Question

Let \(\mathbf{u}\) be an arbitrary fixed unit vector and show that any vector \(\mathbf{b}\) satisfies $$b^{2}=(\mathbf{u} \cdot \mathbf{b})^{2}+(\mathbf{u} \times \mathbf{b})^{2}$$ Explain this result in words, with the help of a picture.

Short answer

This equation shows the magnitude of a vector is the sum of the squared magnitudes of its projections parallel and perpendicular to a unit vector.

Step-by-step solution

Understand the Problem

We need to show that for any vector \( \mathbf{b} \), the equation \( b^{2}=(\mathbf{u} \cdot \mathbf{b})^{2}+(\mathbf{u} \times \mathbf{b})^{2} \) holds true, where \( \mathbf{u} \) is a unit vector.

Recall Vector Magnitude and Definitions

Recall that the magnitude of a vector \( \mathbf{b} \) is given by \( b = \| \mathbf{b} \| \) and is equal to \( \sqrt{\mathbf{b} \cdot \mathbf{b}} \). Also, the dot product \( \mathbf{u} \cdot \mathbf{b} \) measures projection along \( \mathbf{u} \), and the magnitude of the cross product \( \| \mathbf{u} \times \mathbf{b} \| \) measures the component orthogonal to \( \mathbf{u} \).

Express \( \mathbf{b} \) in Terms of \( \mathbf{u} \)

Any vector \( \mathbf{b} \) can be decomposed into components parallel and perpendicular to \( \mathbf{u} \). Let \( \mathbf{b}_{\parallel} = (\mathbf{u} \cdot \mathbf{b}) \mathbf{u} \) and \( \mathbf{b}_{\perp} = \mathbf{b} - \mathbf{b}_{\parallel} \).

Use Pythagorean Theorem in Vector Form

Since \( \mathbf{b} = \mathbf{b}_{\parallel} + \mathbf{b}_{\perp} \), by orthogonality, \( b^{2} = \| \mathbf{b}_{\parallel} \|^{2} + \| \mathbf{b}_{\perp} \|^{2} \). Here, \( \| \mathbf{b}_{\parallel} \| = |\mathbf{u} \cdot \mathbf{b}| \) and \( \| \mathbf{b}_{\perp} \| = \| \mathbf{u} \times \mathbf{b} \| \).

Plug into the Original Equation

Plugging these into the equation gives \[ \mathbf{b}^{2} = (\mathbf{u} \cdot \mathbf{b})^{2} + (\mathbf{u} \times \mathbf{b})^{2} \], thus proving the statement.

Key concepts

Dot Product

The dot product, often referred to as the scalar product, is a way to multiply two vectors to produce a scalar. It is calculated by multiplying the magnitudes of the two vectors and the cosine of the angle between them. The formula is:* \( \mathbf{u} \cdot \mathbf{b} = \|\mathbf{u}\| \times \|\mathbf{b}\| \times \cos(\theta) \)* In simpler terms, the dot product * measures how much one vector extends in the direction of another, * acts like a form of projection measuring lengths.* For unit vectors * like \(\mathbf{u}\),\( \mathbf{u} \cdot \mathbf{b} \) simplifies to just \(\cos(\theta)\) of the components of \(\mathbf{b}\), * in this case also corresponds to \(\mathbf{b}\)'s projection onto \(\mathbf{u}\).* If the vectors are perpendicular, the dot product is zero.* It helps identify the component of \( \mathbf{b} \) along \( \mathbf{u} \) as \((\mathbf{u} \cdot \mathbf{b})\).

Cross Product

The cross product is another way to multiply vectors, but the result is a new vector, not a scalar. It measures how vectors differ from each other in 3D space and considers the area of the parallelogram they span. The cross product formula is as follows:* \( \mathbf{u} \times \mathbf{b} = \|\mathbf{u}\| \times \|\mathbf{b}\| \times \sin(\theta) \mathbf{n} \)* \(\mathbf{n}\) is a unit vector perpendicular to the plane formed by \(\mathbf{u}\) and \(\mathbf{b}\).* This equation results in: * a vector orthogonally encompassing width, * orientation given by right-hand rule.* The magnitude \( \|\mathbf{u} \times \mathbf{b}\| \) corresponds to area information; it also * embodies the component perpendicular to \( \mathbf{u} \).* If vectors are parallel, the cross product yields zero.* Cross product gives crucial insight into vector orientation in space.

Vector Decomposition

Vector decomposition entails breaking down a vector into orthogonal components. This enables us to express any vector as a sum of vectors in distinct directions.* For any vector \( \mathbf{b} \), it can be broken down into two components: * Parallel to a given unit vector \( \mathbf{u} \): \( \mathbf{b}_{\parallel} = (\mathbf{u} \cdot \mathbf{b}) \mathbf{u} \). * Perpendicular to \( \mathbf{u} \): \( \mathbf{b}_{\perp} = \mathbf{b} - \mathbf{b}_{\parallel} \).* The parallel component is the projection on \( \mathbf{u} \), while the perpendicular part * is what remains of \( \mathbf{b} \) after extracting parallel influence.* This method benefits resolving vectors along predetermined axes, * often used for problem solving in physics.

Pythagorean Theorem in Vector Form

The Pythagorean Theorem is a cornerstone in vector calculations, offering a way to sum vector components orthogonally.* Given a vector \( \mathbf{b} \), it acts as * \( \mathbf{b} = \mathbf{b}_{\parallel} + \mathbf{b}_{\perp} \).* Using this for magnitude assessment, * \( b^{2} = \|\mathbf{b}\|^{2} \rightarrow \|\mathbf{b}_{\parallel}\|^{2} + \|\mathbf{b}_{\perp}\|^{2} \) * This aligns with familiar Pythagorean identity in right triangles.* Applied specifically to vectors, it allows splitting terms like: * \( (\mathbf{u} \cdot \mathbf{b})^{2} \) * \( (\mathbf{u} \times \mathbf{b})^{2} \)* This equation tells us how naturally vectors * sum their squared partsencompassing measure of size for both components.