Problem 40 - Algebra and Trigonometry

Question

\(37-40\) Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$ \mathbf{u}=\langle- 6,6\rangle, \quad \mathbf{v}=\langle- 2,-1\rangle $$

Step-by-Step Solution

Answer

|u| = 6√2, |v| = √5, |2u| = 12√2, |(1/2)v| = √5/2, |u+v| = √89, |u-v| = √65, |u|-|v| = 6√2 - √5.

Step 1: Calculate |u|

The magnitude of a vector, \( \mathbf{u} = \langle x, y \rangle \), is given by the formula \( |\mathbf{u}| = \sqrt{x^2 + y^2} \). For \( \mathbf{u} = \langle -6, 6 \rangle \), we have:\[|\mathbf{u}| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\]

Step 2: Calculate |v|

Use the same formula to find the magnitude of \( \mathbf{v} = \langle -2, -1 \rangle \):\[|\mathbf{v}| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\]

Step 3: Calculate |2u|

The magnitude of \( 2 \mathbf{u} \) (where \( c \) is a scalar, \( |c\mathbf{u}| = |c||\mathbf{u}| \)) is twice the magnitude of \( \mathbf{u} \). Hence:\[|2\mathbf{u}| = 2|\mathbf{u}| = 2 \cdot 6\sqrt{2} = 12\sqrt{2}\]

Step 4: Calculate |(1/2)v|

For the vector \( \frac{1}{2} \mathbf{v} \), its magnitude is half the magnitude of \( \mathbf{v} \):\[\left| \frac{1}{2} \mathbf{v} \right| = \frac{1}{2} |\mathbf{v}| = \frac{1}{2} \sqrt{5} = \frac{\sqrt{5}}{2}\]

Step 5: Calculate |u+v|

First, find \( \mathbf{u} + \mathbf{v} \):\[\mathbf{u} + \mathbf{v} = \langle -6 + (-2), 6 + (-1) \rangle = \langle -8, 5 \rangle\]Then calculate the magnitude:\[|\mathbf{u} + \mathbf{v}| = \sqrt{(-8)^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89}\]

Step 6: Calculate |u-v|

First, find \( \mathbf{u} - \mathbf{v} \):\[\mathbf{u} - \mathbf{v} = \langle -6 - (-2), 6 - (-1) \rangle = \langle -4, 7 \rangle\]Then calculate the magnitude:\[|\mathbf{u} - \mathbf{v}| = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}\]

Step 7: Calculate |u|-|v|

Subtract the magnitudes found in Steps 1 and 2:\[|\mathbf{u}| - |\mathbf{v}| = 6\sqrt{2} - \sqrt{5}\]

Key Concepts

Unit Vectors

A unit vector is a special type of vector with a magnitude of 1. Its primary use is to indicate direction, without affecting the scaling of a component. Finding a unit vector in the same direction as a given vector is done by dividing each component of the vector by its magnitude. For instance, if we have a vector \( \mathbf{a} = \langle x, y \rangle \), the unit vector \( \mathbf{a}_{\text{unit}} \) is calculated by:\[ \mathbf{a}_{\text{unit}} = \frac{1}{|\mathbf{a}|}\langle x, y \rangle \]Here, \(|\mathbf{a}|\) is the magnitude of vector \(\mathbf{a}\). This operation preserves the direction of \(\mathbf{a}\) while scaling its magnitude to 1. In practical applications, unit vectors are often used to simplify vector calculations because they allow isolation of direction from magnitude. This is particularly useful in physics and engineering where direction is crucial for understanding forces, velocities, and other directional phenomena.

Vector Addition

Vector addition is a fundamental operation allowing the combination of two or more vectors into a single resultant vector. For vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), the resulting vector \( \mathbf{u} + \mathbf{v} \) is:\[ \mathbf{u} + \mathbf{v} = \langle u_1 + v_1, u_2 + v_2 \rangle \]
It’s important because vectors can represent physical quantities like displacement, force, or velocity. Thus, vector addition helps predict outcomes when combined forces or movements occur.

• Example: If \( \mathbf{u} = \langle -6, 6 \rangle \) and \( \mathbf{v} = \langle -2, -1 \rangle \), their sum is \( \langle -8, 5 \rangle \).

In visual terms, if you place the start of one vector at the end of the other, the resultant vector completes the triangle or parallelogram formed. This graphical representation helps understand how multiple forces might act on an object simultaneously.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar, which is a real number. This operation changes the magnitude of the vector but not its direction, unless the scalar is negative, in which case the direction is reversed. Given a vector \( \mathbf{u} = \langle u_1, u_2 \rangle \) and a scalar \( c \), the product \( c\mathbf{u} \) is:\[ c\mathbf{u} = \langle cu_1, cu_2 \rangle \]
Scalar multiplication is particularly useful in resizing vectors, such as scaling up or down the force magnitude in physics. It also aids in determining vectors’ projections and reflections.

• Example: For \( 2\mathbf{u} \) where \( \mathbf{u} = \langle -6, 6 \rangle \), results in \( \langle -12, 12 \rangle \).
• If we take \( \frac{1}{2}\mathbf{v} \), where \( \mathbf{v} = \langle -2, -1 \rangle \), the result is \( \langle -1, -0.5 \rangle \).

This flexibility allows scalar multiplication to model a wide range of real-world scenarios where resizing vector magnitudes is necessary.

Vector Subtraction

Vector subtraction involves finding the difference between two vectors and is carried out by subtracting their corresponding components. For vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), the difference \( \mathbf{u} - \mathbf{v} \) is:\[ \mathbf{u} - \mathbf{v} = \langle u_1 - v_1, u_2 - v_2 \rangle \]
This operation is often used to find the relative position or distance between two points represented by vectors.

• Example: If \( \mathbf{u} = \langle -6, 6 \rangle \) and \( \mathbf{v} = \langle -2, -1 \rangle \), then \( \mathbf{u} - \mathbf{v} = \langle -4, 7 \rangle \).

Visually, vector subtraction can be thought of as adding the negative of a vector. By placing the tail of \( \mathbf{v} \) at the head of \( \mathbf{u} \), the resultant vector points from \( \mathbf{v} \) to \( \mathbf{u} \), clearly indicating the directional difference between the two.

Geometry

In the context of vector operations, geometry plays a crucial role. Vectors not only represent quantities with magnitude and direction but also describe geometric transformations and relationships. Understanding concepts like vector magnitude, addition, and subtraction involves geometric reasoning.

• Magnitudes reveal the "length" or "size" of a vector.
• Addition and subtraction describe movements or displacements from one point to another.

Furthermore, vectors can describe geometric elements such as lines and planes. In two dimensions, they help in visualizing positions and forces, while in higher dimensions they become essential for modeling spaces and transformations. Overall, studying vectors geometrically provides insight into both simple and complex spatial relationships, allowing for deeper understanding and application of their properties in various fields.