Problem 18 - APEX Calculus

Question

Find \(\|\vec{u}\|,\|\vec{v}\|,\|\vec{u}+\vec{v}\|\) and \(\|\vec{u}-\vec{v}\|\) \(\vec{u}=\langle-3,2,2\rangle, \quad \vec{v}=\langle 1,-1,1\rangle\)

Step-by-Step Solution

Answer

\(\|\vec{u}\| = \sqrt{17}, \|\vec{v}\| = \sqrt{3}, \|\vec{u} + \vec{v}\| = \sqrt{14}, \|\vec{u} - \vec{v}\| = \sqrt{26}.\)

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

The magnitude of a vector \(\vec{u} = \langle a, b, c \rangle\) is calculated using the formula: \(\|\vec{u}\| = \sqrt{a^2 + b^2 + c^2}\). For \(\vec{u} = \langle -3, 2, 2 \rangle\), compute: \[\|\vec{u}\| = \sqrt{(-3)^2 + 2^2 + 2^2} = \sqrt{9 + 4 + 4} = \sqrt{17}\].

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

Using the same formula, \(\|\vec{v}\| = \sqrt{1^2 + (-1)^2 + 1^2}\). For \(\vec{v} = \langle 1, -1, 1 \rangle\), compute: \[\|\vec{v}\| = \sqrt{1 + 1 + 1} = \sqrt{3}\].

Step 3: Calculate the Sum \(\vec{u} + \vec{v}\)

To find the vector sum \(\vec{u} + \vec{v}\), add corresponding components: \(\langle -3, 2, 2 \rangle + \langle 1, -1, 1 \rangle = \langle -3+1, 2-1, 2+1 \rangle = \langle -2, 1, 3 \rangle\).

Step 4: Calculate the Magnitude of \(\vec{u} + \vec{v}\)

Now, find the magnitude of \(\vec{u} + \vec{v} = \langle -2, 1, 3 \rangle\) using the formula: \(\|\vec{u} + \vec{v}\| = \sqrt{(-2)^2 + 1^2 + 3^2}\). This results in \[\|\vec{u} + \vec{v}\| = \sqrt{4 + 1 + 9} = \sqrt{14}\].

Step 5: Calculate the Difference \(\vec{u} - \vec{v}\)

To find the vector difference \(\vec{u} - \vec{v}\), subtract corresponding components: \(\langle -3, 2, 2 \rangle - \langle 1, -1, 1 \rangle = \langle -3-1, 2+1, 2-1 \rangle = \langle -4, 3, 1 \rangle\).

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

Now, find the magnitude of \(\vec{u} - \vec{v} = \langle -4, 3, 1 \rangle\) using the same formula: \(\|\vec{u} - \vec{v}\| = \sqrt{(-4)^2 + 3^2 + 1^2}\). Compute this to get: \[\|\vec{u} - \vec{v}\| = \sqrt{16 + 9 + 1} = \sqrt{26}\].

Key Concepts

Vector Addition

Vector addition is the process of combining two or more vectors to form a single vector. Imagine you have two vectors, \( \vec{u} \) and \( \vec{v} \), positioned graphically in 3D space. Each vector represents a movement from an initial point to a terminal point. When you add these vectors, you essentially place them tip to tail and find the resultant by drawing a vector from the start of the first vector to the end of the last.
To add vectors, you add their corresponding components:

• \( \vec{u} = \langle -3, 2, 2 \rangle \)
• \( \vec{v} = \langle 1, -1, 1 \rangle \)

So to find \( \vec{u} + \vec{v} \), you calculate:
\[ \vec{u} + \vec{v} = \langle -3 + 1, 2 - 1, 2 + 1 \rangle = \langle -2, 1, 3 \rangle \]This computation results in the vector sum \( \vec{u} + \vec{v} = \langle -2, 1, 3 \rangle \), which is also a 3D vector itself.

Vector Subtraction

Vector subtraction is similar to vector addition but rather than adding, you subtract the components of one vector from another. Consider again the vectors \( \vec{u} \) and \( \vec{v} \). If vector addition is visualized by placing vectors tip to tail, vector subtraction can be imagined as adding the negative of one vector.
To subtract two vectors, \( \vec{u} \) and \( \vec{v} \), do the following:

• Subtract the corresponding components of \( \vec{v} \) from \( \vec{u} \):
• \( \vec{u} - \vec{v} = \langle -3 - 1, 2 + 1, 2 - 1 \rangle = \langle -4, 3, 1 \rangle \)

The vector \( \vec{u} - \vec{v} = \langle -4, 3, 1 \rangle \) is the result of this operation. It can also be described as a 3D vector indicating a displacement.

3D Vectors

3D vectors are mathematical objects that hold three components representing directions along the x, y, and z axes in three-dimensional space. These components symbolize coordinates or movements in their respective directions.
For example:

• Vector \( \vec{u} = \langle -3, 2, 2 \rangle \) has components -3 in the x-direction, 2 in the y-direction, and 2 in the z-direction.
• Vector \( \vec{v} = \langle 1, -1, 1 \rangle \) navigates 1 unit along the x-axis, -1 unit along the y-axis, and 1 unit along the z-axis.

These vectors can be visualized as arrows or lines that stretch from the origin \( \langle 0, 0, 0 \rangle \) to their terminal point corresponding to their component values. The length or magnitude of these vectors can be calculated to understand the total "size" or "extent" of their directional influence, using the formula \( \|\vec{a}\| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

Problem Solving in Calculus

Problem-solving in calculus often involves understanding and manipulating vectors, especially when working in three dimensions. Calculus problems can include finding the magnitude of vectors or performing vector operations like addition and subtraction.
For instance, to solve for the magnitude of the vector \( \vec{u} + \vec{v} \), you first need to understand that this involves calculating:

• \( \|\vec{u} + \vec{v}\| = \sqrt{(-2)^2 + 1^2 + 3^2} = \sqrt{14} \)

Improvement advice for problem-solving includes:

• Breaking problems into smaller parts.
• Systematically applying vector formulas.
• Visualizing vectors to better grasp their interactions.

Grasping these fundamental aspects can make tackling calculus problems involving vectors much more straightforward.