Problem 6 - 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 \(\mathbf{u}\). $$\mathbf{v}=-\mathbf{i}+\mathbf{j}, \quad \mathbf{u}=\sqrt{2} \mathbf{i}+\sqrt{3} \mathbf{j}+2 \mathbf{k}$$

Step-by-Step Solution

Answer

a. \(-\sqrt{2} + \sqrt{3}, \sqrt{2}, 3\); b. \(\frac{-\sqrt{2} + \sqrt{3}}{3\sqrt{2}}\); c. \(\frac{-\sqrt{2} + \sqrt{3}}{\sqrt{2}}\); d. Projected vector obtained is: \(\left( \frac{\sqrt{2} - \sqrt{3}}{2} \mathbf{i} + \frac{-\sqrt{2} + \sqrt{3}}{2} \mathbf{j} \right)\).

Step 1: Calculate the Dot Product 𝒗 · 𝒖

The dot product of two vectors \( \mathbf{v} = a_1 \mathbf{i} + b_1 \mathbf{j} + c_1 \mathbf{k} \) and \( \mathbf{u} = a_2 \mathbf{i} + b_2 \mathbf{j} + c_2 \mathbf{k} \) is given by \( \mathbf{v} \cdot \mathbf{u} = a_1 a_2 + b_1 b_2 + c_1 c_2 \). For \( \mathbf{v} = -\mathbf{i} + \mathbf{j} \) and \( \mathbf{u} = \sqrt{2}\mathbf{i} + \sqrt{3}\mathbf{j} + 2\mathbf{k} \), we find:\( \mathbf{v} \cdot \mathbf{u} = (-1)(\sqrt{2}) + (1)(\sqrt{3}) + (0)(2) = -\sqrt{2} + \sqrt{3} \).

Step 2: Calculate the Magnitude of 𝒗

The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is given by \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \). For \( \mathbf{v} = -\mathbf{i} + \mathbf{j} \), we have:\( |\mathbf{v}| = \sqrt{(-1)^2 + (1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \).

Step 3: Calculate the Magnitude of 𝒖

The magnitude of \( \mathbf{u} = \sqrt{2}\mathbf{i} + \sqrt{3}\mathbf{j} + 2\mathbf{k} \) is:\( |\mathbf{u}| = \sqrt{(\sqrt{2})^2 + (\sqrt{3})^2 + 2^2} = \sqrt{2 + 3 + 4} = \sqrt{9} = 3 \).

Step 4: Calculate the Cosine of the Angle between 𝒗 and 𝒖

The cosine of the angle \( \theta \) between two vectors is given by:\( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \).Substitute the values we found:\( \cos \theta = \frac{-\sqrt{2} + \sqrt{3}}{\sqrt{2} \cdot 3} = \frac{-\sqrt{2} + \sqrt{3}}{3\sqrt{2}} \).

Step 5: Calculate the Scalar Component of 𝒖 in the Direction of 𝒗

The scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by:\( \text{Scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \).Using our results from previous steps:\( \text{Scalar component} = \frac{-\sqrt{2} + \sqrt{3}}{\sqrt{2}} \).

Step 6: Calculate the Vector Projection of 𝒖 onto 𝒗

The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is determined by:\( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v} \).First, find \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} = \frac{-\sqrt{2} + \sqrt{3}}{2} \).Then, multiply with \( \mathbf{v} = -\mathbf{i} + \mathbf{j} \):\( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{-\sqrt{2} + \sqrt{3}}{2} \right)(-\mathbf{i} + \mathbf{j}) \).Distribute this scalar:\( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( -\frac{-\sqrt{2} + \sqrt{3}}{2} \mathbf{i} + \frac{-\sqrt{2} + \sqrt{3}}{2} \mathbf{j} \right) \).

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is about calculating its length or size in space. This is done using a formula that involves the vector's components. For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude is found using:

• \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \)

To see this in action, consider the vector \( \mathbf{v} = -\mathbf{i} + \mathbf{j} \). Here:

• \( a = -1 \), \( b = 1 \), and \( c = 0 \)
• Magnitude: \( \sqrt{(-1)^2 + (1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \)

This method gives us a clear measure of how far the vector "travels," and it's particularly useful in determining distances and comparing vector sizes.

Calculating the magnitude is a foundational step in many vector operations, such as finding directions or projecting vectors. It simplifies further steps like understanding angles or projecting one vector onto another.

Cosine of the Angle Between Vectors

The cosine of the angle between two vectors reveals how similar or different their directions are. For two vectors \( \mathbf{v} \) and \( \mathbf{u} \), it's computed by dividing their dot product by the product of their magnitudes. The formula is:

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

Calculating this provides insight into whether the vectors point in the same, opposite, or perpendicular directions.
For example, when storing previous calculations:

• \( \mathbf{v} \cdot \mathbf{u} = -\sqrt{2} + \sqrt{3} \)
• \( |\mathbf{v}| = \sqrt{2} \)
• \( |\mathbf{u}| = 3 \)

It yields:

• \( \cos \theta = \frac{-\sqrt{2} + \sqrt{3}}{3\sqrt{2}} \)

A cosine approaching 1 indicates vectors moving in very similar directions.
The range from -1 to 1 also helps identify whether they are more perpendicular or in exact opposite directions.

Scalar Component of a Vector

When dealing with vectors, sometimes we're interested in how much of one vector goes in the direction of another. This is the scalar component, which measures that amount. It is found by taking the dot product of the vectors and dividing it by the magnitude of the vector giving direction:

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

This component is particularly useful when projecting vectors or understanding their influence in certain directions. In practice, with values:

• \( \mathbf{v} \cdot \mathbf{u} = -\sqrt{2} + \sqrt{3} \)
• \( |\mathbf{v}| = \sqrt{2} \)

You get:

• \( \text{Scalar component} = \frac{-\sqrt{2} + \sqrt{3}}{\sqrt{2}} \)

This calculation is crucial for applications such as balancing forces or determining the "useful" vector quantity in physical models.

Vector Projection

Vector projection is like casting the shadow of one vector onto another vector. It gives both a direction and a magnitude along that direction. For vector \( \mathbf{u} \) onto \( \mathbf{v} \), the projection formula is:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v} \)

This step ensures we're looking at the "part" of one vector that aligns with another. Here's how you apply it:1. Calculate \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \)2. Multiply this by \( \mathbf{v} \)For our example:

• \( \mathbf{v} \cdot \mathbf{u} = -\sqrt{2} + \sqrt{3} \), \( |\mathbf{v}|^2 = 2 \)
• \( \frac{-\sqrt{2} + \sqrt{3}}{2} \)
• Projection: \( \left( \frac{-\sqrt{2} + \sqrt{3}}{2} \right)(-\mathbf{i} + \mathbf{j}) \)

This result is key in applications such as physics and engineering, where determining the influence of directions or simplifying vector analyses is often needed.