Problem 7 - 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 \(\mathbf{u}\). $$\mathbf{v}=5 \mathbf{i}+\mathbf{j}, \quad \mathbf{u}=2 \mathbf{i}+\sqrt{17} \mathbf{j}$$

Step-by-Step Solution

Answer

The dot product is \(10 + \sqrt{17}\), magnitudes are \(\sqrt{26}\) and \(\sqrt{21}\). The cosine, scalar, and projection depend on these values.

Step 1: Dot Product

To find the dot product \( \mathbf{v} \cdot \mathbf{u} \), we use the formula:\[ \mathbf{v} \cdot \mathbf{u} = v_1 u_1 + v_2 u_2 \]Given \( \mathbf{v} = 5 \mathbf{i} + \mathbf{j} \) and \( \mathbf{u} = 2 \mathbf{i} + \sqrt{17} \mathbf{j} \), the components are:- \( v_1 = 5, v_2 = 1 \)- \( u_1 = 2, u_2 = \sqrt{17} \)Substitute these into the formula:\[ 5 \times 2 + 1 \times \sqrt{17} = 10 + \sqrt{17} \]

Step 2: Magnitudes of Vectors

To find \( |\mathbf{v}| \) and \( |\mathbf{u}| \), use the formula for the magnitude of a vector:\[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2} \text{ and } |\mathbf{u}| = \sqrt{u_1^2 + u_2^2} \]Calculate \( |\mathbf{v}| \):\[ |\mathbf{v}| = \sqrt{5^2 + 1^2} = \sqrt{26} \]Calculate \( |\mathbf{u}| \):\[ |\mathbf{u}| = \sqrt{2^2 + (\sqrt{17})^2} = \sqrt{21} \]

Step 3: Cosine of the Angle

The cosine of the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{u} \) is given by:\[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \]Using the dot product from Step 1 and magnitudes from Step 2:\[ \cos(\theta) = \frac{10 + \sqrt{17}}{\sqrt{26} \times \sqrt{21}} \]

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

The scalar component is given by:\[ \text{scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \]Using \( \mathbf{v} \cdot \mathbf{u} = 10 + \sqrt{17} \) and \( |\mathbf{v}| = \sqrt{26} \):\[ \text{scalar component} = \frac{10 + \sqrt{17}}{\sqrt{26}} \]

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

The vector projection of \( \mathbf{u} \) on \( \mathbf{v} \) is given by:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v} \]First, compute \( \mathbf{v} \cdot \mathbf{v} = 5^2 + 1^2 = 26 \),then:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{10 + \sqrt{17}}{26} \right) (5 \mathbf{i} + \mathbf{j}) \]

Key Concepts

Dot Product

The dot product is a powerful operation between two vectors, resulting in a scalar value, which essentially represents the magnitude of one vector in the direction of another. To calculate the dot product, you multiply the corresponding components of each vector and sum the results.
For vectors \( \mathbf{v} = 5 \mathbf{i} + \mathbf{j} \) and \( \mathbf{u} = 2 \mathbf{i} + \sqrt{17} \mathbf{j} \), the dot product \( \mathbf{v} \cdot \mathbf{u} \) is calculated as:

• Multiply \( v_1 \) by \( u_1 \): \( 5 \times 2 = 10 \)
• Multiply \( v_2 \) by \( u_2 \): \( 1 \times \sqrt{17} = \sqrt{17} \)
• Add these results: \( 10 + \sqrt{17} \)

Resulting in a dot product value of \( 10 + \sqrt{17} \). Understanding the dot product is crucial for finding angles between vectors and projections.

Magnitude of a Vector

The magnitude of a vector, also known as its length or norm, is calculated using the Pythagorean theorem and provides a sense of how long the vector is in a coordinate system.
For any vector \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} \), its magnitude is determined using:

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

For \( \mathbf{v} \), with components \( 5 \) and \( 1 \), calculate:

• \( |\mathbf{v}| = \sqrt{5^2 + 1^2} = \sqrt{26} \)

And for \( \mathbf{u} \), with components \( 2 \) and \( \sqrt{17} \):

• \( |\mathbf{u}| = \sqrt{2^2 + (\sqrt{17})^2} = \sqrt{21} \)

Magnitude tells us how "big" or "small" a vector is and is crucial for normalizing vectors and calculating angles.

Cosine of the Angle

The cosine of the angle \( \theta \) between two vectors \( \mathbf{v} \) and \( \mathbf{u} \) reveals how aligned or perpendicular the vectors are with respect to each other. It's found using the formula:

• \( \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}||\mathbf{u}|} \)

By inputting our previous results:

• \( \mathbf{v} \cdot \mathbf{u} = 10 + \sqrt{17} \)
• \( |\mathbf{v}| = \sqrt{26}, |\mathbf{u}| = \sqrt{21} \)

We find:

• \( \cos(\theta) = \frac{10 + \sqrt{17}}{\sqrt{26} \times \sqrt{21}} \)

This ratio tells us the relative orientation of the vectors. A cosine of 1 indicates they’re perfectly aligned; -1, perfectly opposed; and 0, perfectly perpendicular.

Scalar Component

The scalar component of one vector in the direction of another gives the "shadow" or "footprint" of that vector along the direction of the other. This concept is pivotal when you need to find how much of a vector's magnitude points in a specific direction.
It is represented by:

• \( \text{scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \)

Plugging in our prior calculations:

• \( \mathbf{v} \cdot \mathbf{u} = 10 + \sqrt{17} \)
• \( |\mathbf{v}| = \sqrt{26} \)

We calculate:

• \( \text{scalar component} = \frac{10 + \sqrt{17}}{\sqrt{26}} \)

This scalar gives a simplified measure of how much \( \mathbf{u} \) extends in the direction of \( \mathbf{v} \).

Vector Projection

The vector projection is the representation of one vector onto another, showing how one vector linearly "falls" onto another. It's a vector itself, which essentially tells us how much of the vector \( \mathbf{u} \) is pointing in the direction of \( \mathbf{v} \).
For vector projection:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v} \)

With \( \mathbf{v} \cdot \mathbf{u} = 10 + \sqrt{17} \) and \( \mathbf{v} \cdot \mathbf{v} = 26 \), the expression becomes:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{10 + \sqrt{17}}{26} \right) (5 \mathbf{i} + \mathbf{j}) \)

Projection transforms the multidimensional relationship into a simple linear perspective, offering insights into problems like force distributions and directionality in physics and engineering.