Problem 1 - 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 \(\mathbf{u}\). $$\mathbf{v}=2 \mathbf{i}-4 \mathbf{j}+\sqrt{5} \mathbf{k}, \quad \mathbf{u}=-2 \mathbf{i}+4 \mathbf{j}-\sqrt{5} \mathbf{k}$$

Step-by-Step Solution

Answer

a. -25, 5, 5; b. -1; c. -5; d. \(-2\mathbf{i} + 4\mathbf{j} - \sqrt{5}\mathbf{k}\).

Step 1: Calculate the Dot Product of Vectors

To find \( \mathbf{v} \cdot \mathbf{u} \), calculate the product of each component of \( \mathbf{v} \) and \( \mathbf{u} \), then sum the results. This follows the formula \( \mathbf{v} \cdot \mathbf{u} = v_i \cdot u_i + v_j \cdot u_j + v_k \cdot u_k \). Applying this, we have: \[\begin{align*}\mathbf{v} \cdot \mathbf{u} &= (2)(-2) + (-4)(4) + (\sqrt{5})(-\sqrt{5}) \&= -4 - 16 - 5 \&= -25.\end{align*}\]

Step 2: Calculate the Magnitude of \(\mathbf{v}\)

To find \(|\mathbf{v}|\), apply the formula \(|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\):\[\begin{align*}|\mathbf{v}| &= \sqrt{2^2 + (-4)^2 + (\sqrt{5})^2} \&= \sqrt{4 + 16 + 5} \&= \sqrt{25} \&= 5.\end{align*}\]

Step 3: Calculate the Magnitude of \(\mathbf{u}\)

Similarly, for \(|\mathbf{u}|\), apply the formula:\[\begin{align*}|\mathbf{u}| &= \sqrt{(-2)^2 + 4^2 + (-\sqrt{5})^2} \&= \sqrt{4 + 16 + 5} \&= \sqrt{25} \&= 5.\end{align*}\]

Step 4: Find the Cosine of the Angle Between the Vectors

Using the dot product and magnitudes, the formula for cosine of the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{u} \) is:\[\cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|\cdot |\mathbf{u}|}\]Substitute the known values:\[\begin{align*}\cos \theta &= \frac{-25}{5 \times 5} \&= \frac{-25}{25} \&= -1.\end{align*}\]

Step 5: Calculate the Scalar Component of \( \mathbf{u} \) in the Direction of \( \mathbf{v} \)

The scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by:\[\text{Scalar component of } \mathbf{u} \text{ in the direction of } \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|}\]Using our results:\[\begin{align*}\text{Scalar component} &= \frac{-25}{5} \&= -5.\end{align*}\]

Step 6: Calculate the Vector Projection of \( \mathbf{u} \) onto \( \mathbf{v} \)

The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \mathbf{v}\]Using the known values:\[\begin{align*}\text{proj}_{\mathbf{v}} \mathbf{u} &= \frac{-25}{5^2} \mathbf{v} \&= \frac{-25}{25} (2\mathbf{i} - 4\mathbf{j} + \sqrt{5}\mathbf{k}) \&= - (2\mathbf{i} - 4\mathbf{j} + \sqrt{5}\mathbf{k}) \&= -2\mathbf{i} + 4\mathbf{j} - \sqrt{5}\mathbf{k}.\end{align*}\]

Key Concepts

Dot Product

When dealing with vectors in calculus, the **dot product** is one way to measure how much two vectors align with each other. The dot product, also known as the scalar product, combines two vectors to produce a single number, or scalar. This is calculated by multiplying corresponding components of the vectors and then summing those products. For example, if you have two vectors \( \mathbf{v} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{u} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the dot product is \( \mathbf{v} \cdot \mathbf{u} = a_1b_1 + a_2b_2 + a_3b_3 \). The result is a scalar indicating how much one vector extends in the direction of the other.

• In our exercise, with vectors \( \mathbf{v} \) and \( \mathbf{u} \), the dot product is calculated as \(-25\).
• This negative result suggests that the vectors point in opposite directions in some sense.

Magnitude of a Vector

The **magnitude** of a vector, also known as its length or norm, describes how long the vector is. In vector calculus, the magnitude is always a positive value or zero (for a zero vector). It is calculated using the formula \(|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \), where \(x, y, z\) are the components of the vector. This formula comes from the Pythagorean theorem and can be thought of as a three-dimensional extension of it.

• For vector \( \mathbf{v} = 2 \mathbf{i} - 4 \mathbf{j} + \sqrt{5} \mathbf{k} \), the magnitude is \(5\).
• Similarly, the magnitude of vector \( \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \) is also \(5\).

This result signifies the vectors have the same length though they might not point in the same direction.

Vector Projection

The **projection** of one vector onto another gives a new vector that shows how much of one vector lies in the direction of another. The formula for projecting vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{\lvert \mathbf{v} \rvert^2} \mathbf{v} \). This scale factor \( \frac{\mathbf{v} \cdot \mathbf{u}}{\lvert \mathbf{v} \rvert^2} \) tells us how much to "stretch" or "shrink" \( \mathbf{v} \) to line up with \( \mathbf{u} \).

• For our problem, using vectors \( \mathbf{v} \) and \( \mathbf{u} \), the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) works out to be \(-2\mathbf{i} + 4\mathbf{j} - \sqrt{5}\mathbf{k} \), which is interestingly \( \mathbf{u} \) itself because of the proportion between the vectors' magnitudes and the dot product.

This means \( \mathbf{u} \) retains the same direction but perfectly aligned along \( \mathbf{v} \), if \( \mathbf{u} \) were slid onto \( \mathbf{v} \).

Cosine of the Angle Between Vectors

The angle between two vectors is fundamental in understanding their spatial relationship. The **cosine of the angle** between vectors \( \mathbf{v} \) and \( \mathbf{u} \) can be found using the dot product formula. It is expressed as \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{\lvert \mathbf{v} \rvert \cdot \lvert \mathbf{u} \rvert} \). This formula results from rearranging the law of cosines for vector spaces.

• In this exercise, the cosine of the angle \( \theta \) between vectors \( \mathbf{v} \) and \( \mathbf{u} \) is \(-1\).
• This result means the vectors are pointing in exactly opposite directions, forming an angle of 180 degrees between them.

Such a result underscores the usefulness of vector calculations in determining not only magnitude but directionality in 3D space.