Problem 5 - Thomas Calculus

Question

In Exercises \(1-8,\) find a. \(\mathbf{v} \cdot \mathbf{u},|\mathbf{v}|,|\mathbf{u}|\) b. the cosine of the angle between \(\mathbf{v}\) and \(\mathbf{u}\) c. the scalar component of \(\mathbf{u}\) in the direction of \(\mathbf{v}\) d. the vector projv \(\mathbf{u}\) . $$ \mathbf{v}=5 \mathbf{j}-3 \mathbf{k}, \quad \mathbf{u}=\mathbf{i}+\mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

Dot product is 2. Cosine of angle is \( \frac{2}{\sqrt{102}} \). Scalar component is \( \frac{2}{\sqrt{34}} \). Projection is \( \frac{5}{17} \mathbf{j} - \frac{3}{17} \mathbf{k} \).

Step 1: Calculating Dot Product

Determine the dot product of \( \mathbf{v} \cdot \mathbf{u} \) where \( \mathbf{v} = 5\mathbf{j} - 3\mathbf{k} \) and \( \mathbf{u} = \mathbf{i} + \mathbf{j} + \mathbf{k} \). Use the formula for dot product: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). Thus, \( \mathbf{v} \cdot \mathbf{u} = 0\cdot1 + 5\cdot1 + (-3)\cdot1 = 5 - 3 = 2 \).

Step 2: Finding Magnitudes

Calculate the magnitudes \( |\mathbf{v}| \) and \( |\mathbf{u}| \). For \( |\mathbf{v}| \), use \( |\mathbf{v}| = \sqrt{0^2 + 5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \). For \( |\mathbf{u}| \), use \( |\mathbf{u}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).

Step 3: Calculating Cosine of the Angle

Use the dot product and magnitudes to find the cosine of the angle between \( \mathbf{v} \) and \( \mathbf{u} \). Use the formula \( \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \). So, \( \cos(\theta) = \frac{2}{\sqrt{34} \cdot \sqrt{3}} = \frac{2}{\sqrt{102}} \).

Step 4: Scalar Component of \( \mathbf{u} \) in \( \mathbf{v} \) Direction

Find the scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) using the formula: \( \text{scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \). Thus, \( \frac{2}{\sqrt{34}} \).

Step 5: Vector Projection of \( \mathbf{u} \) on \( \mathbf{v} \)

Calculate the projection of \( \mathbf{u} \) on \( \mathbf{v} \) with \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v} \). Substitute the known values to get \( \left( \frac{2}{34} \right)(5\mathbf{j} - 3\mathbf{k}) = \frac{10}{34} \mathbf{j} - \frac{6}{34} \mathbf{k} \). Simplify to get \( \frac{5}{17} \mathbf{j} - \frac{3}{17} \mathbf{k} \).

Key Concepts

Vector Magnitude

Calculating the magnitude of a vector is an essential skill when studying vectors. The magnitude represents the "length" or size of a vector and is denoted by vertical bars, such as \(|\mathbf{v}|\). To find the magnitude of a three-dimensional vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), use the formula:
\[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]
In our example, the vector \(\mathbf{v}\) is \(5\mathbf{j} - 3\mathbf{k}\). Compute its magnitude by substituting the coefficients:
\[ |\mathbf{v}| = \sqrt{0^2 + 5^2 + (-3)^2} = \sqrt{34} \]
The same method applies to vector \(\mathbf{u} = \mathbf{i} + \mathbf{j} + \mathbf{k}\), resulting in:
\[ |\mathbf{u}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
This process helps in various calculations, such as determining the angle between vectors or comparing their sizes.

Cosine of Angle Between Vectors

The cosine of the angle between two vectors helps to understand their relative direction. It is determined by using the dot product and the magnitudes of the vectors. The formula is given by:
\[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \]
In our example, the dot product \(\mathbf{v} \cdot \mathbf{u}\) was calculated as 2, and we previously found \(|\mathbf{v}| = \sqrt{34}\) and \(|\mathbf{u}| = \sqrt{3}\). Plug these values into the formula:
\[ \cos(\theta) = \frac{2}{\sqrt{34} \cdot \sqrt{3}} = \frac{2}{\sqrt{102}} \]
Understanding the cosine value reveals the type of angle between the vectors:

• If \(\cos(\theta) = 1\), the vectors are parallel.
• If \(\cos(\theta) = 0\), the vectors are perpendicular.
• If \(\cos(\theta) = -1\), the vectors are directly opposite.

This insight is useful in physics, engineering, and various fields where directionality is key.

Scalar Component

The scalar component of one vector in the direction of another is a crucial concept in vector mathematics. It shows how much of one vector "lies" in the direction of another. To find the scalar component of vector \(\mathbf{u}\) in the direction of \(\mathbf{v}\), use:
\[ \text{scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \]
Given \(\mathbf{v} \cdot \mathbf{u} = 2\) and \(|\mathbf{v}| = \sqrt{34}\), the scalar component becomes:
\[ \frac{2}{\sqrt{34}} \]
This measurement translates the dot product into a scalar value, showing how much of \(\mathbf{u}\) aligns with \(\mathbf{v}\). It is used in physics to determine work done by forces or in projections of vectors.

Vector Projection

The vector projection quantifies the component of one vector in the direction of another vector, represented as another vector. This concept is vital for tasks involving relative positioning and alignment in space.
To project vector \(\mathbf{u}\) onto \(\mathbf{v}\), use the formula:
\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v} \]
Substituting known values \(\mathbf{v} \cdot \mathbf{u} = 2\) and \(|\mathbf{v}|^2 = 34\), along with \(\mathbf{v} = 5\mathbf{j} - 3\mathbf{k}\), we compute:
\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{2}{34} \right)(5\mathbf{j} - 3\mathbf{k}) = \frac{10}{34} \mathbf{j} - \frac{6}{34} \mathbf{k} \]
Upon simplifying, this becomes:
\[ \frac{5}{17} \mathbf{j} - \frac{3}{17} \mathbf{k} \]
Understanding vector projections helps in applications where the relative length along another vector's direction is important, such as shadows on a surface or force components in mechanics.