Question

Find the scalar projection of \(\mathbf{u}=5 \mathbf{i}+5 \mathbf{j}+2 \mathbf{k}\) on \(\mathbf{v}=-\sqrt{5} \mathbf{i}+\sqrt{5} \mathbf{j}+\mathbf{k}\).

Short answer

The scalar projection is \( \frac{2 \sqrt{11}}{11} \).

Step-by-step solution

Identify Formula for Scalar Projection

The scalar projection of a vector \( \mathbf{u} \) onto a vector \( \mathbf{v} \) is found using the formula: \[ \text{proj}_{\mathbf{v}} (\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} \] where \( \mathbf{u} \cdot \mathbf{v} \) is the dot product and \( \| \mathbf{v} \| \) is the magnitude of \( \mathbf{v} \).

Calculate the Dot Product \( \mathbf{u} \cdot \mathbf{v} \)

The dot product of \( \mathbf{u} = 5 \mathbf{i} + 5 \mathbf{j} + 2 \mathbf{k} \) and \( \mathbf{v} = -\sqrt{5} \mathbf{i} + \sqrt{5} \mathbf{j} + \mathbf{k} \) is calculated as follows: \[ \mathbf{u} \cdot \mathbf{v} = (5)(-\sqrt{5}) + (5)(\sqrt{5}) + (2)(1) = -5\sqrt{5} + 5\sqrt{5} + 2 = 2 \]

Determine the Magnitude \( \| \mathbf{v} \| \)

The magnitude of vector \( \mathbf{v} \) is given as: \[ \| \mathbf{v} \| = \sqrt{(-\sqrt{5})^2 + (\sqrt{5})^2 + (1)^2} = \sqrt{5 + 5 + 1} = \sqrt{11} \]

Compute the Scalar Projection

Using the dot product and magnitude, compute the scalar projection: \[ \text{proj}_{\mathbf{v}} (\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} = \frac{2}{\sqrt{11}} \]

Simplify the Scalar Projection

To simplify the result, multiply the numerator and the denominator by \( \sqrt{11} \): \[ \text{proj}_{\mathbf{v}} (\mathbf{u}) = \frac{2 \sqrt{11}}{11} \]

Key concepts

Scalar Projection

When we talk about the scalar projection of one vector onto another, we're describing how much of one vector goes in the direction of another. Imagine shining a light from above one vector onto another; the shadow that it casts is called the scalar projection.

This concept uses both the dot product and the magnitude of a vector. To find the scalar projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \), you use the formula:

\[ \text{proj}_{\mathbf{v}} (\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} \]

- **\( \mathbf{u} \cdot \mathbf{v} \)** is the dot product.
- **\( \| \mathbf{v} \| \)** is the magnitude or length of vector \( \mathbf{v} \).

This formula helps you find how much of \( \mathbf{u} \) goes along the direction of \( \mathbf{v} \). It's very handy in physics and engineering, showing how vectors align or interact with one another.

Dot Product

The dot product, also known as the scalar product, is a crucial operation in vector calculus. It gives us a single number, a scalar, that reveals a lot about the relationship between two vectors.

To compute the dot product of two vectors \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) and \( \mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \), you use the formula:

\[ \mathbf{u} \cdot \mathbf{v} = ax + by + cz \]

This process involves multiplying corresponding components of each vector and then adding them together.

In our example, \( \mathbf{u} = 5 \mathbf{i} + 5 \mathbf{j} + 2 \mathbf{k} \) and \( \mathbf{v} = -\sqrt{5} \mathbf{i} + \sqrt{5} \mathbf{j} + \mathbf{k} \). So, the dot product is:

- \( 5\times(-\sqrt{5}) + 5\times(\sqrt{5}) + 2\times 1 = 2 \)

This reveals that though the vectors somewhat "cancel out," there is still a slight overlap in their direction, as reflected by the resulting number.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, is a measure of how long the vector is. It is calculated using the Pythagorean theorem by squaring each element, summing them, and taking the square root.

For a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), its magnitude is:

\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \]

In the original problem, \( \mathbf{v} = -\sqrt{5} \mathbf{i} + \sqrt{5} \mathbf{j} + 1\mathbf{k} \), and the computation for its magnitude is:

- \( \sqrt{(-\sqrt{5})^2 + (\sqrt{5})^2 + 1^2} = \sqrt{11} \)

The importance of the magnitude lies in its usage for normalization, to compute the scalar projection, or even in understanding trajectory and force in physics.

Calculus Problem Solving

Solving problems in vector calculus can seem daunting, but approaching them step-by-step makes them more manageable. Let's see a breakdown of tackling such problems.

- **Understand the Problem:** Identify what you need, such as finding scalar projections or the magnitude of a vector.

- **Collect Formulas:** Know your formulas and what they compute, like the dot product or the magnitude.

- **Plug and Solve:** Input the values carefully, simplifying as needed. In our exercise, we computed the dot product, found the magnitude, and used these to find the scalar projection.

- **Simplify:** Always aim for the simplest form of your answer, as we did by multiplying by \( \sqrt{11} \) to tidy up the result.

This systematic approach ensures clarity and accuracy, helping you not only solve practice problems but also apply these concepts in real-world scenarios like physics, engineering, and computer science.