Problem 17 - Essential Calculus Early Transcendentals

Question

\(15-17\) Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) $$\mathbf{a}=4 \mathbf{i}-3 \mathbf{j}+\mathbf{k}, \quad \mathbf{b}=2 \mathbf{i}-\mathbf{k}$$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 53°.

Step 1: Find the Dot Product of Vectors

Calculate the dot product of \( \mathbf{a} \) and \( \mathbf{b} \), which is defined as: \[\mathbf{a} \cdot \mathbf{b} = (4)(2) + (-3)(0) + (1)(-1)\] Simplify to find the dot product: \[\mathbf{a} \cdot \mathbf{b} = 8 + 0 - 1 = 7\]

Step 2: Calculate the Magnitude of Each Vector

Find the magnitude of \( \mathbf{a} \) using: \[|\mathbf{a}| = \sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26}\]Find the magnitude of \( \mathbf{b} \) using: \[|\mathbf{b}| = \sqrt{2^2 + 0^2 + (-1)^2} = \sqrt{4 + 0 + 1} = \sqrt{5}\]

Step 3: Use the Cosine Formula for Angle Between Vectors

The formula to find the angle \( \theta \) between two vectors is given by: \[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}\]Substitute the known values: \[\cos \theta = \frac{7}{\sqrt{26} \cdot \sqrt{5}}\] This simplifies to: \[\cos \theta = \frac{7}{\sqrt{130}}\]

Step 4: Find the Exact Value of the Angle

Find \( \theta \) by taking the inverse cosine:\[\theta = \cos^{-1}\left(\frac{7}{\sqrt{130}}\right)\]

Step 5: Approximate the Angle to the Nearest Degree

Use a calculator to determine \( \theta \) and approximate it to the nearest degree:\( \theta \approx 53^\circ \).

Key Concepts

Dot Product

The dot product, often known as the scalar product, is a way of multiplying two vectors to receive a scalar, or real number. It essentially measures how much one vector extends in the direction of another. To compute the dot product of two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), use the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

For example, for \( \mathbf{a} = 4 \mathbf{i} - 3 \mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = 2 \mathbf{i} - \mathbf{k} \), you calculate the dot product by multiplying and adding corresponding components: - \((4)(2) = 8\) - \((-3)(0) = 0\) - \((1)(-1) = -1\) Thus, the dot product is \( 8 + 0 - 1 = 7 \). It's a simple yet powerful tool for finding angles between vectors.

Vector Magnitude

Vector magnitude, sometimes referred to as vector length or norm, represents how long a vector is. Imagine it as the distance from the origin to the point defined by the vector's coordinates in space. It's crucial when comparing the size and direction of different vectors. To find the magnitude of a vector \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} \), we use the formula:

• \(|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\)

For \( \mathbf{a} = 4 \mathbf{i} - 3 \mathbf{j} + \mathbf{k} \), the magnitude is: - \( |\mathbf{a}| = \sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26} \) For \( \mathbf{b} = 2 \mathbf{i} - \mathbf{k} \), it is: - \( |\mathbf{b}| = \sqrt{2^2 + 0^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \) These magnitudes are used later to find the angle between the vectors.

Cosine Formula

The cosine formula is used to find the angle between two vectors, which is critical in various applications like physics and engineering. This is done by relating the dot product of two vectors to the magnitudes of those vectors. The formula for the cosine of the angle \( \theta \) between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \)

In our example, with the dot product calculated as 7 and magnitudes \( \sqrt{26} \) from vector \( \mathbf{a} \) and \( \sqrt{5} \) from vector \( \mathbf{b} \): - \( \cos \theta = \frac{7}{\sqrt{26} \cdot \sqrt{5}} = \frac{7}{\sqrt{130}} \) This formula is crucial in gradual steps as we work towards determining the exact angle between the vectors.

Inverse Cosine

When you have the cosine of an angle, the inverse cosine (sometimes called arc cosine) is used to find the actual angle in radians or degrees. It "undoes" the cosine function, essentially finding the angle that has the given cosine value. Once you have \( \cos \theta = \frac{7}{\sqrt{130}} \), find \( \theta \) using the inverse cosine: - \( \theta = \cos^{-1}\left(\frac{7}{\sqrt{130}}\right) \) This process allows you to derive the angle between the two vectors. With a calculator, you approximate \( \theta \) to be about 53 degrees. Inverse cosine is a reverse function calculation, enabling the translation from a ratio back to an angle, helpful in many vector and trigonometric problems.