Problem 27 - University Calculus: Early Transcendentals

Question

Which of the following are always true, and which are not always true? Give reasons for your answers. a. \(|\mathbf{u}|=\sqrt{\mathbf{u} \cdot \mathbf{u}}\) b. \(\mathbf{u} \cdot \mathbf{u}=|\mathbf{u}|\) c. \(\mathbf{u} \times \mathbf{0}=\mathbf{0} \times \mathbf{u}=\mathbf{0}\) d. \(\mathbf{u} \times(-\mathbf{u})=\mathbf{0}\) e. \(\mathbf{u} \times \mathbf{v}=\mathbf{v} \times \mathbf{u}\) f. \(\mathbf{u} \times(\mathbf{v}+\mathbf{w})=\mathbf{u} \times \mathbf{v}+\mathbf{u} \times \mathbf{w}\) g. \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{v}=0\) h. \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}=\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})\)

Step-by-Step Solution

Answer

a, c, d, f, g, h are always true; b, e are not always true.

Step 1: Analyze Expression (a)

For expression (a), \(|\mathbf{u}|=\sqrt{\mathbf{u} \cdot \mathbf{u}}\), it represents the definition of the magnitude of a vector \(\mathbf{u}\). A vector \(\mathbf{u}\) dotted with itself is equal to its magnitude squared, \(\mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2\). Taking the square root gives the magnitude, making this statement always true.

Step 2: Evaluate Expression (b)

Expression (b) \(\mathbf{u} \cdot \mathbf{u}=|\mathbf{u}|\) is tested next. Since \(\mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2\), and this is equal to the square of the magnitude, not the magnitude itself, this is not always true unless \(|\mathbf{u}| = 1\).

Step 3: Review Expression (c)

Expression (c), \(\mathbf{u} \times \mathbf{0} = \mathbf{0} \times \mathbf{u} = \mathbf{0}\), is a property of the cross product. Any vector crossed with the zero vector results in the zero vector. Thus, this statement is always true.

Step 4: Consider Expression (d)

For expression (d), \(\mathbf{u} \times (-\mathbf{u}) = \mathbf{0}\), consider that the cross product of any vector with any scalar multiple of itself, including its negative, is the zero vector. Therefore, this statement is always true.

Step 5: Check Expression (e)

For expression (e), \(\mathbf{u} \times \mathbf{v} = \mathbf{v} \times \mathbf{u}\), note that the cross product is anti-commutative. This means \(\mathbf{u} \times \mathbf{v} = -\mathbf{v} \times \mathbf{u}\); thus, this statement is not always true.

Step 6: Validate Expression (f)

In expression (f), \(\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \mathbf{u} \times \mathbf{v} + \mathbf{u} \times \mathbf{w}\), we recognize the distributive property of the cross product over addition. This statement is always true.

Step 7: Test Expression (g)

Expression (g), \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{v} = 0\), is analyzed next. The cross product \(\mathbf{u} \times \mathbf{v}\) is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\), thus the dot product with \(\mathbf{v}\) is zero. This statement is always true.

Step 8: Inspect Expression (h)

Expression (h), \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\), involves the scalar triple product. This property is known as the scalar triple product identity and is always true.

Key Concepts

Cross Product

The cross product is a way to multiply two vectors in three-dimensional space, resulting in a new vector. This new vector is perpendicular to both of the original vectors. The cross product of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is denoted by \(\mathbf{u} \times \mathbf{v}\). To calculate it, we use the following formula:

• \(\mathbf{u} \times \mathbf{v} = \|\mathbf{u}\|\|\mathbf{v}\|\sin(\theta)\mathbf{n}\)

Here, \(\theta\) is the angle between \(\mathbf{u}\) and \(\mathbf{v}\), and \(\mathbf{n}\) is a unit vector perpendicular to both. The result is zero if \(\mathbf{u}\) and \(\mathbf{v}\) are parallel, explaining why \(\mathbf{u} \times \mathbf{0} = \mathbf{0}\).
Remember that the cross product is anti-commutative, meaning \(\mathbf{u} \times \mathbf{v} = -\mathbf{v} \times \mathbf{u}\), and it obeys the distributive law, illustrated by \(\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \mathbf{u} \times \mathbf{v} + \mathbf{u} \times \mathbf{w}\). These properties are essential to solve many physics and engineering problems.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors to get a scalar. It is a measure of how much one vector goes in the direction of another. For vectors \(\mathbf{u}\) and \(\mathbf{v}\), it is defined as:

• \(\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\|\mathbf{v}\|\cos(\theta)\)

Here, \(\theta\) is the angle between the vectors. If the angle is 90 degrees, the dot product is zero because the vectors are orthogonal. This property is used in calculating perpendicularity and in projections.
One common use is finding vector magnitude since \(\mathbf{u} \cdot \mathbf{u} = \|\mathbf{u}\|^2\). Thus, \(|\mathbf{u}| = \sqrt{\mathbf{u} \cdot \mathbf{u}}\). It's straightforward but crucial for handling 3D vector calculations.

Scalar Triple Product

The scalar triple product involves three vectors and results in a scalar. It is expressed as \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\) or equivalently \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\). Both expressions yield the same result, showcasing the scalar triple product identity.
This product is essential for determining the volume of a parallelepiped formed by three vectors, as it provides the signed volume. A zero scalar triple product indicates the vectors are co-planar, having zero volume.
The scalar triple product is extensively used in physics and engineering for calculating torque and other rotational dynamics properties.

Vector Magnitude

The magnitude of a vector, also known as its length or norm, is a measure of its size. For a vector \(\mathbf{u} = (u_1, u_2, u_3)\), the magnitude is calculated using the formula:

• \(|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}\)

It is always a non-negative number and provides critical information concerning a vector's scale. In various applications, vector magnitude is used to normalize vectors, making their lengths equal to one while preserving direction.
Understanding vector magnitude is foundational for performing operations like dot and cross products and is particularly useful in physics for expressing forces, velocities, and more.