Understanding Vector Magnitudes and Their Relationship

The question asks about the relationship between two non-zero vectors, let's call them \(\vec{A}\) and \(\vec{B}\), when the magnitude of their sum is equal to the magnitude of their difference. We are given the condition:

\(|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|\)

To analyze this condition, we can square both sides of the equation. Squaring the magnitude of a vector is equivalent to taking the dot product of the vector with itself.

\(|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2\)

This can be written in terms of dot products:

\((\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = (\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B})\)

Expanding the Dot Products

Let's expand both sides using the distributive property of the dot product (\(\vec{X} \cdot (\vec{Y} + \vec{Z}) = \vec{X} \cdot \vec{Y} + \vec{X} \cdot \vec{Z}\)) and remembering that the dot product is commutative (\(\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}\)):

Left side: \((\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = \vec{A} \cdot \vec{A} + \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{A} + \vec{B} \cdot \vec{B}\)

Using \(\vec{A} \cdot \vec{A} = |\vec{A}|^2\), \(\vec{B} \cdot \vec{B} = |\vec{B}|^2\), and \(\vec{B} \cdot \vec{A} = \vec{A} \cdot \vec{B}\), the left side becomes:

\(|\vec{A}|^2 + 2(\vec{A} \cdot \vec{B}) + |\vec{B}|^2\)

Right side: \((\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B}) = \vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} - \vec{B} \cdot \vec{A} + \vec{B} \cdot \vec{B}\)

Similarly, using the properties of the dot product, the right side becomes:

\(|\vec{A}|^2 - 2(\vec{A} \cdot \vec{B}) + |\vec{B}|^2\)

Equating and Simplifying the Expression

Now, we set the expanded left side equal to the expanded right side according to the initial condition:

\(|\vec{A}|^2 + 2(\vec{A} \cdot \vec{B}) + |\vec{B}|^2 = |\vec{A}|^2 - 2(\vec{A} \cdot \vec{B}) + |\vec{B}|^2\)

We can subtract \(|\vec{A}|^2\) and \(|\vec{B}|^2\) from both sides of the equation:

\(2(\vec{A} \cdot \vec{B}) = -2(\vec{A} \cdot \vec{B})\)

Now, bring the term from the right side to the left side:

\(2(\vec{A} \cdot \vec{B}) + 2(\vec{A} \cdot \vec{B}) = 0\)

\(4(\vec{A} \cdot \vec{B}) = 0\)

Since 4 is a non-zero constant, the equation implies that the dot product of vectors \(\vec{A}\) and \(\vec{B}\) must be zero:

\(\vec{A} \cdot \vec{B} = 0\)

Interpreting the Dot Product Result

The dot product of two non-zero vectors \(\vec{A}\) and \(\vec{B}\) is given by the formula:

\(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta)\)

where \(|\vec{A}|\) is the magnitude of \(\vec{A}\), \(|\vec{B}|\) is the magnitude of \(\vec{B}\), and \(\theta\) is the angle between the two vectors.

From our calculation, we found that \(\vec{A} \cdot \vec{B} = 0\). So, we have:

\(|\vec{A}| |\vec{B}| \cos(\theta) = 0\)

The problem states that \(\vec{A}\) and \(\vec{B}\) are non-zero vectors, which means \(|\vec{A}| \neq 0\) and \(|\vec{B}| \neq 0\). Therefore, for the equation \(|\vec{A}| |\vec{B}| \cos(\theta) = 0\) to be true, the only possibility is that \(\cos(\theta) = 0\).

The angle \(\theta\) between two vectors is typically in the range \(0^\circ \leq \theta \leq 180^\circ\). The value of \(\theta\) for which \(\cos(\theta) = 0\) in this range is \(\theta = 90^\circ\).

An angle of \(90^\circ\) between two vectors means the vectors are perpendicular.

Evaluating the Options

Let's consider the given options based on our finding that the vectors must be perpendicular:

• The vectors are parallel: If vectors are parallel, the angle is \(0^\circ\) or \(180^\circ\). \(\cos(0^\circ)=1\) and \(\cos(180^\circ)=-1\). The dot product would be \(|\vec{A}||\vec{B}|\) or \(-|\vec{A}||\vec{B}|\), which is not zero for non-zero vectors.
• The vectors are perpendicular: If vectors are perpendicular, the angle is \(90^\circ\). \(\cos(90^\circ)=0\). The dot product is \(|\vec{A}||\vec{B}| \times 0 = 0\), which matches our result.
• The vectors are anti-parallel: This is a specific case of parallel vectors where the angle is \(180^\circ\). As explained above, this does not result in a zero dot product.
• The vectors must be unit vectors: The condition \(\vec{A} \cdot \vec{B} = 0\) depends only on the angle between the non-zero vectors, not their magnitudes. They could be unit vectors, but they could also be vectors of any other magnitude, as long as they are not zero magnitude. So, they don't have to be unit vectors.

Therefore, the condition that the magnitude of the sum of two non-zero vectors is equal to the magnitude of their difference implies that the vectors are perpendicular.

The table clearly shows that \(|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2\) (which means \(|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|\)) only holds true when the vectors are perpendicular, resulting in the dot product being zero.

Revision Table: Vector Properties

Additional Information: Geometric Interpretation

The vectors \((\vec{A} + \vec{B})\) and \((\vec{A} - \vec{B})\) can be visualized as the diagonals of a parallelogram formed by vectors \(\vec{A}\) and \(\vec{B}\).

• \((\vec{A} + \vec{B})\) is the diagonal starting from the origin (tail of \(\vec{A}\) and \(\vec{B}\)).
• \((\vec{A} - \vec{B})\) is the diagonal from the head of \(\vec{B}\) to the head of \(\vec{A}\). Equivalently, if you form the parallelogram with \(\vec{A}\) and \(\vec{B}\), \(\vec{A}-\vec{B}\) connects the head of \(\vec{B}\) to the head of \(\vec{A}\).

The problem states that the magnitudes of these two diagonals are equal. In a parallelogram, the diagonals are equal in length if and only if the parallelogram is a rectangle. A parallelogram formed by vectors \(\vec{A}\) and \(\vec{B}\) is a rectangle if and only if the adjacent sides (\(\vec{A}\) and \(\vec{B}\)) are perpendicular. This geometric interpretation aligns perfectly with the algebraic result derived using the dot product.