Problem 74 - Precalculus

Question

Show that \(|\mathbf{u}-\mathbf{v}|^{2}=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}-2(\mathbf{u} \cdot \mathbf{v})\).

Step-by-Step Solution

Answer

The expression is confirmed: \(|\mathbf{u}-\mathbf{v}|^{2} = |\mathbf{u}|^{2} + |\mathbf{v}|^{2} - 2(\mathbf{u} \cdot \mathbf{v})\).

Step 1: Understand the expression

The given expression is \(|\mathbf{u}-\mathbf{v}|^{2}\), which represents the square of the magnitude of the vector \(\mathbf{u}-\mathbf{v}\). We need to express this in terms of the magnitudes \(|\mathbf{u}|\) and \(|\mathbf{v}|\), and the dot product \(\mathbf{u} \cdot \mathbf{v}\).

Step 2: Invoke the formula for magnitude

The magnitude of a vector \(\mathbf{w}\) is defined as \(|\mathbf{w}|^2=\mathbf{w} \cdot \mathbf{w}\). For our expression, substitute \(\mathbf{w} = \mathbf{u} - \mathbf{v}\), so \(|\mathbf{u}-\mathbf{v}|^2 = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\).

Step 3: Apply the distributive property

Apply the distributive property on the dot product: \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = \mathbf{u} \cdot \mathbf{u} - \mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v}\).

Step 4: Recognize properties of dot product

Remember that the dot product is commutative, so \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\). Therefore, the expression simplifies to \(\mathbf{u} \cdot \mathbf{u} - 2(\mathbf{u} \cdot \mathbf{v}) + \mathbf{v} \cdot \mathbf{v}\).

Step 5: Relate to magnitudes

The terms \(\mathbf{u} \cdot \mathbf{u}\) and \(\mathbf{v} \cdot \mathbf{v}\) are just \(|\mathbf{u}|^2\) and \(|\mathbf{v}|^2\), respectively, since magnitude squared is the vector dotted with itself. Substituting these into the expression gives \(|\mathbf{u}|^2 + |\mathbf{v}|^2 - 2(\mathbf{u} \cdot \mathbf{v})\).

Step 6: Conclusion

We have shown that \(|\mathbf{u}-\mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2(\mathbf{u} \cdot \mathbf{v})\) by expanding and simplifying the expression. This completes the proof.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. This operation involves two vectors, say \( \mathbf{u} \) and \( \mathbf{v} \), and results in a single scalar value. The dot product is calculated by multiplying corresponding components of the vectors and then summing those products.
For vectors \( \mathbf{u} = [u_1, u_2, \ldots, u_n] \) and \( \mathbf{v} = [v_1, v_2, \ldots, v_n] \), the dot product is given by:
\[\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \ldots + u_nv_n\]
It’s important to note:

• Dot product results in a scalar, not a vector.
• The operation is commutative, meaning \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \).
• The dot product is used to find the angle between two vectors and to determine orthogonality.

In our problem, it helps express the relationship between the magnitudes of the vectors \( \mathbf{u} - \mathbf{v} \), \( \mathbf{u} \), and \( \mathbf{v} \). It shows up in the form of \( 2(\mathbf{u} \cdot \mathbf{v}) \), which reflects the correlation between the two vectors.

Vector Subtraction

Vector subtraction is essential in numerous physical and mathematical applications. It involves subtracting the components of one vector from the components of another vector. If you have \( \mathbf{u} = [u_1, u_2, \ldots, u_n] \) and \( \mathbf{v} = [v_1, v_2, \ldots, v_n] \), then the vector subtraction \( \mathbf{u} - \mathbf{v} \) is calculated as:
\[\mathbf{u} - \mathbf{v} = [u_1 - v_1, u_2 - v_2, \ldots, u_n - v_n]\]
Steps to remember:

• Subtract corresponding components of the vectors.
• Resulting vector may point in an entirely new direction.

In the context of this problem, vector subtraction leads to finding the vector \( \mathbf{u} - \mathbf{v} \), whose magnitude squared is analyzed. It's this subtraction that forms the core of the equation \(|\mathbf{u}-\mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2(\mathbf{u}\cdot \mathbf{v})\). Calculating the vector subtraction correctly is crucial for further operations like finding the magnitude of the resulting vector.

Distributive Property

The distributive property is a key rule in arithmetic and algebra. It states that multiplying a number or an expression by a sum is the same as doing each multiplication separately. For vectors, the distributive property applies to the dot product. If \( \mathbf{a}, \mathbf{b}, \) and \( \mathbf{c} \) are vectors, the property states:
\[\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}\]
This property helps in simplifying complex vector expressions by spreading a dot product over addition or subtraction inside the vector.

• Use it to distribute the dot operation across vector subtraction.
• Maintains equal outcomes through simplification of vector expressions.

In our problem, the distributive property is used to expand \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\) into \( \mathbf{u} \cdot \mathbf{u} - 2(\mathbf{u} \cdot \mathbf{v}) + \mathbf{v} \cdot \mathbf{v} \). This is crucial for breaking down and proving the desired equation for magnitudes and dot products.