Problem 4 - University Calculus: Early Transcendentals

Question

Find a. \(\quad \mathbf{v} \cdot \mathbf{u},|\mathbf{v}|,|\mathbf{u}|\) 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 proje \(\mathbf{u}\). $$\mathbf{v}=2 \mathbf{i}+10 \mathbf{j}-11 \mathbf{k}, \quad \mathbf{u}=2 \mathbf{i}+2 \mathbf{j}+\mathbf{k}$$

Step-by-Step Solution

Answer

a. Dot product: 13, |v| = 15, |u| = 3. b. Cosine of angle: 13/45. c. Scalar component: 13/15. d. Vector projection: (26/225)i + (130/225)j - (143/225)k.

Step 1: Calculate Dot Product

The dot product \( \mathbf{v} \cdot \mathbf{u} \) is calculated using the formula: \( \mathbf{v} \cdot \mathbf{u} = v_x u_x + v_y u_y + v_z u_z \). Substituting the given vectors, we have: \( (2)(2) + (10)(2) + (-11)(1) = 4 + 20 - 11 = 13 \). Thus, the dot product is 13.

Step 2: Find Magnitude of Vectors

To find the magnitudes \( |\mathbf{v}| \) and \( |\mathbf{u}| \), use the formula: \( |\mathbf{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \). For \( \mathbf{v} = 2 \mathbf{i} + 10 \mathbf{j} - 11 \mathbf{k} \), \( |\mathbf{v}| = \sqrt{2^2 + 10^2 + (-11)^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15 \). For \( \mathbf{u} = 2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \), \( |\mathbf{u}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \).

Step 3: Calculate Cosine of Angle

The cosine of the angle between \( \mathbf{v} \) and \( \mathbf{u} \) is given by \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} = \frac{13}{15 \times 3} = \frac{13}{45} \).

Step 4: Find Scalar Component

The scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by the projection formula: \( \text{Scalar Component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} = \frac{13}{15} \).

Step 5: Compute Vector Projection

The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is calculated using: \( \text{Proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v} = \left(\frac{13}{15^2} \right) \mathbf{v} = \left(\frac{13}{225} \right) (2 \mathbf{i} + 10 \mathbf{j} - 11 \mathbf{k}) = \left(\frac{26}{225} \right) \mathbf{i} + \left(\frac{130}{225} \right) \mathbf{j} - \left(\frac{143}{225} \right) \mathbf{k} \).

Key Concepts

Dot Product

In the world of vector calculus, the dot product is a fundamental operation. It combines two vectors to produce a scalar, helping us understand various properties such as angles between vectors. Given vectors \( \mathbf{v} = 2 \mathbf{i} + 10 \mathbf{j} - 11 \mathbf{k} \) and \( \mathbf{u} = 2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \), the dot product \( \mathbf{v} \cdot \mathbf{u} \) can be computed using the formula:

• \( \mathbf{v} \cdot \mathbf{u} = v_x u_x + v_y u_y + v_z u_z \)
• Substituting the values from the vectors, we have: \( (2 \cdot 2) + (10 \cdot 2) + (-11 \cdot 1) = 4 + 20 - 11 = 13 \)

The result of this calculation is the scalar value of 13. The dot product is not only useful in calculating angles but is also a measure of how much one vector extends in the direction of another.

Magnitude of Vectors

The magnitude of a vector, also known as its length or norm, is a measure of the vector's size. You can imagine it as the vector's distance from the origin in a vector space. For any vector \( \mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k} \), the magnitude is found using the formula:

• \( |\mathbf{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \)

To find the magnitudes of \( \mathbf{v} \) and \( \mathbf{u} \):

• For \( \mathbf{v} = 2 \mathbf{i} + 10 \mathbf{j} - 11 \mathbf{k} \):\( |\mathbf{v}| = \sqrt{2^2 + 10^2 + (-11)^2} = \sqrt{225} = 15 \)
• For \( \mathbf{u} = 2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \):\( |\mathbf{u}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3 \)

Knowing the magnitude of vectors is crucial in many applications, such as physics and engineering, where vector size and direction heavily influence outcomes.

Cosine of Angle Between Vectors

The cosine of the angle between two vectors gives us valuable information about their relative direction. It tells us whether the vectors are pointing in more or less the same direction or in opposite directions. The formula used to find the cosine of the angle \( \theta \) between vectors is:

• \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \)
• Substituting the prior results, we have \( \cos \theta = \frac{13}{15 \cdot 3} = \frac{13}{45} \)

A positive cosine value indicates that the angle is less than 90 degrees, which means the vectors are more aligned than not. A cosine value that is close to zero indicates orthogonality, meaning the vectors are perpendicular.

Scalar Component

The scalar component of one vector in the direction of another helps determine how much of the first vector is aligned with the second. This can be particularly useful in applications like physics, where forces along a line are a common focus. The formula for the scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is:

• \( \text{Scalar Component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \)
• Substituting values:\( \frac{13}{15} \)

This value represents the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) in terms of magnitude, signifying how much of \( \mathbf{u} \) lies in the direction of \( \mathbf{v} \).

Vector Projection

The vector projection tells us not only how much of one vector points in the direction of another but also provides a vector with this direction and magnitude. It is, in essence, the component of one vector along another. The formula for projecting \( \mathbf{u} \) onto \( \mathbf{v} \) is:

• \( \text{Proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2}\right) \mathbf{v} \)
• Substitute the values, \( \frac{13}{15^2} \) gives \( \left(\frac{13}{225}\right) \mathbf{v} \)
• Expand it with \( \mathbf{v} = 2 \mathbf{i} + 10 \mathbf{j} - 11 \mathbf{k} \)
• This results in: \( \left(\frac{26}{225}\right) \mathbf{i} + \left(\frac{130}{225}\right) \mathbf{j} - \left(\frac{143}{225}\right) \mathbf{k} \)

Vector projections are frequently used when resolving forces or velocities in applied mathematics, especially in physics and engineering.