Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the two-dimensional vectors \(\mathbf{u}\) and \(\mathbf{v}\) in Problems \(9-12\), find the sum \(\mathbf{u}+\mathbf{v}\), the difference \(\mathbf{u}-\mathbf{v}\), and the magnitudes \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\). $$ \mathbf{u}=\langle-1,0\rangle, \mathbf{v}=\langle 3,4\rangle $$

Step-by-Step Solution

Answer

\(\mathbf{u} + \mathbf{v} = \langle 2, 4 \rangle\), \(\mathbf{u} - \mathbf{v} = \langle -4, -4 \rangle\), \(\|\mathbf{u}\| = 1\), \(\|\mathbf{v}\| = 5\).

Step 1: Find the Sum of Vectors

To find the sum of the vectors \( \mathbf{u} \) and \( \mathbf{v} \), add their respective components: \( \mathbf{u} + \mathbf{v} = \langle -1 + 3, 0 + 4 \rangle = \langle 2, 4 \rangle \).

Step 2: Find the Difference of Vectors

To find the difference \( \mathbf{u} - \mathbf{v} \), subtract the components of \( \mathbf{v} \) from \( \mathbf{u} \): \( \mathbf{u} - \mathbf{v} = \langle -1 - 3, 0 - 4 \rangle = \langle -4, -4 \rangle \).

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

To calculate the magnitude of \( \mathbf{u} = \langle -1, 0 \rangle \), use the formula \( \|\mathbf{u}\| = \sqrt{(-1)^2 + 0^2} = \sqrt{1} = 1 \).

Step 4: Calculate the Magnitude of \( \mathbf{v} \)

To find the magnitude of \( \mathbf{v} = \langle 3, 4 \rangle \), use the formula \( \|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

Key Concepts

Sum of Vectors

Adding vectors is a fundamental operation in vector mathematics. The sum of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is found by adding their corresponding components together. For example, if you have \(\mathbf{u} = \langle-1, 0\rangle\) and \(\mathbf{v} = \langle3, 4\rangle\), the process is simple:

• Add the first components: \(-1 + 3 = 2\).
• Add the second components: \(0 + 4 = 4\).

So, the sum \(\mathbf{u} + \mathbf{v}\) is the vector \(\langle 2, 4\rangle\). This process can easily be visualized on a plane, where the vectors form a new vector \(\mathbf{r}\) by joining tail to tip, creating a parallelogram. Understanding this concept helps in predicting the resultant force or direction when two different forces or velocities are being applied together.

Vector Difference

Subtracting one vector from another is analogous to finding a difference in a set of directions or forces. To find the difference \(\mathbf{u} - \mathbf{v}\) for vectors \(\mathbf{u} = \langle-1, 0\rangle\) and \(\mathbf{v} = \langle3, 4\rangle\), you would:

• Subtract the first components: \(-1 - 3 = -4\).
• Subtract the second components: \(0 - 4 = -4\).

Thus, the result is \(\langle -4, -4 \rangle\). Geometrically, this is equivalent to flipping the vector \(\mathbf{v}\) around, effectively reversing it and then adding it to \(\mathbf{u}\). It's a vector operation that shows changes in direction or quantity, often used in physics to determine net displacements or changes in velocity.

Magnitude of a Vector

The magnitude of a vector provides important information about its length or size, regardless of its direction. When calculating the magnitude of vectors \(\mathbf{u}\) and \(\mathbf{v}\), the Pythagorean Theorem is used but in the vector context:- For \(\mathbf{u} = \langle -1, 0 \rangle\), use the formula: \( \|\mathbf{u}\| = \sqrt{(-1)^2 + 0^2} = \sqrt{1} = 1\). This tells us that the length of \(\mathbf{u}\) is 1 unit.
- For \(\mathbf{v} = \langle 3, 4 \rangle\), compute: \( \|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\). Here, \(\mathbf{v}\) is 5 units long.Understanding magnitude is crucial as it gives a scalar representation of the vector, which is widely used in diverse fields like physics to describe speed, force, and other vector quantities.

Two-dimensional Vectors

Two-dimensional vectors are vectors that exist in a plane, incorporating both a direction and a magnitude. They are often described using two components, typically labeled as \(x\) and \(y\). In our example, \(\mathbf{u} = \langle -1, 0 \rangle\) and \(\mathbf{v} = \langle 3, 4 \rangle\), these components are understood as the horizontal and vertical influences, respectively.- The vector \(\mathbf{u}\) can be visualized as pointing left one unit along the x-axis, staying neutral on the y-axis.- The vector \(\mathbf{v}\) moves three units on the x-axis and four units on the y-axis, forming a right-angled triangle.Two-dimensional vectors are fundamental in applications such as navigation, where they help identify position and direction. Whether plotting the course of a moving object or resolving forces into perpendicular directions, these vectors provide useful insights into relative positioning and movement.