Problem 10 - 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 \(\|\mathbf{v}\|\). $$ \mathbf{u}=\langle 0,0\rangle, \mathbf{v}=\langle-3,4\rangle $$

Step-by-Step Solution

Answer

The sum is \( \langle -3, 4 \rangle \), the difference is \( \langle 3, -4 \rangle \), \( \| \mathbf{u} \| = 0 \), and \( \| \mathbf{v} \| = 5 \).

Step 1: Find the Sum of Vectors

To find the sum of vectors \( \mathbf{u} + \mathbf{v} \), simply add their corresponding components.The vectors are \( \mathbf{u} = \langle 0, 0 \rangle \) and \( \mathbf{v} = \langle -3, 4 \rangle \).\[ \mathbf{u} + \mathbf{v} = \langle 0 + (-3), 0 + 4 \rangle = \langle -3, 4 \rangle \]

Step 2: Find the Difference of Vectors

To calculate the difference \( \mathbf{u} - \mathbf{v} \), subtract the components of \( \mathbf{v} \) from \( \mathbf{u} \).\[ \mathbf{u} - \mathbf{v} = \langle 0 - (-3), 0 - 4 \rangle = \langle 0 + 3, 0 - 4 \rangle = \langle 3, -4 \rangle \]

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

The magnitude of a vector \( \langle a, b \rangle \) is given by \( \| \mathbf{u} \| = \sqrt{a^2 + b^2} \).For \( \mathbf{u} = \langle 0, 0 \rangle \), the magnitude is:\[ \| \mathbf{u} \| = \sqrt{0^2 + 0^2} = \sqrt{0} = 0 \]

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

Similarly, the magnitude of \( \mathbf{v} = \langle -3, 4 \rangle \) is calculated as follows:\[ \| \mathbf{v} \| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Key Concepts

Vector Addition

Vector addition is a crucial operation when working with vectors. It involves combining two or more vectors to produce a resultant vector. This operation is performed by adding the corresponding components of the vectors together. For instance, given two-dimensional vectors \(\mathbf{u} = \langle u_1, u_2 \rangle\) and \(\mathbf{v} = \langle v_1, v_2 \rangle\), the sum \(\mathbf{u} + \mathbf{v}\) becomes \(\langle u_1 + v_1, u_2 + v_2 \rangle\).
In the case of vectors \(\mathbf{u} = \langle 0, 0 \rangle\) and \(\mathbf{v} = \langle -3, 4 \rangle\), adding these results in \(\langle 0 + (-3), 0 + 4 \rangle = \langle -3, 4 \rangle\).
This calculation shows that, even starting from zero, adding another vector with distinct values moves us to its specific location in the vector plane.

• Vector addition modifies overall direction and magnitude.
• The result can depict new displacement if these vectors represent forces or velocities.

Understanding vector addition is foundational in physics and engineering, especially when combining forces or navigating using displacement.

Vector Subtraction

Vector subtraction, much like addition, involves component-wise operations between vectors. Instead of adding the corresponding components, subtraction involves taking the components of one vector away from another. 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 determined as \(\langle u_1 - v_1, u_2 - v_2 \rangle\).
With the vectors \(\mathbf{u} = \langle 0, 0 \rangle\) and \(\mathbf{v} = \langle -3, 4 \rangle\), the subtraction results in \(\langle 0 - (-3), 0 - 4 \rangle = \langle 3, -4 \rangle\).
This result effectively reverses the direction of vector \(\mathbf{v}\) and repositions vector \(\mathbf{u}\) accordingly.

• Vector subtraction can be interpreted as finding the vector required to go from \(\mathbf{v}\) to \(\mathbf{u}\).
• It is essential in determining relative velocities and positions in physics.

Grasping vector subtraction expands analytical skills in displacement and transition evaluation.

Magnitude of a Vector

The magnitude of a vector can be thought of as its length or size, giving a scalar representation of its distance from the origin in a vector space. For a vector \(\mathbf{a} = \langle a_1, a_2 \rangle\), the magnitude \( \| \mathbf{a} \| \) is computed using the Pythagorean theorem: \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \).
Using this formula, the magnitude of \(\mathbf{u} = \langle 0, 0 \rangle\) is straightforward, yielding \( \sqrt{0^2 + 0^2} = 0 \).
On the other hand, for \(\mathbf{v} = \langle -3, 4 \rangle\), it calculates to \( \sqrt{(-3)^2 + 4^2} = 5 \).

• This measurement helps understand a vector's capacity in influencing or representing parameter space.
• The magnitude is critical in physics for quantifying elements like speed and force magnitude without considering direction.

Recognizing the magnitude of vectors aids in visualizing and measuring shifts within multidimensional models.

Two-Dimensional Vectors

Two-dimensional vectors are pivotal in fields such as physics, engineering, and computer graphics. They have two parts, often represented as \(\langle x, y \rangle\), and can describe various phenomena, including directional movements and forces along an XY-plane. The beauty of two-dimensional vectors lies in their ability to represent motion and trajectory in a plane.

When you address problems involving two-dimensional vectors, you work primarily with horizontal and vertical dimensions. This gives rise to practical applications like projectile motion in physics or representing screens in graphic designs.

• The XY components \(x\) and \(y\) solidify the vector’s representation, where each signifies a distinct directional influence.
• They simplify the visualization of concepts such as direction and triangle laws used in navigation and structural design.

Understanding two-dimensional vectors serves as the first step for tackling multi-dimensional analysis and problem-solving effectively.