consider a rhombus ABCD

a) find the resultant of vector AB + vector AD and vector AB - vector AD? (cosine rule)
b) What will be the value of the dot product of vector AB + vector AD and vector AB - vector AD always be? (always zero)
c) Is this value of the dot product of vector AB - vector AD dot product vector AB + vector AD (commutative property)

Can someone please check if the answers in the brackets are correct!!!

a) To find the resultant of vector AB + vector AD and vector AB - vector AD, we can use the cosine rule. The cosine rule states that for any triangle with sides a, b, and c and corresponding angles A, B, and C, the relationship between the sides and angles is given by the following equation:

c^2 = a^2 + b^2 - 2abcos(C)

In our case, let's consider vector AB + vector AD as vector R1 and vector AB - vector AD as vector R2. The magnitude of vector R1 can be found using the formula:

|R1|^2 = |AB|^2 + |AD|^2 + 2|AB||AD|cosθ

where |AB| and |AD| are the magnitudes of vectors AB and AD respectively, and θ is the angle between them.

Similarly, the magnitude of vector R2 can be found using the formula:

|R2|^2 = |AB|^2 + |AD|^2 - 2|AB||AD|cosθ

Since the two vectors have the same magnitudes |AB| and |AD|, and the angle between them is also the same, we can simplify the equations as:

|R1|^2 = 2|AB|^2 + 2|AD|^2 + 2|AB||AD|cosθ

|R2|^2 = 2|AB|^2 + 2|AD|^2 - 2|AB||AD|cosθ

Therefore, the resultant of vector AB + vector AD is given by |R1| = √(2|AB|^2 + 2|AD|^2 + 2|AB||AD|cosθ), and the resultant of vector AB - vector AD is given by |R2| = √(2|AB|^2 + 2|AD|^2 - 2|AB||AD|cosθ).

b) The dot product of vector AB + vector AD and vector AB - vector AD can be found using the formula:

AB · AB + AB · (-AD) + AD · AB + AD · (-AD)

The dot product of a vector with itself is the square of its magnitude, so we can simplify the equation as:

|AB|^2 + AB · (-AD) + AD · AB + |AD|^2

The dot product of vector AB with vector -AD is equal to the dot product of vector -AB with vector AD since scalar multiplication does not change the dot product.

Therefore, the expression becomes:

|AB|^2 - AB · AD + AD · AB + |AD|^2

Using the commutative property of addition, we can rewrite the expression as:

(|AB|^2 + AD · AB) + (|AD|^2 - AB · AD)

This can be further simplified as:

(|AB|^2 + AD · AB) - (AB · AD - |AD|^2)

AB · AD is equal to AD · AB since the dot product is commutative. Therefore, the expression becomes:

(|AB|^2 + AD · AB) - (AD · AB - |AD|^2)

AD · AB cancels out, resulting in:

(|AB|^2 + AD · AB) - (AD · AB - |AD|^2) = |AB|^2 + AD · AB - AD · AB + |AD|^2

The terms AD · AB and -AD · AB cancel each other out, giving:

(|AB|^2 + |AD|^2) + (|AD|^2 + |AB|^2)

This simplifies to:

2|AB|^2 + 2|AD|^2

Therefore, the dot product of vector AB + vector AD and vector AB - vector AD is always equal to 2|AB|^2 + 2|AD|^2.

c) The dot product of vector AB - vector AD and vector AB + vector AD is given by:

(AB - AD) · (AB + AD)

Using the distributive property, we can expand this expression as:

AB · AB + AB · AD - AD · AB - AD · AD

The terms AB · AD and -AD · AB cancel each other out, resulting in:

AB · AB - AD · AD

This is equivalent to the value -AB · AB + AD · AD since the dot product is commutative.

Therefore, the dot product of vector AB - vector AD and vector AB + vector AD is equal to -AB · AB + AD · AD.

In conclusion, the value of the dot product of vector AB - vector AD and vector AB + vector AD is not equal to the value of the dot product of vector AB + vector AD and vector AB - vector AD. Therefore, it does not satisfy the commutative property of dot product.