Problem 53 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the value of \(k\) such that the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal. $$\begin{array}{l} \mathbf{u}=\mathbf{i}+4 \mathbf{j} \\ \mathbf{v}=7 k \mathbf{i}-5 \mathbf{j} \end{array}$$

Step-by-Step Solution

Answer

The value of \(k\) such that the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal is \(k = 20 / 7\).

Step 1: Representation of vectors

The vectors \(\mathbf{u}\) and \(\mathbf{v}\) can be represented as \(\mathbf{u}=\mathbf{i}+4 \mathbf{j}\) and \(\mathbf{v}=7 k \mathbf{i}-5 \mathbf{j}\) respectively.

Step 2: Dot Product of vectors

The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\| \cos \theta\), where \(\theta\) is the angle between \(\mathbf{u}\) and \(\mathbf{v}\). We know that if \(\theta=90^{\circ}\), then \(\mathbf{u} \cdot \mathbf{v}=0\)

Step 3: Calculation for k

So we take the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\) and equate to zero to find the value of \(k\). \(\mathbf{u} \cdot \mathbf{v}=(\mathbf{i}+4 \mathbf{j}) \cdot (7 k \mathbf{i}-5 \mathbf{j})=0\).\nOn simplifying we get \(7k -20 = 0\). Solving for \(k\) we get \(k = 20 / 7\)

Key Concepts

Dot Product of Vectors

The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar quantity. It is a fundamental operation in vector algebra and is not only easy to calculate but also has important geometrical interpretations. Given two vectors \(\mathbf{a}\) and \(\mathbf{b}\), the dot product is calculated by multiplying corresponding components of the vectors and then summing those products together. The formula for the dot product is given by \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \cdots + a_nb_n\), where \(a_i\) and \(b_i\) are components of the vectors \(\mathbf{a}\) and \(\mathbf{b}\), respectively.

The geometric interpretation of the dot product involves the angle \(\theta\) between the vectors. In this interpretation, the dot product equals the product of the magnitudes of the two vectors and the cosine of the angle between them: \(\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\| \cos \theta\). This characteristic leads us directly to understanding vector orthogonality—a concept you'll see is core to many vector problems.

Vector Orthogonality

When two vectors are described as orthogonal, it means they are perpendicular to each other. This is a key concept in vector algebra and has many applications in different fields such as computer graphics, physics, and engineering.

How do we determine if vectors are orthogonal? The answer lies in the dot product. If the dot product of two vectors is zero, \(\mathbf{a} \cdot \mathbf{b} = 0\), the vectors are orthogonal, implying they form a \(90^\circ\) angle with each other. In the given exercise, the vectors \(\mathbf{u}\) and \(\mathbf{v}\) will be orthogonal when their dot product equals zero. This simple yet powerful property simplifies the process of finding such vectors or verifying their orthogonality.

Solving for Scalar Multiple in Vectors

Often in vector algebra, we are tasked with finding a scalar multiple that satisfies a given condition. The term 'scalar multiple' refers to a quantity by which we multiply a vector, affecting its magnitude but not its direction. In the exercise, we're looking for a scalar \(k\) such that when \(\mathbf{v}\) is multiplied by \(7k\), the resulting vector is orthogonal to \(\mathbf{u}\).

To find \(k\), we take the dot product of the two vectors and set it equal to zero because we know from the property of orthogonality that this must be true for perpendicular vectors. Following the expression \(\mathbf{u} \cdot \mathbf{v} = 0\), we plug in the components of the vectors, including the unknown scalar multiple, and solve for it. This leads us directly to the scalar \(k\) that will make the two vectors orthogonal. Understanding how to manipulate scalar multiples in vectors is crucial for tasks such as this and forms an integral part of learning about vector spaces and transformations.