Sure, let's analyze each option step by step to determine which one represents the distributive axiom of scalar multiplication over vector addition in vector spaces.

### Option A: $ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v} $
- This option states that if you multiply a scalar $ a $ by the sum of two vectors $ \vec{u} $ and $ \vec{v} $, it is the same as multiplying the scalar $ a $ by each vector individually and then adding the results.
- This is indeed the distributive property of scalar multiplication over vector addition.

### Option B: $ a(\vec{u} \cdot \vec{v}) = (a\vec{u}) \cdot \vec{v} $
- This option involves the dot product (denoted by $ \cdot $) of two vectors $ \vec{u} $ and $ \vec{v} $.
- The distributive property of scalar multiplication over vector addition does not involve the dot product, so this is not the correct axiom.

### Option C: $ a\vec{u} + b\vec{v} = (a + b)(\vec{u} + \vec{v}) $
- This option suggests that the sum of two scalar multiplications is equal to the scalar sum times the vector sum.
- This is not a correct representation of the distributive property of scalar multiplication over vector addition.

### Option D: $ a\vec{u} \cdot \vec{v} = a(\vec{u} \cdot \vec{v}) $
- This option also involves the dot product and states that multiplying a scalar $ a $ by the dot product of two vectors is the same as taking the dot product of the scalar-multiplied vector and the other vector.
- This is not the distributive property of scalar multiplication over vector addition.

### Conclusion
The correct option that represents the distributive axiom of scalar multiplication over vector addition in vector spaces is:

**Option A: $ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v} $**

Question

Sure, let's analyze each option step by step to determine which one represents the distributive axiom of scalar multiplication over vector addition in vector spaces.

### Option A: $ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v} $
- This option states that if you multiply a scalar $ a $ by the sum of two vectors $ \vec{u} $ and $ \vec{v} $, it is the same as multiplying the scalar $ a $ by each vector individually and then adding the results.
- This is indeed the distributive property of scalar multiplication over vector addition.

### Option B: $ a(\vec{u} \cdot \vec{v}) = (a\vec{u}) \cdot \vec{v} $
- This option involves the dot product (denoted by $ \cdot $) of two vectors $ \vec{u} $ and $ \vec{v} $.
- The distributive property of scalar multiplication over vector addition does not involve the dot product, so this is not the correct axiom.

### Option C: $ a\vec{u} + b\vec{v} = (a + b)(\vec{u} + \vec{v}) $
- This option suggests that the sum of two scalar multiplications is equal to the scalar sum times the vector sum.
- This is not a correct representation of the distributive property of scalar multiplication over vector addition.

### Option D: $ a\vec{u} \cdot \vec{v} = a(\vec{u} \cdot \vec{v}) $
- This option also involves the dot product and states that multiplying a scalar $ a $ by the dot product of two vectors is the same as taking the dot product of the scalar-multiplied vector and the other vector.
- This is not the distributive property of scalar multiplication over vector addition.

### Conclusion
The correct option that represents the distributive axiom of scalar multiplication over vector addition in vector spaces is:

**Option A: $ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v} $**

Knowee AI · Accepted Answer

Sure, let's analyze each option step by step to determine which one represents the distributive axiom of scalar multiplication over vector addition in vector spaces.

### Option A: $ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v} $
- This option states that if you multiply a scalar $ a $ by the sum of two vectors $ \vec{u} $ and $ \vec{v} $, it is the same as multiplying the scalar $ a $ by each vector individually and then adding the results.
- This is indeed the distributive property of scalar multiplication over vector addition.

### Option B: $ a(\vec{u} \cdot \vec{v}) = (a\vec{u}) \cdot \vec{v} $
- This option involves the dot product (denoted by $ \cdot $) of two vectors $ \vec{u} $ and $ \vec{v} $.
- The distributive property of scalar multiplication over vector addition does not involve the dot product, so this is not the correct axiom.

### Option C: $ a\vec{u} + b\vec{v} = (a + b)(\vec{u} + \vec{v}) $
- This option suggests that the sum of two scalar multiplications is equal to the scalar sum times the vector sum.
- This is not a correct representation of the distributive property of scalar multiplication over vector addition.

### Option D: $ a\vec{u} \cdot \vec{v} = a(\vec{u} \cdot \vec{v}) $
- This option also involves the dot product and states that multiplying a scalar $ a $ by the dot product of two vectors is the same as taking the dot product of the scalar-multiplied vector and the other vector.
- This is not the distributive property of scalar multiplication over vector addition.

### Conclusion
The correct option that represents the distributive axiom of scalar multiplication over vector addition in vector spaces is:

**Option A: $ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v} $**

Question

Solution

Option A: $a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v}$

Option B: $a(\vec{u} \cdot \vec{v}) = (a\vec{u}) \cdot \vec{v}$

Option C: $a\vec{u} + b\vec{v} = (a + b)(\vec{u} + \vec{v})$

Option D: $a\vec{u} \cdot \vec{v} = a(\vec{u} \cdot \vec{v})$

Conclusion

Similar Questions

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Question

Solution

Option A: a(u⃗+v⃗)=au⃗+av⃗ a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v} a(u+v)=au+av

Option B: a(u⃗⋅v⃗)=(au⃗)⋅v⃗ a(\vec{u} \cdot \vec{v}) = (a\vec{u}) \cdot \vec{v} a(u⋅v)=(au)⋅v

Option C: au⃗+bv⃗=(a+b)(u⃗+v⃗) a\vec{u} + b\vec{v} = (a + b)(\vec{u} + \vec{v}) au+bv=(a+b)(u+v)

Option D: au⃗⋅v⃗=a(u⃗⋅v⃗) a\vec{u} \cdot \vec{v} = a(\vec{u} \cdot \vec{v}) au⋅v=a(u⋅v)