Question

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are known nonzero vectors in \(\mathbb{R}^{3}\). a. Prove that the equation \(\mathbf{u} \times \mathbf{z}=\mathbf{v}\) has a nonzero solution \(\mathbf{z}\) if and only if \(\mathbf{u} \cdot \mathbf{v}=0 .\) (Hint: Take the dot product of both sides with v.) b. Explain this result geometrically.

Short answer

**Short answer question:** Explain the relationship between the dot product of two vectors and the existence of a nonzero solution for the equation \(\mathbf{u} \times \mathbf{z} = \mathbf{v}\). **Short answer:** The existence of a nonzero solution for the equation \(\mathbf{u} \times \mathbf{z} = \mathbf{v}\) is ensured if and only if the dot product of the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is zero. Geometrically, this means that the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal to each other.

Step-by-step solution

Part a: Prove the given condition using algebraic manipulation

We want to prove that if the equation \(\mathbf{u} \times \mathbf{z} = \mathbf{v}\) has a nonzero solution \(\mathbf{z}\), then \(\mathbf{u} \cdot \mathbf{v} = 0\), and vice versa. 1. Assume that the equation \(\mathbf{u} \times \mathbf{z} = \mathbf{v}\) has a nonzero solution \(\mathbf{z}\). Take the dot product of both sides of this equation with \(\mathbf{v}\): $$ (\mathbf{u} \times \mathbf{z}) \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{v}. $$ 2. Use the property \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a} \): $$ (\mathbf{u} \cdot \mathbf{v})\mathbf{z} - (\mathbf{z} \cdot \mathbf{v})\mathbf{u} = \mathbf{v} \cdot \mathbf{v}. $$ 3. Since the solution \(\mathbf{z}\) is nonzero, the equation can not be satisfied for all values of \(\mathbf{z}\), unless the coefficients of \(\mathbf{u}\) and \(\mathbf{z}\) are both zero. Thus, we have: $$ (\mathbf{u} \cdot \mathbf{v}) = 0 $$ and $$ (\mathbf{z} \cdot \mathbf{v}) = 0. $$ 4. Conversely, assume that \(\mathbf{u} \cdot \mathbf{v} = 0\). Our goal is to find a nonzero solution \(\mathbf{z}\) such that \(\mathbf{u} \times \mathbf{z} = \mathbf{v}\). We can see that since \(\mathbf{u} \cdot \mathbf{v} = 0\), \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal (forming a right angle). Therefore, \(\mathbf{z}\) must be a vector perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\). This can be achieved by finding a vector \(\mathbf{z} = \mathbf{u} \times \mathbf{w}\) for some vector \(\mathbf{w}\), because the cross product of two vectors is always orthogonal to both of the original vectors. By substituting \(\mathbf{z}\) into the original equation, we get: $$ \mathbf{u} \times (\mathbf{u} \times \mathbf{w}) = \mathbf{v}. $$ 5. Now the equation is true if and only if we can find a vector \(\mathbf{w}\) such that \(\mathbf{u} \times (\mathbf{u} \times \mathbf{w}) = \mathbf{v}\). We know that the cross product is distributive over vector addition and scalar multiplication, so we can write this equation in the following form: $$ (\mathbf{u} \cdot (\mathbf{u} \times \mathbf{w}))\mathbf{u} - (\mathbf{u} \times (\mathbf{u} \times \mathbf{w}))\mathbf{w} = \mathbf{v}. $$ Since \(\mathbf{u} \cdot \mathbf{v} = 0\), the left-hand side of the equation is equal to $$ -(\mathbf{u} \times (\mathbf{u} \times \mathbf{w}))\mathbf{w}. $$ 6. To find the vector \(\mathbf{w}\) such that the equation is satisfied, we can simply set this expression equal to \(\mathbf{v}\), and solve for \(\mathbf{w}\): $$ (\mathbf{u} \times (\mathbf{u} \times \mathbf{w}))\mathbf{w} = -\mathbf{v}. $$ 7. We have shown that if \(\mathbf{u} \cdot \mathbf{v} = 0\), there exists a nonzero solution \(\mathbf{z} = \mathbf{u} \times \mathbf{w}\) for the given equation. This completes the proof of the given condition.

Part b: Explain the result geometrically

Geometrically, the result can be explained as follows: - When \(\mathbf{u} \cdot \mathbf{v} = 0\), the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal to each other. In this case, the equation \(\mathbf{u} \times \mathbf{z} = \mathbf{v}\) has a nonzero solution \(\mathbf{z}\) because the cross product of \(\mathbf{u}\) and \(\mathbf{z}\) must be orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). We can find a vector \(\mathbf{z}\) that is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\), making the given equation true. - Conversely, if the equation \(\mathbf{u} \times \mathbf{z} = \mathbf{v}\) has a nonzero solution \(\mathbf{z}\), it means that there exists a vector \(\mathbf{z}\) such that its cross product with \(\mathbf{u}\) is equal to \(\mathbf{v}\). Since the cross product of two vectors is orthogonal to both of the original vectors, \(\mathbf{u}\) and \(\mathbf{v}\) must be orthogonal, meaning their dot product is zero (\(\mathbf{u} \cdot \mathbf{v} = 0\)). In conclusion, the given equation has a nonzero solution if and only if the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) is zero, which geometrically means that the vectors are orthogonal to each other.

Key concepts

Vector Multiplication

In mathematics, particularly in vector calculus, vector multiplication refers to several different ways in which vectors can be multiplied with each other. The two most common types of multiplication involving vectors are the dot product and the cross product. The dot product is a way to find the scalar (numerical) product of two vectors, showing what portion of one vector goes in the direction of another. The cross product, on the other hand, results in a new vector that is perpendicular to both vectors being multiplied.

Understanding vector multiplication is important in fields like physics and engineering, where vectors represent quantities such as forces or velocities. Multiplying vectors can help determine the angle between them, calculate work (in the case of the dot product), or find a vector that is orthogonal to a plane defined by two other vectors (in the case of the cross product).

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is symbolized by a dot (\text{·}).

Mathematically, given two vectors \textbf{a} and \textbf{b}, the dot product is defined as:
\[\textbf{a} \text{·} \textbf{b} = |\textbf{a}| |\textbf{b}| \text{cos}(\theta)\]
where |\textbf{a}| and |\textbf{b}| are the magnitudes (lengths) of the vectors and \(\theta\) is the angle between them. If the dot product is zero, it indicates that the vectors are orthogonal, meaning they are at a right angle to each other. This property is key in proving certain geometric and algebraic conditions involving vectors, as demonstrated in the exercise and solution above.

Vector Orthogonality

When two vectors are orthogonal, they are perpendicular to each other. This concept is central in various mathematical and physical contexts, as orthogonal vectors have a dot product of zero.

In three-dimensional space, two non-zero vectors \textbf{a} and \textbf{b} are orthogonal if and only if:\[\textbf{a} \text{·} \textbf{b} = 0\]
Orthogonality is a fundamental concept when dealing with projections, decompositions, and even in solving equations involving vectors. For instance, the given problem uses the principle of orthogonality to establish the conditions under which the cross product equation \textbf{u} \text{×} \textbf{z} = \textbf{v} has a non-zero solution.

Geometric Interpretation of Cross Product

The geometric interpretation of the cross product is visually rich and aids in understanding concepts in vector calculus and physics. Given two vectors \textbf{a} and \textbf{b} in three-dimensional space, the cross product \textbf{a} \text{×} \textbf{b} is a vector that is perpendicular to both \textbf{a} and \textbf{b}.

The magnitude of the cross product vector is given by the area of the parallelogram that \textbf{a} and \textbf{b} span. The formula for the magnitude is:\[|\textbf{a} \text{×} \textbf{b}| = |\textbf{a}| |\textbf{b}| \text{sin}(\theta)\]
where \(\theta\) is the angle between \textbf{a} and \textbf{b}. The direction of the cross product follows the right-hand rule, which is a convention that assigns a direction to the resultant vector. This geometric perspective adds an intuitive layer to the cross product, as seen in the explanation provided for the step-by-step solution: the vector \textbf{z} can geometrically be visualized as being perpendicular to both \textbf{u} and \textbf{v}, thereby giving rise to a vector that fulfills the cross product equation.