Problem 15 - Thomas Calculus

Question

Direction angles and direction cosines The direction angles \(\alpha, \beta,\) and \(\gamma\) of a vector \(\mathbf{v}=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\) are defined as follows: \(\alpha\) is the angle between \(\mathbf{v}\) and the positive \(x\) -axis \((0 \leq \alpha \leq \pi)\) \(\beta\) is the angle between \(\mathbf{v}\) and the positive \(y\) -axis \((0 \leq \beta \leq \pi)\) \(\gamma\) is the angle between \(\mathbf{v}\) and the positive \(z\) -axis \((0 \leq \gamma \leq \pi)\) a. Show that $$\cos \alpha=\frac{a}{|\mathbf{v}|}, \quad \cos \beta=\frac{b}{|\mathbf{v}|}, \quad \cos \gamma=\frac{c}{|\mathbf{v}|}$$ and \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 .\) These cosines are called the direction cosines of \(\mathbf{v} .\) b. Unit vectors are built from direction cosines Show that if \(\mathbf{v}=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\) is a unit vector, then \(a, b,\) and \(c\) are the direction cosines of \(\mathbf{v} .\)

Step-by-Step Solution

Answer

Direction cosines \(\cos \alpha, \cos \beta, \cos \gamma\) are \(\frac{a}{|\mathbf{v}|}, \frac{b}{|\mathbf{v}|}, \frac{c}{|\mathbf{v}|}\), summing to 1, while unit vectors have components as direction cosines.

Step 1: Understanding Direction Cosines

We need to find expressions for the cosines of the angles between the vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) and each of the coordinate axes. These are the direction cosines denoted by \( \cos \alpha, \cos \beta, \) and \( \cos \gamma \) for the angles \( \alpha, \beta, \) and \( \gamma \) respectively.

Step 2: Calculating Magnitude of \(\mathbf{v}\)

Calculate the magnitude of the vector \( \mathbf{v} \). The magnitude \(|\mathbf{v}|\) is given by the formula: \[|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}.\]

Step 3: Find Direction Cosine \(\cos \alpha\)

The direction cosine \( \cos \alpha \) is the cosine of the angle between the vector \( \mathbf{v} \) and the x-axis. By definition, it can be expressed as \[\cos \alpha = \frac{a}{|\mathbf{v}|} = \frac{a}{\sqrt{a^2 + b^2 + c^2}}.\]

Step 4: Find Direction Cosine \(\cos \beta\)

The direction cosine \( \cos \beta \) is the cosine of the angle between the vector \( \mathbf{v} \) and the y-axis. By definition, it can be expressed as \[\cos \beta = \frac{b}{|\mathbf{v}|} = \frac{b}{\sqrt{a^2 + b^2 + c^2}}.\]

Step 5: Find Direction Cosine \(\cos \gamma\)

The direction cosine \( \cos \gamma \) is the cosine of the angle between the vector \( \mathbf{v} \) and the z-axis. By definition, it can be expressed as \[\cos \gamma = \frac{c}{|\mathbf{v}|} = \frac{c}{\sqrt{a^2 + b^2 + c^2}}.\]

Step 6: Verify Direction Cosines Identity

Show that the sum of the squares of the direction cosines equals one:\[\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \left(\frac{a}{|\mathbf{v}|}\right)^2 + \left(\frac{b}{|\mathbf{v}|}\right)^2 + \left(\frac{c}{|\mathbf{v}|}\right)^2 = \frac{a^2 + b^2 + c^2}{a^2 + b^2 + c^2} = 1.\]

Step 7: Show \(\mathbf{v}\) as Unit Vector Uses Direction Cosines

A unit vector has a magnitude of 1. So if \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is a unit vector, \[\sqrt{a^2 + b^2 + c^2} = 1.\]Substitute \( a, b, c \) into the expressions for the direction cosines, each will then simplify to the components themselves, hence \( a, b, \) and \( c \) are the direction cosines.

Key Concepts

Direction Cosines

Direction cosines are an essential concept when understanding vector orientation in space. They help in expressing how a vector direction aligns with the coordinate axes. For any vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the direction cosines \( \cos \alpha, \cos \beta, \) and \( \cos \gamma \) represent the cosine of angles between the vector \( \mathbf{v} \) and the \( x, y, \) and \( z \) axes respectively.
To find these, you calculate as follows:

• \( \cos \alpha = \frac{a}{|\mathbf{v}|} = \frac{a}{\sqrt{a^2 + b^2 + c^2}} \)
• \( \cos \beta = \frac{b}{|\mathbf{v}|} = \frac{b}{\sqrt{a^2 + b^2 + c^2}} \)
• \( \cos \gamma = \frac{c}{|\mathbf{v}|} = \frac{c}{\sqrt{a^2 + b^2 + c^2}} \)

A very important aspect of direction cosines is their identity: \[\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\]This tells us that the vector's orientation is perfectly described within this 3D space.

Vector Mathematics

Vector mathematics is all about understanding quantities that have both magnitude and direction. In the 3D coordinate system, vectors are typically expressed in terms of their components along the standard basis vectors: \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \). Thus, any vector \( \mathbf{v} \) can be written as \( a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \).
The magnitude (or length) of a vector \( \mathbf{v} \) is given by the formula \(|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}\). This helps determine how long the vector is in the spatial context.
Additionally, vectors have direction, which is often described using angles with the coordinate axes, leading us to direction angles and cosines. Calculating dot products and cross products are also methods used to find angles and relationships between vectors in vector mathematics.

Unit Vectors

A unit vector is a vector that has a magnitude of exactly 1. They are primarily used to indicate direction. For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the requirements for it to be a unit vector is \( \sqrt{a^2 + b^2 + c^2} = 1 \). This ensures the vector solely represents direction.
If a vector is a unit vector, each of its components \( a, b, \) and \( c \) also serves as the direction cosines with the respective coordinate axes. This allows transformation of any vector into a unit vector through normalization, calculated by dividing each component by the vector's magnitude.
Unit vectors are foundational in physics and engineering as they help define precise directions in scalar fields and rotational dynamics.