Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

If both \(\mathbf{u} \times \mathbf{v}=\mathbf{0}\) and \(\mathbf{u} \cdot \mathbf{v}=0\), what can you conclude about u or \(\mathbf{v}\) ?

Step-by-Step Solution

Answer

At least one of the vectors must be the zero vector.

Step 1: Understanding the Problem

We need to determine what information we can infer about the vectors \(\mathbf{u}\) and \(\mathbf{v}\) given that \(\mathbf{u} \times \mathbf{v} = \mathbf{0}\) and \(\mathbf{u} \cdot \mathbf{v} = 0\). The first condition tells us that the vectors are parallel, while the second condition indicates they are orthogonal.

Step 2: Analyzing the Cross Product Condition

The cross product \(\mathbf{u} \times \mathbf{v} = \mathbf{0}\) implies that the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are either parallel or one of them is the zero vector. This is because the magnitude of the cross product is zero, which occurs when the sine of the angle between them is zero.

Step 3: Analyzing the Dot Product Condition

The dot product \(\mathbf{u} \cdot \mathbf{v} = 0\) indicates that the vectors are orthogonal. This is because the dot product is zero if the cosine of the angle between the vectors is zero, implying that the vectors are at a right angle (90 degrees) to each other.

Step 4: Combining Both Conditions

Since the vectors are both parallel and orthogonal simultaneously, the only mathematical possibility is that at least one of the vectors must be the zero vector. This satisfies both conditions, as the zero vector is technically both parallel to and orthogonal to any vector.

Key Concepts

Cross Product

The cross product is an operation between two vectors, say \( \mathbf{u} \) and \( \mathbf{v} \). It results in a new vector that is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \). The magnitude of this new vector is equivalent to the area of the parallelogram with sides defined by \( \mathbf{u} \) and \( \mathbf{v} \). This operation is specific to three-dimensional space.

• The cross product \( \mathbf{u} \times \mathbf{v} = \mathbf{0} \) suggests the sine of the angle \( \theta \) between \( \mathbf{u} \) and \( \mathbf{v} \) is zero.
• This implies the vectors are either parallel or one of the vectors is the zero vector.

Remember that parallel vectors have no defined perpendicular vector, resulting in a zero cross product. On the other hand, the zero vector's cross product with any vector will also be zero, as it has no direction. So in essence, a zero cross product points out a very specific relationship between two vectors as specified above.

Dot Product

The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is a scalar, and it measures the magnitude of one vector projected onto the other. Represented mathematically, it is:\[\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta)\]

• If \( \cos(\theta) = 0 \), then \( \mathbf{u} \cdot \mathbf{v} = 0 \), which means the vectors are orthogonal.
• Orthogonal vectors meet at a right angle, and have no apparent projection of one onto the other.

This characteristic makes the dot product useful for determining angles between vectors. In cases where it is zero, the vectors do not "lean" towards each other. They are perfectly perpendicular, standing independent in direction.

Orthogonal Vectors

Orthogonal vectors are two vectors that meet at a right angle. Think of them like the coordinate axes \( x \) and \( y \) in the Cartesian plane—completely independent directions from each other.

• Mathematically, orthogonality is detected via a dot product that equals zero, as the cosine of 90 degrees is zero.
• Being orthogonal implies there is no component of one vector in the direction of the other.

In the exercise scenario where both the cross and dot products equal zero, it seems like a contradiction because parallel vectors aren’t usually orthogonal. The resolution lies in recognizing the zero vector as the agent satisfying both conditions. The zero vector is unique because it doesn't "point" anywhere and can be considered simultaneously parallel and orthogonal to every vector.