Problem 78 - Precalculus with Calculus Previews

Question

\(\mathbf{u}=\langle 1,-1\rangle\) and \(\mathbf{v}=\langle 2,6\rangle\). Find the indicated number. $$ \operatorname{comp}_{\mathbf{u}} \mathbf{v} $$

Step-by-Step Solution

Answer

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

Step 1: Recall the formula for the scalar projection

The scalar projection of \( \mathbf{v} \) on \( \mathbf{u} \) is given by the formula \( \operatorname{comp}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||} \). To solve this, we need to calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \) and the magnitude of \( \mathbf{u} \), denoted as \( ||\mathbf{u}|| \).

Step 2: Calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \)

The dot product is calculated as follows: \( \mathbf{u} \cdot \mathbf{v} = 1 \cdot 2 + (-1) \cdot 6 = 2 - 6 = -4 \). So, \( \mathbf{u} \cdot \mathbf{v} = -4 \).

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

The magnitude of \( \mathbf{u} \), \( ||\mathbf{u}|| \), is calculated using the formula \( ||\mathbf{u}|| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \).

Step 4: Compute the scalar projection

Now, substitute the values into the formula: \( \operatorname{comp}_{\mathbf{u}} \mathbf{v} = \frac{-4}{\sqrt{2}} \). Simplify the expression to get: \( \operatorname{comp}_{\mathbf{u}} \mathbf{v} = -2\sqrt{2} \).

Key Concepts

Dot Product

The dot product is an essential concept in vector algebra and is often used in various applications involving vectors. It provides a way to multiply two vectors to get a scalar value rather than another vector.
To compute the dot product, you take two vectors with equal dimensions and sum the products of their corresponding components. For example, given two vectors \( \mathbf{u} = \langle a, b \rangle \) and \( \mathbf{v} = \langle c, d \rangle \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is given by:
\[ \mathbf{u} \cdot \mathbf{v} = a \cdot c + b \cdot d \]
This operation gives you a single number, which is a scalar. In our exercise, we had the vectors \( \mathbf{u} = \langle 1, -1 \rangle \) and \( \mathbf{v} = \langle 2, 6 \rangle \). The dot product was calculated as \( 1 \cdot 2 + (-1) \cdot 6 = 2 - 6 = -4 \), resulting in \( \mathbf{u} \cdot \mathbf{v} = -4 \).
The dot product is important in finding angles between vectors or in computing projections, which involves both magnitude and direction of vectors.

Magnitude of a Vector

The magnitude of a vector is a measure of its length or size. It is calculated using the Pythagorean theorem and represents the distance of the vector from the origin in a given coordinate system.
Given a vector \( \mathbf{u} = \langle a, b \rangle \), the magnitude is calculated using:
\[ ||\mathbf{u}|| = \sqrt{a^2 + b^2} \]
This formula calculates the Euclidean norm of the vector, which gives you a positive scalar that quantifies how long the vector is.

• In our specific exercise, we had the vector \( \mathbf{u} = \langle 1, -1 \rangle \).
• The magnitude calculation was \( ||\mathbf{u}|| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \).

Understanding the magnitude is crucial not just in geometry, but also in physics and engineering, where it plays a role in determining forces, velocities, and other directional quantities.

Scalar Projection

The scalar projection, or the component of one vector onto another, tells you how much of one vector "lies in the direction" of the other vector.
To find the scalar projection of a vector \( \mathbf{v} \) on a vector \( \mathbf{u} \), you use the formula:
\[ \operatorname{comp}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||} \]
This formula helps you determine the length of the projection of vector \( \mathbf{v} \) onto \( \mathbf{u} \). It's different from the vector projection, which would include both magnitude and direction.

• The calculated dot product of \( \mathbf{u} \) and \( \mathbf{v} \) was \(-4\).
• The magnitude of \( \mathbf{u} \) was \( \sqrt{2} \).
• Substituting these values, the scalar projection was \( \frac{-4}{\sqrt{2}} = -2\sqrt{2} \).

Understanding scalar projection is essential in fields like physics, where projecting forces onto certain axes simplifies the analysis of forces acting on objects.