of the vectors is,

b×c=det(x^yz^110101)

Solve the above dot product,

b×c=x^(1−0)−y^(1−0)+z^(0−1)=x^−y^−z^ (IV)

Conclusion:

Therefore, The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Answer 7

Chapter 1, Problem 1.1P

To determine

The value of b+c, 5b+2c, b⋅c,and b×c.

Answer 8

Expert Solution & Answer

Answer to Problem 1.1P

The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The sum of the given vectors is,

b+c=(x^+y^)+(x^+z^)=2x^+y^+z (I)

The sum of the given two vectors is 2x^+y^+z.

The sum of the given two vectors to find 5b+2c is,

5b+c=5(x^+y^)+2(x^+z^)=x^(5+2)+5y^+2z^=7x^+5y^+2z^ (II)

The value of b⋅c is,

b⋅c=(x^+y^)⋅(x^+z^)=x^⋅x^+x^⋅z^+y^⋅z^=1+0+0+0=1 (III)

The vector or cross product of the vectors is,

b×c=det(x^yz^110101)

Solve the above dot product,

b×c=x^(1−0)−y^(1−0)+z^(0−1)=x^−y^−z^ (IV)

Conclusion:

Therefore, The value of b+c is 2x^+y^+z^_, 5b+2c is 7x^+5y^+2z^_ , b⋅c is 1 and b×c is x^−y^−z^_

Answer 12

Ch. 1 - Prob. 1.1P Ch. 1 - Prob. 1.2P Ch. 1 - Prob. 1.3P Ch. 1 - Prob. 1.4P Ch. 1 - Prob. 1.5P Ch. 1 - Prob. 1.6P Ch. 1 - Prob. 1.7P Ch. 1 - Prob. 1.8P Ch. 1 - Prob. 1.9P Ch. 1 - Prob. 1.10P

The value of b + c, 5b + 2 c, b ⋅ c,and b × c .

Videos

The value of b+c, 5b+2c, b⋅c,and b×c.

Answer to Problem 1.1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 1 Solutions