The value of | A → × B → | and compared with | A → | | B → | . The angle between the vectors A → and B → .

Question

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Answer 1

Question

Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(a)

Expert Solution

Explanation of Solution

Given data:

A→=4i→, B→=6i→+6j→

Formula Used:

Consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i→, B→=6i→+6j→A→×B→=(4i→)×(6i→+6j→)=(4i→×6i→)+(4i→×6j→)=0+24k^=24k^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6j^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|24k^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6j^)=24(i^⋅i^)+24(i^⋅j^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45o .

(b)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(b)

Expert Solution

Explanation of Solution

Given data:

A→=4i^, B→=6i^+6k^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i^, B→=6i^+6k^A→×B→=(4i^)×(6i^+6k^)=(4i^×6i→)+(4i^×6k^)=0+24(−j^)=−24j^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6k^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|−24j^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6k^)=24(i^⋅i^)+24(i^⋅k^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45° .

(c)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(c)

Expert Solution

Explanation of Solution

Given data:

A→=2i^+3j^, B→=3i^+2j^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=2i^+3j^, B→=3i^+2j^A→×B→=6(i^×i^)+4(i^×j^)+9(j^×i^)+6(j^×j^)=6(0)+4k^−9k^+6(0)=−5k^

The magnitude of vector A→ and B→ are

|A→|=(2i^)2+(3j^)2+02=13|B→|=(3i^)2+(2j^)2+02=13|A→||B→|=1313=13|A→×B→|(|A→||B→|)=|−5k^|13=0.385

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=(2i^+3j^)⋅(3i^+2j^)=6(i^⋅i^)+6(i^⋅k^)=6(1)+6(1)=12

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(1213)=23o

Conclusion:

The angle between the vectors A→ and B→ is 23° .

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Answer 2

Question

Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(a)

Expert Solution

Explanation of Solution

Given data:

A→=4i→, B→=6i→+6j→

Formula Used:

Consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i→, B→=6i→+6j→A→×B→=(4i→)×(6i→+6j→)=(4i→×6i→)+(4i→×6j→)=0+24k^=24k^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6j^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|24k^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6j^)=24(i^⋅i^)+24(i^⋅j^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45o .

(b)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(b)

Expert Solution

Explanation of Solution

Given data:

A→=4i^, B→=6i^+6k^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i^, B→=6i^+6k^A→×B→=(4i^)×(6i^+6k^)=(4i^×6i→)+(4i^×6k^)=0+24(−j^)=−24j^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6k^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|−24j^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6k^)=24(i^⋅i^)+24(i^⋅k^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45° .

(c)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(c)

Expert Solution

Explanation of Solution

Given data:

A→=2i^+3j^, B→=3i^+2j^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=2i^+3j^, B→=3i^+2j^A→×B→=6(i^×i^)+4(i^×j^)+9(j^×i^)+6(j^×j^)=6(0)+4k^−9k^+6(0)=−5k^

The magnitude of vector A→ and B→ are

|A→|=(2i^)2+(3j^)2+02=13|B→|=(3i^)2+(2j^)2+02=13|A→||B→|=1313=13|A→×B→|(|A→||B→|)=|−5k^|13=0.385

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=(2i^+3j^)⋅(3i^+2j^)=6(i^⋅i^)+6(i^⋅k^)=6(1)+6(1)=12

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(1213)=23o

Conclusion:

The angle between the vectors A→ and B→ is 23° .

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Answer 3

Question

Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(a)

Expert Solution

Explanation of Solution

Given data:

A→=4i→, B→=6i→+6j→

Formula Used:

Consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i→, B→=6i→+6j→A→×B→=(4i→)×(6i→+6j→)=(4i→×6i→)+(4i→×6j→)=0+24k^=24k^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6j^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|24k^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6j^)=24(i^⋅i^)+24(i^⋅j^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45o .

(b)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(b)

Expert Solution

Explanation of Solution

Given data:

A→=4i^, B→=6i^+6k^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i^, B→=6i^+6k^A→×B→=(4i^)×(6i^+6k^)=(4i^×6i→)+(4i^×6k^)=0+24(−j^)=−24j^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6k^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|−24j^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6k^)=24(i^⋅i^)+24(i^⋅k^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45° .

(c)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(c)

Expert Solution

Explanation of Solution

Given data:

A→=2i^+3j^, B→=3i^+2j^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=2i^+3j^, B→=3i^+2j^A→×B→=6(i^×i^)+4(i^×j^)+9(j^×i^)+6(j^×j^)=6(0)+4k^−9k^+6(0)=−5k^

The magnitude of vector A→ and B→ are

|A→|=(2i^)2+(3j^)2+02=13|B→|=(3i^)2+(2j^)2+02=13|A→||B→|=1313=13|A→×B→|(|A→||B→|)=|−5k^|13=0.385

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=(2i^+3j^)⋅(3i^+2j^)=6(i^⋅i^)+6(i^⋅k^)=6(1)+6(1)=12

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(1213)=23o

Conclusion:

The angle between the vectors A→ and B→ is 23° .

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Answer 4

Question

Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(a)

Expert Solution

Explanation of Solution

Given data:

A→=4i→, B→=6i→+6j→

Formula Used:

Consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i→, B→=6i→+6j→A→×B→=(4i→)×(6i→+6j→)=(4i→×6i→)+(4i→×6j→)=0+24k^=24k^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6j^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|24k^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6j^)=24(i^⋅i^)+24(i^⋅j^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45o .

(b)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(b)

Expert Solution

Explanation of Solution

Given data:

A→=4i^, B→=6i^+6k^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i^, B→=6i^+6k^A→×B→=(4i^)×(6i^+6k^)=(4i^×6i→)+(4i^×6k^)=0+24(−j^)=−24j^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6k^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|−24j^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6k^)=24(i^⋅i^)+24(i^⋅k^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45° .

(c)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(c)

Expert Solution

Explanation of Solution

Given data:

A→=2i^+3j^, B→=3i^+2j^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=2i^+3j^, B→=3i^+2j^A→×B→=6(i^×i^)+4(i^×j^)+9(j^×i^)+6(j^×j^)=6(0)+4k^−9k^+6(0)=−5k^

The magnitude of vector A→ and B→ are

|A→|=(2i^)2+(3j^)2+02=13|B→|=(3i^)2+(2j^)2+02=13|A→||B→|=1313=13|A→×B→|(|A→||B→|)=|−5k^|13=0.385

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=(2i^+3j^)⋅(3i^+2j^)=6(i^⋅i^)+6(i^⋅k^)=6(1)+6(1)=12

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(1213)=23o

Conclusion:

The angle between the vectors A→ and B→ is 23° .

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Answer 5

Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(a)

Expert Solution

Explanation of Solution

Given data:

A→=4i→, B→=6i→+6j→

Formula Used:

Consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i→, B→=6i→+6j→A→×B→=(4i→)×(6i→+6j→)=(4i→×6i→)+(4i→×6j→)=0+24k^=24k^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6j^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|24k^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6j^)=24(i^⋅i^)+24(i^⋅j^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45o .

(b)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(b)

Expert Solution

Explanation of Solution

Given data:

A→=4i^, B→=6i^+6k^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i^, B→=6i^+6k^A→×B→=(4i^)×(6i^+6k^)=(4i^×6i→)+(4i^×6k^)=0+24(−j^)=−24j^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6k^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|−24j^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6k^)=24(i^⋅i^)+24(i^⋅k^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45° .

(c)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(c)

Expert Solution

Explanation of Solution

Given data:

A→=2i^+3j^, B→=3i^+2j^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=2i^+3j^, B→=3i^+2j^A→×B→=6(i^×i^)+4(i^×j^)+9(j^×i^)+6(j^×j^)=6(0)+4k^−9k^+6(0)=−5k^

The magnitude of vector A→ and B→ are

|A→|=(2i^)2+(3j^)2+02=13|B→|=(3i^)2+(2j^)2+02=13|A→||B→|=1313=13|A→×B→|(|A→||B→|)=|−5k^|13=0.385

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=(2i^+3j^)⋅(3i^+2j^)=6(i^⋅i^)+6(i^⋅k^)=6(1)+6(1)=12

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(1213)=23o

Conclusion:

The angle between the vectors A→ and B→ is 23° .

Answer 6

Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(a)

Expert Solution

Explanation of Solution

Given data:

A→=4i→, B→=6i→+6j→

Formula Used:

Consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i→, B→=6i→+6j→A→×B→=(4i→)×(6i→+6j→)=(4i→×6i→)+(4i→×6j→)=0+24k^=24k^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6j^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|24k^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6j^)=24(i^⋅i^)+24(i^⋅j^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45o .

Answer 7

Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

Answer 8

(a)

Expert Solution

Explanation of Solution

Given data:

A→=4i→, B→=6i→+6j→

Formula Used:

Consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i→, B→=6i→+6j→A→×B→=(4i→)×(6i→+6j→)=(4i→×6i→)+(4i→×6j→)=0+24k^=24k^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6j^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|24k^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6j^)=24(i^⋅i^)+24(i^⋅j^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45o .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(b)

Expert Solution

Explanation of Solution

Given data:

A→=4i^, B→=6i^+6k^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i^, B→=6i^+6k^A→×B→=(4i^)×(6i^+6k^)=(4i^×6i→)+(4i^×6k^)=0+24(−j^)=−24j^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6k^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|−24j^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6k^)=24(i^⋅i^)+24(i^⋅k^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45° .

Answer 13

(b)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

Answer 14

(b)

Expert Solution

Explanation of Solution

Given data:

A→=4i^, B→=6i^+6k^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=4i^, B→=6i^+6k^A→×B→=(4i^)×(6i^+6k^)=(4i^×6i→)+(4i^×6k^)=0+24(−j^)=−24j^

The magnitude of vector A→ and B→ are

|A→|=(4i^)2+02+02=4|B→|=(6i^)2+(6k^)2+02=62|A→||B→|=4(62)=242|A→×B→|(|A→||B→|)=|−24j^|242=0.707

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=4i^⋅(6i^+6k^)=24(i^⋅i^)+24(i^⋅k^)=24

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(24242)=cos−1(12)=45o

Conclusion:

The angle between the vectors A→ and B→ is 45° .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

(c)

Expert Solution

Explanation of Solution

Given data:

A→=2i^+3j^, B→=3i^+2j^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=2i^+3j^, B→=3i^+2j^A→×B→=6(i^×i^)+4(i^×j^)+9(j^×i^)+6(j^×j^)=6(0)+4k^−9k^+6(0)=−5k^

The magnitude of vector A→ and B→ are

|A→|=(2i^)2+(3j^)2+02=13|B→|=(3i^)2+(2j^)2+02=13|A→||B→|=1313=13|A→×B→|(|A→||B→|)=|−5k^|13=0.385

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=(2i^+3j^)⋅(3i^+2j^)=6(i^⋅i^)+6(i^⋅k^)=6(1)+6(1)=12

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(1213)=23o

Conclusion:

The angle between the vectors A→ and B→ is 23° .

Answer 19

(c)

To determine

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

Answer 20

(c)

Expert Solution

Explanation of Solution

Given data:

A→=2i^+3j^, B→=3i^+2j^

Formula used:

We consider vectors A→ and B→ are perpendicular the vector product of two vectors, that is,

|A→×B→|=|A→||B→|sinϕ=|A→||B→|sin90o|A→×B→||A→||B→|=1 .....(1)

Using scalar product of two vectors we can find angle between the vectors.

|A→⋅B→|=|A→||B→|cosϕϕ=cos−1(A→⋅B→|A→||B→|)

Calculation:

The vector product of vector A→ and B→ is

A→=2i^+3j^, B→=3i^+2j^A→×B→=6(i^×i^)+4(i^×j^)+9(j^×i^)+6(j^×j^)=6(0)+4k^−9k^+6(0)=−5k^

The magnitude of vector A→ and B→ are

|A→|=(2i^)2+(3j^)2+02=13|B→|=(3i^)2+(2j^)2+02=13|A→||B→|=1313=13|A→×B→|(|A→||B→|)=|−5k^|13=0.385

Comparing with equation (1) , the vector A→ and B→ are not perpendicular.

The scalar product of vector A→ and B→ is

A→⋅B→=(2i^+3j^)⋅(3i^+2j^)=6(i^⋅i^)+6(i^⋅k^)=6(1)+6(1)=12

The angle between the vectors A→ and B→ is

ϕ=cos−1(A→⋅B→|A→||B→|)=cos−1(1213)=23o

Conclusion:

The angle between the vectors A→ and B→ is 23° .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Ch. 10 - Prob. 1P Ch. 10 - Prob. 2P Ch. 10 - Prob. 3P Ch. 10 - Prob. 4P Ch. 10 - Prob. 5P Ch. 10 - Prob. 6P Ch. 10 - Prob. 7P Ch. 10 - Prob. 8P Ch. 10 - Prob. 9P Ch. 10 - Prob. 10P

The value of | A → × B → | and compared with | A → | | B → | . The angle between the vectors A → and B → .

Videos

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

Explanation of Solution

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

Explanation of Solution

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions

The value of | A → × B → | and compared with | A → | | B → | . The angle between the vectors A → and B → .

Videos

To Calculate:The value of |A→×B→| and compared with |A→||B→| . The angle between the vectors A→ and B→ .

Explanation of Solution

To Calculate: The value of |A→×B→| and compared with |A→||B→| . The angle between the vectors A→ and B→ .

Explanation of Solution

To Calculate: The value of |A→×B→| and compared with |A→||B→| . The angle between the vectors A→ and B→ .

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions

To Calculate:The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .

To Calculate: The value of |A→×B→| and compared with |A→||B→| .

The angle between the vectors A→ and B→ .