Problem 28 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the angle \(\theta\) between the vectors. $$\begin{aligned} &\mathbf{u}=4 \mathbf{j}\\\ &\mathbf{v}=-9 \mathbf{i} \end{aligned}$$

Step-by-Step Solution

Answer

The angle \(\theta\) between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is 90 degrees.

Step 1: Express/Write down the vectors

The vectors \(\mathbf{u}\) = \(4 \mathbf{j}\) and \(\mathbf{v}\) = \(-9 \mathbf{i}\) can be written in \(\mathbf{i}\), \(\mathbf{j}\) notation as \(\mathbf{u} = 0 \mathbf{i}+ 4 \mathbf{j}\) and \(\mathbf{v} = -9 \mathbf{i}+ 0 \mathbf{j}\) respectively.

Step 2: Compute the dot product of the vectors

The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\), is given as \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos(\theta) \). So, the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) is \( \mathbf{u} \cdot \mathbf{v} = 0*(-9) + 4*0 = 0.\)

Step 3: Calculate the magnitudes of the given vectors

The magnitude of a vector \(\mathbf{a}\) = \(a_i \mathbf{i} + a_j \mathbf{j}\) is given by \(|\mathbf{a}| = \sqrt{a_i^2 + a_j^2}\). Therefore, the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\) are \(|\mathbf{u}| = \sqrt{0^2 + 4^2}= 4\) and \(|\mathbf{v}| = \sqrt{(-9)^2 + 0^2} = 9\) respectively.

Step 4: Find the angle between the vectors

The angle \(\theta\) between \(\mathbf{u}\) and \(\mathbf{v}\) can be calculated with the formula \(\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\). Substituting the dot product and magnitudes from previous steps, the cosine of the angle is \(\cos(\theta) = \frac{0}{4*9} = 0\). This gives \(\theta = \cos^{-1}(0) = 90^{\circ}\).

Key Concepts

Dot Product

The dot product is a way to multiply two vectors that results in a scalar. This operation is also known as the "scalar product." For two-dimensional vectors, like those given in the exercise

1. \(\mathbf{u} = 0 \mathbf{i} + 4 \mathbf{j}\) and \(\mathbf{v} = -9 \mathbf{i} + 0 \mathbf{j}\),
2. the dot product \(\mathbf{u} \cdot \mathbf{v} = (0)(-9) + (4)(0) = 0\).

This formula works by multiplying the corresponding components of each vector.

• The result indicates that the vectors are orthogonal, or at right angles to each other, if the dot product is zero.
• In general, the dot product can also be calculated using the formula \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos(\theta) \), where \(\theta\) is the angle between the vectors.

In this case, because the result is zero, it confirms that the angle between \(\mathbf{u}\) and \(\mathbf{v}\) is 90 degrees.

Magnitude of a Vector

The magnitude of a vector can be thought of as its length or size, and it is always a non-negative scalar. For a two-dimensional vector, say \(\mathbf{a} = a_i \mathbf{i} + a_j \mathbf{j}\), the magnitude \(|\mathbf{a}|\) is calculated using the formula

• \(|\mathbf{a}| = \sqrt{a_i^2 + a_j^2}\).

Let's see how it applies to the vectors from the exercise:

1. For \(\mathbf{u} = 0 \mathbf{i} + 4 \mathbf{j}\), the magnitude is \(|\mathbf{u}| = \sqrt{0^2 + 4^2} = 4\).
2. For \(\mathbf{v} = -9 \mathbf{i} + 0 \mathbf{j}\), the magnitude is \(|\mathbf{v}| = \sqrt{(-9)^2 + 0^2} = 9\).

These magnitudes are useful in computing further vector operations, such as finding angles between vectors.

Trigonometric Functions

Trigonometric functions are fundamental in connecting the geometric and algebraic sides of vectors.
To find the angle \(\theta\) between two vectors, the cosine function is particularly useful. It relates the dot product and magnitudes to the angle with the formula:

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

Substituting the values from the exercise gives:

1. The dot product \(\mathbf{u} \cdot \mathbf{v} = 0\).
2. The magnitudes are \(|\mathbf{u}| = 4\) and \(|\mathbf{v}| = 9\).

Thus, \(\cos(\theta) = \frac{0}{4 \times 9} = 0\).
Using the inverse cosine function \(\cos^{-1}\), we find that \(\theta = \cos^{-1}(0) = 90^{\circ}\).
This confirms the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are indeed perpendicular.