Problem 91 - Precalculus with Limits

Question

THINK ABOUT IT What can be said about the vectors \(\mathbf{u}\) and \(\mathbf{v}\) under each condition? (a) The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) equals \(\mathbf{u}\). (b) The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) equals \(\mathbf{0}\).

Step-by-Step Solution

Answer

(a) Vector \( \mathbf{u} \) and \( \mathbf{v} \) are parallel to each other. (b) Vector \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal (perpendicular) to each other.

Step 1: Analyzing the First Condition

For condition (a) where the projection of vector \( \mathbf{u} \) onto \( \mathbf{v} \) equals \( \mathbf{u} \), we can infer that vector \( \mathbf{u} \) entirely lies in the direction of vector \( \mathbf{v} \). The only way vector \( \mathbf{u} \) could be completely projected onto \( \mathbf{v} \) would be if \( \mathbf{u} \) and \( \mathbf{v} \) were parallel, i.e., \( \mathbf{u} \) could be obtained by multiplying \( \mathbf{v} \) by some scalar. Hence, they must be parallel vectors.

Step 2: Analyzing the Second Condition

For condition (b) where the projection of vector \( \mathbf{u} \) onto \( \mathbf{v} \) equals \( \mathbf{0} \), we can infer that no part of vector \( \mathbf{u} \) lies in the direction of \( \mathbf{v} \). This is only possible if \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal (perpendicular), i.e., the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) equals 0. The projection being \( \mathbf{0} \) confirms this. Hence, they must be orthogonal vectors.

Key Concepts

Parallel Vectors

When two vectors are described as parallel, this means that they are running in exactly the same direction or are identical in their orientation, although not necessarily the same in magnitude. For two vectors to be parallel, one vector must be a scalar multiple of the other. Simply put, if you can multiply one vector by a single number (a scalar) to get the other vector, those vectors are parallel.

In the context of the exercise, when the vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) results in \( \mathbf{u} \), it indicates that \( \mathbf{u} \) aligns with \( \mathbf{v} \) in such a way that it lies entirely on the line \ extended by \( \mathbf{v} \), confirming their parallelism. This scenario embodies the fundamental definition of parallel vectors in a graphical sense.

Orthogonal Vectors

Vectors are orthogonal to each other if they meet at a right angle, also known as being perpendicular. The significance of orthogonal vectors in mathematics cannot be understated, as they form the basis of many complex vector operations and are integral to topics such as vector projections.

Orthogonality is purely about the angle between the vectors. Hence, when the projection of one vector onto another is the zero vector (\( \mathbf{0} \)), it suggests that there is no component of one vector in the direction of the other. As evidenced in step 2 of the exercise, the only situation where this occurs naturally is when the vectors are orthogonal. The dot product of orthogonal vectors will always result in zero, irrespective of their magnitudes, making this a powerful test for orthogonality.

Vector Dot Product

Understanding the dot product—also known as the scalar product—is crucial, as it is a fundamental operation in vector calculus. The dot product takes two vectors and returns a single number (a scalar); this number represents the magnitude of the projection of one vector onto another when the vectors are aligned at their tails.

The formula for the dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\, |\mathbf{b}|\, \cos(\theta) \), where \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes (lengths) of the vectors, and \( \theta \) is the angle between them. If the vectors are orthogonal, the dot product is zero because \( \cos(90^\circ) = 0 \). As seen in the exercise, a zero dot product signifies perpendicular vectors.

Scalar Multiplication of Vectors

Scalar multiplication involves multiplying a vector by a scalar (a real number), resulting in a new vector that has the same (or opposite, if the scalar is negative) direction but a different magnitude. This operation is pivotal in stretching or compressing vectors and is essential for creating parallel vectors.

The scalar multiplication of a vector \( \mathbf{v} \) with a scalar \( k \) is represented as \( k\mathbf{v} \). In our exercise scenario, where the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) equals \( \mathbf{u} \), the implication is that \( \mathbf{u} \) could be obtained through scalar multiplication of \( \mathbf{v} \) by a certain scalar. This concept is integral to understanding how vectors can be transformed while maintaining their direction, a crucial aspect of linear transformations.