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.