Let u and v be unit vectors. If w is a vector such that w + (w x u) = v, then prove that |(u x v) .

Q: Wait, why can't we just divide by $\vec{u}$ to isolate $\vec{w}$?

In vector algebra, there is no operation called 'division' by a vector. You cannot divide by a vector because it has both magnitude and direction. Instead, we use dot and cross products to 'project' or 'rotate' the equation into a form where we can isolate the variable. Think of dot and cross products as tools to extract specific components of the vector.

Q: How do I know whether to take the dot product or the cross product first?

It is a strategic choice. The dot product is a scalar operator—it reduces the dimensionality of the equation and helps us find the component of $\vec{w}$ parallel to $\vec{u}$. The cross product is a vector operator—it helps us find the components perpendicular to $\vec{u}$. Using both gives us a complete picture of $\vec{w}$.

Q: Why does the scalar triple product simplify so beautifully at the end?

That is the elegance of geometry! When you expand the dot product, terms like $(\vec{u} \times \vec{v}) \cdot \vec{v}$ vanish because the cross product is orthogonal to its constituents. You are left only with the magnitude of the cross product, which is the geometric definition of the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$.

Question

Let $\vec{u}$ and $\vec{v}$ be unit vectors. If $\vec{w}$ is a vector such that $\vec{w} + (\vec{w} 	imes \vec{u}) = \vec{v}$, then prove that $|(\vec{u} 	imes \vec{v}) \cdot \vec{w}| \le 1/2$ and that the equality holds if and only if $\vec{u}$ is perpendicular to $\vec{v}$.

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into the Unknown

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are unraveling a mystery.

We are presented with two unit vectors, $\vec{u}$ and $\vec{v}$, and a third, mysterious vector $\vec{w}$ defined by the relationship:

$$\vec{w} + (\vec{w} 	imes \vec{u}) = \vec{v}$$

Our goal is to prove that the scalar triple product $|(\vec{u} 	imes \vec{v}) \cdot \vec{w}|$ is bounded by $\frac{1}{2}$, and to understand the condition under which this bound is reached.

### Phase 1: The Filtering Strategy

When you see an equation like $\vec{w} + (\vec{w} 	imes \vec{u}) = \vec{v}$, your first instinct should be to isolate $\vec{w}$. We use the tools of projection by taking the dot product of the entire equation with $\vec{u}$.

The dot product acts as a scalar filter. Distributing the dot product, we obtain:

$$(\vec{w} \cdot \vec{u}) + ((\vec{w} 	imes \vec{u}) \cdot \vec{u}) = \vec{v} \cdot \vec{u}$$

Now, look closely at the second term. By the definition of the cross product, $\vec{w} 	imes \vec{u}$ is a vector perpendicular to both $\vec{w}$ and $\vec{u}$. Therefore, its dot product with $\vec{u}$ must be zero.

The equation collapses beautifully into:

$$\vec{w} \cdot \vec{u} = \vec{v} \cdot \vec{u}$$

We have successfully found the component of $\vec{w}$ along $\vec{u}$.

### Phase 2: The Rotation Strategy

Now that we have the parallel component, we need the perpendicular one. We take the cross product of the original equation with $\vec{u}$ from the left:

$$\vec{u} 	imes (\vec{w} + \vec{w} 	imes \vec{u}) = \vec{u} 	imes \vec{v}$$

Distributing this, we get:

$$\vec{u} 	imes \vec{w} + \vec{u} 	imes (\vec{w} 	imes \vec{u}) = \vec{u} 	imes \vec{v}$$

This looks intimidating, but we apply the Vector Triple Product identity, often called the BAC-CAB rule: $\vec{A} 	imes (\vec{B} 	imes \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$.

Applying this to $\vec{u} 	imes (\vec{w} 	imes \vec{u})$, we get $(\vec{u} \cdot \vec{u})\vec{w} - (\vec{u} \cdot \vec{w})\vec{u}$. Since $\vec{u}$ is a unit vector, $\vec{u} \cdot \vec{u} = 1$.

Substituting our previous result $\vec{u} \cdot \vec{w} = \vec{u} \cdot \vec{v}$, the expression becomes:

$$\vec{w} - (\vec{u} \cdot \vec{v})\vec{u}$$

### Phase 3: The Synthesis

We are almost there. We have:

$$\vec{u} 	imes \vec{w} + \vec{w} - (\vec{u} \cdot \vec{v})\vec{u} = \vec{u} 	imes \vec{v}$$

From the original equation, we know $\vec{w} 	imes \vec{u} = \vec{v} - \vec{w}$, which implies $\vec{u} 	imes \vec{w} = \vec{w} - \vec{v}$. Substituting this back, we get:

$$(\vec{w} - \vec{v}) + \vec{w} - (\vec{u} \cdot \vec{v})\vec{u} = \vec{u} 	imes \vec{v}$$

Combining the terms, we find:

$$2\vec{w} = \vec{v} + (\vec{u} 	imes \vec{v}) + (\vec{u} \cdot \vec{v})\vec{u}$$

We have isolated $\vec{w}$.

### Phase 4: The Grand Finale

Finally, we evaluate the scalar triple product $(\vec{u} 	imes \vec{v}) \cdot \vec{w}$. Substituting our expression for $\vec{w}$, we get:

$$\frac{1}{2} (\vec{u} 	imes \vec{v}) \cdot [\vec{v} + (\vec{u} 	imes \vec{v}) + (\vec{u} \cdot \vec{v})\vec{u}]$$

Distributing the dot product, we see that $(\vec{u} 	imes \vec{v}) \cdot \vec{v} = 0$ and $(\vec{u} 	imes \vec{v}) \cdot \vec{u} = 0$ because the cross product is orthogonal to its components. We are left with:

$$\frac{1}{2} |\vec{u} 	imes \vec{v}|^2$$

Since $|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}|\sin	heta = \sin	heta$, the expression is $\frac{1}{2} \sin^2	heta$. Since $0 \le \sin^2	heta \le 1$, the value is bounded by $\frac{1}{2}$.

The maximum occurs when $\sin^2	heta = 1$, which means $	heta = \frac{\pi}{2}$, or $\vec{u} \perp \vec{v}$. We have arrived at the truth through the elegance of vector algebra.

Animated Solution for Mathematics - Vector Algebra: Let $u$ and $v$ be unit vectors. If $w$ is a vector such that $w + (w \times u) = v$ , then prove that $∣ (u \times v) \cdot w ∣ \leq 1/2$ and that the equality holds if and only if $u$ is perpendicular to $v$ .

Visualized Solution

Understanding the Given Equation

Taking Dot Product with $u$

Simplifying the Dot Product

Taking Cross Product with $u$

Distributing the Cross Product

Applying Vector Triple Product

Simplifying the Triple Product

Substituting Back into the Equation

Solving for Vector $w$

Calculating the Scalar Triple Product

Evaluating the Dot Products

Simplifying to Trigonometric Form

Final Inequality and Equality Condition

The Sigma Insight: Vector Triple Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

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