Problem 3 - 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 \(\mathbf{u}\) . $$ \mathbf{v}=10 \mathbf{i}+11 \mathbf{j}-2 \mathbf{k}, \quad \mathbf{u}=3 \mathbf{j}+4 \mathbf{k} $$

Step-by-Step Solution

Answer

a. \( \mathbf{v} \cdot \mathbf{u} = 25 \), \(|\mathbf{v}| = 15 \), \(|\mathbf{u}| = 5 \) b. \( \cos \theta = \frac{1}{3} \) c. Scalar component = \( \frac{5}{3} \) d. proj\(_{\mathbf{v}} \mathbf{u} = \frac{10}{9}\mathbf{i} + \frac{11}{9}\mathbf{j} - \frac{2}{9}\mathbf{k} \)

Step 1: Calculate the Dot Product

To find \( \mathbf{v} \cdot \mathbf{u} \), use the formula: \( \mathbf{v} \cdot \mathbf{u} = v_1u_1 + v_2u_2 + v_3u_3 \). Substituting the given vectors: \( \mathbf{v} = 10\mathbf{i} + 11\mathbf{j} - 2\mathbf{k} \) and \( \mathbf{u} = 3\mathbf{j} + 4\mathbf{k} \). This gives: \( 0 \times 0 + 11 \times 3 + (-2) \times 4 = 0 + 33 - 8 = 25 \).

Step 2: Calculate the Magnitudes

First, find \( |\mathbf{v}| \) using \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \). Substituting: \( |\mathbf{v}| = \sqrt{10^2 + 11^2 + (-2)^2} = \sqrt{100 + 121 + 4} = \sqrt{225} = 15 \). Next, find \( |\mathbf{u}| \) using: \( |\mathbf{u}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5 \).

Step 3: Calculate the Cosine of the Angle

To find the cosine of the angle between \( \mathbf{v} \) and \( \mathbf{u} \), use the formula: \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \). Substituting the values: \( \cos \theta = \frac{25}{15 \times 5} = \frac{25}{75} = \frac{1}{3} \).

Step 4: Find the Scalar Component

The scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \). Substituting the values: \( \frac{25}{15} = \frac{5}{3} \).

Step 5: Calculate the Vector Projection

The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is given by \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2}\right) \mathbf{v} \). Using the values: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{25}{225}\right) \mathbf{v} = \frac{1}{9} (10\mathbf{i} + 11\mathbf{j} - 2\mathbf{k}) = \frac{10}{9}\mathbf{i} + \frac{11}{9}\mathbf{j} - \frac{2}{9}\mathbf{k} \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector calculus, also known as the scalar product. It combines two vectors to return a single number (a scalar). This operation helps in determining the angle between the vectors, amongst other applications. The formula for the dot product of two vectors \( \mathbf{v} \) and \( \mathbf{u} \) is: \[\mathbf{v} \cdot \mathbf{u} = v_1u_1 + v_2u_2 + v_3u_3\]For example, with vectors \( \mathbf{v} = 10 \mathbf{i} + 11 \mathbf{j} - 2 \mathbf{k} \) and \( \mathbf{u} = 3 \mathbf{j} + 4 \mathbf{k} \), the dot product includes only the components that are multiplied in the same direction. Thus, we compute:

• \(10 \times 0 = 0 \)
• \(11 \times 3 = 33 \)
• \((-2) \times 4 = -8 \)

Adding these values gives the dot product \( 33 - 8 = 25 \). The dot product is pivotal because it contributes to understanding not just the relation between two vectors but also their alignment.

Magnitude of a Vector

The magnitude of a vector provides a measure of its length. It is crucial in calculating other vector-related metrics like the cosine of the angle between vectors and projections. The magnitude of a vector \( \mathbf{v} = (v_1, v_2, v_3) \) is found using the Pythagorean theorem:\[|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]For instance, for \( \mathbf{v} = 10 \mathbf{i} + 11 \mathbf{j} - 2 \mathbf{k} \), the magnitude is:\[|\mathbf{v}| = \sqrt{10^2 + 11^2 + (-2)^2} = \sqrt{100 + 121 + 4} = \sqrt{225} = 15\]and for \( \mathbf{u} = 3 \mathbf{j} + 4 \mathbf{k} \):\[|\mathbf{u}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]The magnitude is essentially the "size" or "length" of the vector in space, used to determine how large or small a vector is.

Cosine of the Angle Between Vectors

The cosine of the angle between two vectors helps to illustrate how closely aligned the two vectors are. It varies from -1 to 1, where a value of 1 denotes that the vectors point in exactly the same direction, while -1 indicates they point in exactly opposite directions. Neutral values close to zero suggest orthogonality or perpendicular vectors. This important concept is calculated as follows:\[\cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|}\]Substituting values from our example, the cosine is:\[\cos \theta = \frac{25}{15 \times 5} = \frac{25}{75} = \frac{1}{3}\]This means the angle between vectors \( \mathbf{v} \) and \( \mathbf{u} \) is neither too sharp (acute) nor too wide (obtuse), suggesting that they are not purely aligned or opposite.

Scalar Component of a Vector

The scalar component of a vector \( \mathbf{u} \) in the direction of another vector \( \mathbf{v} \) represents the length of the projection of one vector onto the other. It tells us how much of one vector is in the direction of the other. The formula used is:\[\text{Scalar Component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|}\]Applying this to our case with \( \mathbf{v} \) and \( \mathbf{u} \):\[\text{Scalar Component} = \frac{25}{15} = \frac{5}{3}\]This result signifies that \( \mathbf{u} \) has a portion of its length equivalent to \( \frac{5}{3} \) in the direction of \( \mathbf{v} \). It's a crucial value for geometric interpretations and transformations.

Vector Projection

The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) gives a vector that shows how \( \mathbf{u} \) would look if it were "projected" onto \( \mathbf{v} \), acting like a shadow cast on \( \mathbf{v} \). This operation captures the component of one vector parallel to another. The formula for the vector projection is:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2}\right) \mathbf{v}\]Using our vectors, the projection will be:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{25}{225}\right) \mathbf{v} = \frac{1}{9} (10\mathbf{i} + 11\mathbf{j} - 2\mathbf{k})\]Simplifying this expression, the vector projection becomes:

• \( \frac{10}{9} \mathbf{i} \)
• \( \frac{11}{9} \mathbf{j} \)
• \( -\frac{2}{9} \mathbf{k} \)

This indicates how much of \( \mathbf{u} \) moves in the direction of \( \mathbf{v} \), and is vital in various applied mathematics and physics problems.