Problem 18 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the dot product to find the magnitude of \(\mathbf{u}.\) $$\mathbf{u}=\langle-8,15\rangle$$

Step-by-Step Solution

Answer

The magnitude of the vector \(\mathbf{u}=\langle-8,15\rangle\) is \(\sqrt{289} = 17\).

Step 1: Understanding Dot product

The dot product of a vector with itself is the square of its magnitude. Given a vector \(\mathbf{u}=\langle u_1,u_2\rangle\), its dot product with itself is \(u_1^2+u_2^2\).

Step 2: Finding dot product for given vector

The given vector is \(\mathbf{u}=\langle-8,15\rangle\). The dot product for this vector with itself is \((-8)^2 + (15)^2 = 64 + 225 = 289\).

Step 3: Finding magnitude of vector

The magnitude of the vector is the square root of its dot product with itself. Therefore, the magnitude of the vector \(\mathbf{u}=\sqrt{289}\).

Key Concepts

Vector Magnitude

The magnitude of a vector, often referred to as its length, is a measurement of how long the vector is. A vector typically has both a direction and a magnitude.
To find the magnitude of a vector, you can use the formula that involves its components. For a vector \[ \mathbf{u} = \langle u_1, u_2 \rangle \],its magnitude \( |\mathbf{u}| \) is calculated using the Euclidean norm:\[ |\mathbf{u}| = \sqrt{u_1^2 + u_2^2} \].
In simpler terms, you square each component of the vector, sum them up, and take the square root of the result. This gives a straightforward way to measure the size of a vector in a two-dimensional plane.
In our exercise, the vector given is \( \mathbf{u} = \langle -8, 15 \rangle \), which boils down to finding \( \sqrt{(-8)^2 + 15^2} \). Despite the negative component, since a square of a negative number is positive, the length or magnitude will always be a non-negative number.

Vectors

Vectors are entities that have both magnitude and direction. They are quite different from scalar quantities, which merely have magnitude. Vectors can be represented in space using directed line segments. These are often written in angle brackets, such as \( \langle x, y \rangle \) for two-dimensional spaces.
Vectors can be added, subtracted, and multiplied (either by a number, known as a scalar multiplication, or with each other using operations like the dot product).

• A key operation on vectors is the dot product. It is a way to multiply two vectors to get a scalar.quantity. This is especially useful in finding the magnitude or length of a vector, or in determining the angle between two vectors.


• The dot product of a vector with itself is key in calculating the squared magnitude, giving us a shortcut for calculating its length.

Vectors are fundamental components in physics and engineering because they can concisely describe quantities that have both size and direction, such as forces, velocities, and more.

Squared Magnitude

The squared magnitude of a vector is a special terminology for the dot product of the vector with itself.
If you have a vector \( \mathbf{u} = \langle u_1, u_2 \rangle \), then the squared magnitude is found using:\[ u_1^2 + u_2^2 \].
This squared magnitude helps in quickly finding the magnitude without directly taking the square root, which can be useful for comparative purposes in calculations.
For the vector \( \mathbf{u} = \langle -8, 15 \rangle \), its squared magnitude ends up being:\[ 64 + 225 = 289 \].
Understanding this concept is critical in vector algebra as it simplifies many computations and shows how a seemingly complex vector's length can be boiled down to a simple sum of squares. The squared magnitude also helps in contexts like physics, where it can be used for determining the energy of a system without calculating the exact direction.