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Magazine Chapter 11: Problem 38
Show that if \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in 3 -space, then $$ \|\mathbf{u} \times \mathbf{v}\|^{2}=\|\mathbf{u}\|^{2}\|\mathbf{v}\|^{2}-(\mathbf{u} \cdot \mathbf{v})^{2} $$ $$ \begin{array}{l}{\text { [Note: This result is sometimes called Lagrange's }} \\\ {\text { identity. }}\end{array} $$
Short Answer
Expert verified
The identity \( \|\mathbf{u} \times \mathbf{v}\|^2 = \|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - (\mathbf{u} \cdot \mathbf{v})^2 \) holds.
Step by step solution
01
Understand the Cross Product
Recall that the cross product \( \mathbf{u} \times \mathbf{v} \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is a vector orthogonal to both \( \mathbf{u} \) and \( \mathbf{v} \). The magnitude of the cross product is given by \[ \| \mathbf{u} \times \mathbf{v} \| = \| \mathbf{u} \| \cdot \| \mathbf{v} \| \cdot \sin \theta \] where \( \theta \) is the angle between the vectors.
02
Express the Magnitude Squared
Square both sides of the magnitude of the cross product formula to find \[ \| \mathbf{u} \times \mathbf{v} \|^2 = (\| \mathbf{u} \| \cdot \| \mathbf{v} \| \cdot \sin \theta)^2 = \| \mathbf{u} \|^2 \| \mathbf{v} \|^2 \sin^2 \theta. \]
03
Use Identity for \( \sin^2 \theta \)
Recall the identity \( \sin^2 \theta = 1 - \cos^2 \theta \). Substitute this into the expression for \( \| \mathbf{u} \times \mathbf{v} \|^2 \) to get \[ \| \mathbf{u} \times \mathbf{v} \|^2 = \| \mathbf{u} \|^2 \| \mathbf{v} \|^2 (1 - \cos^2 \theta). \]
04
Relate \( \cos \theta \) to the Dot Product
The cosine of the angle \( \theta \) between two vectors can be expressed using the dot product as \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \cdot \| \mathbf{v} \|} \). Thus, \( \cos^2 \theta = \left(\frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \cdot \| \mathbf{v} \|}\right)^2 = \frac{(\mathbf{u} \cdot \mathbf{v})^2}{\| \mathbf{u} \|^2 \| \mathbf{v} \|^2} \).
05
Substitute and Simplify
Substitute the expression for \( \cos^2 \theta \) from Step 4 into the equation from Step 3: \[ \| \mathbf{u} \times \mathbf{v} \|^2 = \| \mathbf{u} \|^2 \| \mathbf{v} \|^2 \left(1 - \frac{(\mathbf{u} \cdot \mathbf{v})^2}{\| \mathbf{u} \|^2 \| \mathbf{v} \|^2}\right). \] Simplify this to get \[ \| \mathbf{u} \times \mathbf{v} \|^2 = \| \mathbf{u} \|^2 \| \mathbf{v} \|^2 - (\mathbf{u} \cdot \mathbf{v})^2.\]
06
Conclusion
This is Lagrange's identity, which shows the relation between the magnitudes of the cross product and dot product of two vectors. It verifies that the square of the magnitude of the cross product is equal to the product of the squares of the magnitudes of the individual vectors minus the square of their dot product.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross Product
The cross product, denoted by \( \mathbf{u} \times \mathbf{v} \), is a vector operation that takes two vectors in three-dimensional space and results in a new vector that is perpendicular to both. It's crucial for determining the direction in this orthogonal plane. Unlike the dot product, which gives a scalar value, the cross product yields a vector.
- The magnitude of the resulting vector from a cross product not only depends on the magnitudes of the input vectors \( \mathbf{u} \) and \( \mathbf{v} \), but also on the sine of the angle \( \theta \) between them.
- The formula for its magnitude is \( \| \mathbf{u} \times \mathbf{v} \| = \| \mathbf{u} \| \cdot \| \mathbf{v} \| \cdot \sin \theta \).
Dot Product
The dot product, noted as \( \mathbf{u} \cdot \mathbf{v} \), provides a way to multiply two vectors, resulting in a scalar. This operation is often used to find the angle between two vectors and to understand their alignment. It is defined as the sum of the products of the corresponding components of the vectors.
- The formula is \( \mathbf{u} \cdot \mathbf{v} = \| \mathbf{u} \| \cdot \| \mathbf{v} \| \cdot \cos \theta \).
- This highlights that, unlike the cross product, the result is a single number rather than a vector.
- The dot product is maximum when the vectors are aligned (\( \cos \theta = 1 \)) and zero if the vectors are perpendicular (\( \cos \theta = 0 \)).
Vector Magnitudes
Vector magnitudes, or the lengths of vectors, are fundamental in understanding a vector's size. The magnitude of a vector \( \mathbf{u} \) is represented by \( \| \mathbf{u} \| \) and is found using the Pythagorean theorem.
- If a vector \( \mathbf{u} \) has components \( (x, y, z) \), then its magnitude is \( \| \mathbf{u} \| = \sqrt{x^2 + y^2 + z^2} \).
- This magnitude is always a non-negative number, providing a scalar value that shows how much one vector differs in length from another.
- Magnitude is crucial for normalizing a vector, which is turning it into a unit vector with a magnitude of one.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for any value of the occurring variables. They are indispensable in simplifying expressions and solving equations involving angles. One notable identity used in vector mathematics is:
- \( \sin^2 \theta + \cos^2 \theta = 1 \), which can be rearranged to \( \sin^2 \theta = 1 - \cos^2 \theta \).
- This identity is particularly useful for converting expressions involving trigonometric functions through substitution, as seen in the derivation of Lagrange's identity for vectors.
