Problem 23 - Algebra and Trigonometry

Question

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$ \mathbf{u}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j} $$

Step-by-Step Solution

Answer

\(|\mathbf{u}| = \sqrt{5}\), \(|\mathbf{v}| = \sqrt{13}\), \(|2\mathbf{u}| = 2\sqrt{5}\), \(\left|\frac{1}{2}\mathbf{v}\right| = \frac{\sqrt{13}}{2}\), \(|\mathbf{u}+\mathbf{v}| = \sqrt{26}\), \(|\mathbf{u}-\mathbf{v}| = \sqrt{10}\), \(|\mathbf{u}| - |\mathbf{v}| = \sqrt{5} - \sqrt{13}\)."

Step 1: Calculate Magnitude of \( \mathbf{u} \)

Given the vector \( \mathbf{u} = 2 \mathbf{i} + \mathbf{j} \), calculate its magnitude using the formula for the magnitude of a vector \( |\mathbf{u}| = \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the components of the vector. Thus, we have:\[|\mathbf{u}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\]

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

The vector \( \mathbf{v} = 3 \mathbf{i} - 2 \mathbf{j} \). To find its magnitude:\[|\mathbf{v}| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\]

Step 3: Calculate Magnitude of \( 2\mathbf{u} \)

The vector \( 2\mathbf{u} = 2(2\mathbf{i} + \mathbf{j}) = 4\mathbf{i} + 2\mathbf{j} \). Find its magnitude:\[|2\mathbf{u}| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\]

Step 4: Calculate Magnitude of \( \frac{1}{2}\mathbf{v} \)

The vector \( \frac{1}{2}\mathbf{v} = \frac{1}{2}(3\mathbf{i} - 2\mathbf{j}) = \frac{3}{2}\mathbf{i} - \mathbf{j} \). Find its magnitude:\[\left|\frac{1}{2}\mathbf{v}\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-1)^2} = \sqrt{\frac{9}{4} + 1} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}\]

Step 5: Calculate Magnitude of \( \mathbf{u} + \mathbf{v} \)

The vector sum \( \mathbf{u} + \mathbf{v} = (2\mathbf{i} + \mathbf{j}) + (3\mathbf{i} - 2\mathbf{j}) = 5\mathbf{i} - \mathbf{j} \). Calculate:\[|\mathbf{u} + \mathbf{v}| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}\]

Step 6: Calculate Magnitude of \( \mathbf{u} - \mathbf{v} \)

The vector difference \( \mathbf{u} - \mathbf{v} = (2\mathbf{i} + \mathbf{j}) - (3\mathbf{i} - 2\mathbf{j}) = -\mathbf{i} + 3\mathbf{j} \). Calculate:\[|\mathbf{u} - \mathbf{v}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}\]

Step 7: Calculate \( |\mathbf{u}| - |\mathbf{v}| \)

Compute the difference of the magnitudes:\[|\mathbf{u}| - |\mathbf{v}| = \sqrt{5} - \sqrt{13}\]This is simplified to leave as-is since further simplification would involve approximate decimal values.

Key Concepts

Vector Addition

Vector addition is a fundamental operation when dealing with vectors. It involves combining two vectors to form a resultant vector. Consider vectors \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \) and \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \). When you add these vectors, you simply add their corresponding components:

• \( \mathbf{i} \) components: \( 2 + 3 = 5 \)
• \( \mathbf{j} \) components: \( 1 - 2 = -1 \)

Thus, the resultant vector is \( \mathbf{u} + \mathbf{v} = 5\mathbf{i} - \mathbf{j} \). The magnitude of this resultant vector is calculated using the formula for vector magnitude, resulting in \( |\mathbf{u} + \mathbf{v}| = \sqrt{26} \). This process highlights how vector addition combines directions and magnitudes.

Vector Subtraction

Vector subtraction involves finding the difference between two vectors, and it's essentially the inverse of vector addition. To subtract vector \( \mathbf{v} \) from \( \mathbf{u} \), you subtract their corresponding components:

• \( \mathbf{i} \) components: \( 2 - 3 = -1 \)
• \( \mathbf{j} \) components: \( 1 - (-2) = 3 \)

This means the resulting vector is \( \mathbf{u} - \mathbf{v} = -\mathbf{i} + 3\mathbf{j} \). The magnitude of the resultant vector is \( |\mathbf{u} - \mathbf{v}| = \sqrt{10} \). By understanding vector subtraction, you can determine how one vector navigates relative to another.

Scalar Multiplication

Scalar multiplication refers to the multiplication of a vector by a scalar (a real number). For a vector \( \mathbf{u} \), multiplying by a scalar \( 2 \) gives you \( 2\mathbf{u} = 4\mathbf{i} + 2\mathbf{j} \). Each component of the vector is multiplied by the scalar, stretching or contracting the vector's magnitude. Similarly, a scalar like \( \frac{1}{2} \) applied to \( \mathbf{v} \) gives \( \frac{1}{2}\mathbf{v} = \frac{3}{2}\mathbf{i} - \mathbf{j} \).
After scalar multiplication, the magnitude of the vector changes proportionally. For example, the magnitude of \( 2\mathbf{u} \) becomes \( 2\sqrt{5} \) and \( \left|\frac{1}{2}\mathbf{v}\right| \) becomes \( \frac{\sqrt{13}}{2} \). This technique is integral in scaling vectors for various applications.

Vector Components

Vectors have components that describe their direction and magnitude in space, commonly expressed in terms of unit vectors like \( \mathbf{i} \) and \( \mathbf{j} \) in two dimensions. For example, vector \( \mathbf{u} = 2 \mathbf{i} + \mathbf{j} \) comprises components \( 2 \) along the \( \mathbf{i} \) axis and \( 1 \) along the \( \mathbf{j} \) axis. Similarly, vector \( \mathbf{v} = 3 \mathbf{i} - 2 \mathbf{j} \) has components \( 3 \) and \( -2 \) respectively.
Breaking vectors into components makes it easier to perform operations such as addition or subtraction. You can manipulate each dimension independently before combining them to find the resultant vector or magnitude. The beauty of vectors lies in their ability to represent both magnitude and direction through their components, making them versatile in expressing physical quantities like force and velocity.