Problem 38 - 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 \(|\mathbf{u}|-|\mathbf{v}|\) $$ \mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}, \quad \mathbf{v}=\mathbf{i}-2 \mathbf{j} $$

Step-by-Step Solution

Answer

\(|\mathbf{u}| = \sqrt{13}, |\mathbf{v}| = \sqrt{5}, |2\mathbf{u}| = 2\sqrt{13}, \left|\frac{1}{2}\mathbf{v}\right| = \frac{\sqrt{5}}{2}, |\mathbf{u} + \mathbf{v}| = \sqrt{2}, |\mathbf{u} - \mathbf{v}| = \sqrt{34}, |\mathbf{u}| - |\mathbf{v}| = \sqrt{13} - \sqrt{5}\).

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

The magnitude of a vector \(\mathbf{u} = a\mathbf{i} + b\mathbf{j}\) is calculated using \( |\mathbf{u}| = \sqrt{a^2 + b^2} \). For \(\mathbf{u} = -2\mathbf{i} + 3\mathbf{j}\), we have:\[ |\mathbf{u}| = \sqrt{(-2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13} \].

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

The magnitude of a vector \(\mathbf{v} = c\mathbf{i} + d\mathbf{j}\) is found similarly: \( |\mathbf{v}| = \sqrt{c^2 + d^2} \). For \(\mathbf{v} = \mathbf{i} - 2\mathbf{j}\), we calculate:\[ |\mathbf{v}| = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \].

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

Scaling a vector by a constant scales the magnitude by the absolute value of the constant. Thus, \(|2\mathbf{u}| = 2|\mathbf{u}|\). Calculating, we get:\[ |2\mathbf{u}| = 2 \times \sqrt{13} = 2\sqrt{13} \].

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

Similarly, \(|\frac{1}{2}\mathbf{v}| = \frac{1}{2}|\mathbf{v}|\). Thus:\[ \left| \frac{1}{2}\mathbf{v} \right| = \frac{1}{2} \times \sqrt{5} = \frac{\sqrt{5}}{2} \].

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

Add the vectors \(\mathbf{u} = -2\mathbf{i} + 3\mathbf{j}\) and \(\mathbf{v} = \mathbf{i} - 2\mathbf{j}\) to get \(\mathbf{u} + \mathbf{v} = (-2 + 1)\mathbf{i} + (3 - 2)\mathbf{j} = -\mathbf{i} + \mathbf{j}\). The magnitude is:\[ |\mathbf{u} + \mathbf{v}| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \].

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

Subtract \(\mathbf{v}\) from \(\mathbf{u}\): \(\mathbf{u} - \mathbf{v} = (-2 - 1)\mathbf{i} + (3 + 2)\mathbf{j} = -3\mathbf{i} + 5\mathbf{j}\). The magnitude is:\[ |\mathbf{u} - \mathbf{v}| = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \].

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

Simply subtract the magnitude of \(\mathbf{v}\) from the magnitude of \(\mathbf{u}\):\[ |\mathbf{u}| - |\mathbf{v}| = \sqrt{13} - \sqrt{5} \].

Key Concepts

Vector Addition

Vector addition is the process of combining two or more vectors to get a resultant vector. When you add vectors, you're combining their line of action and magnitudes. Consider vectors \( \mathbf{u} = -2 \mathbf{i} + 3 \mathbf{j} \) and \( \mathbf{v} = \mathbf{i} - 2 \mathbf{j} \). To add these, you simply add the corresponding components of each vector.

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

The resultant vector is \( \mathbf{u} + \mathbf{v} = -\mathbf{i} + \mathbf{j} \). To find its magnitude, apply the Pythagorean theorem: \[ | \mathbf{u} + \mathbf{v}| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \] This result shows how two separate vector movements create a new pathway in the 2D plane.

Vector Subtraction

Vector subtraction is similar to addition, but instead, you find the difference between the corresponding components. Consider \( \mathbf{u} = -2 \mathbf{i} + 3 \mathbf{j} \) and \( \mathbf{v} = \mathbf{i} - 2 \mathbf{j} \), and subtract \( \mathbf{v} \) from \( \mathbf{u} \).

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

The resultant vector is \( \mathbf{u} - \mathbf{v} = -3 \mathbf{i} + 5 \mathbf{j} \). To find its magnitude, calculate:\[ | \mathbf{u} - \mathbf{v}| = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \] Understanding vector subtraction is crucial for determining net displacements and changes in vector quantities in physics and engineering.

Scaling Vectors

Scaling a vector means multiplying it by a scalar (a number), which stretches or shrinks the vector. To understand scaling, consider the vector \( \mathbf{u} = -2 \mathbf{i} + 3 \mathbf{j} \). If you scale it by 2, you simply multiply each component by 2:

- \(2 \times (-2) = -4\)- \(2 \times 3 = 6\)

So, the new vector is \( 2\mathbf{u} = -4\mathbf{i} + 6\mathbf{j} \), and its magnitude is:\[ | 2\mathbf{u}| = 2|\mathbf{u}| = 2\sqrt{13} \] When scaling by a fraction, such as \(\frac{1}{2}\) for \(\mathbf{v} = \mathbf{i} - 2\mathbf{j} \), each component is halved: \[ \frac{1}{2}\mathbf{v} = \frac{1}{2}\mathbf{i} - \frac{1}{2}\mathbf{j} = \frac{1}{2}(\mathbf{i} - 2\mathbf{j}) \] And its magnitude is:\[ \left| \frac{1}{2}\mathbf{v} \right| = \frac{1}{2} \sqrt{5} \] Scaling is essential when dealing with proportional growth (or shrinkage) in physics or computer graphics.

Vector Operations

Vector operations are arithmetic operations you can perform on vectors, including addition, subtraction, and scaling. They transform simple lines of action (direction along a path) into complex behavior and outcomes. Understanding vector operations is foundational in physics, engineering, and computer science to describe and calculate changes in position, force, velocity, and more.

• Vector Addition: Combines vectors to yield a resultant vector.
• Vector Subtraction: Finds the difference, resulting in a new vector showing the net vector displacement.
• Scaling: Multiplies a vector's magnitude by a constant, effectively changing its length while preserving direction.

By mastering these operations, students can solve complex problems involving movement and forces in reality or simulations. They lay the groundwork for calculus-based physics and fields like robotics, where describing motion accurately is crucial.