Problem 12 - APEX Calculus

Question

Let \(\vec{u}=\langle 1,1,-1\rangle\) and \(\vec{v}=\langle 2,1,2\rangle .\) (a) Find \(\vec{u}+\vec{v}, \vec{u}-\vec{v}, \pi \vec{u}-\sqrt{2} \vec{v} .\) (b) Sketch the above vectors on the same axes, along with \(\vec{u}\) and \(\vec{v} .\) (c) Find \(\vec{x}\) where \(\vec{u}+\vec{x}=\vec{v}+2 \vec{x}\).

Step-by-Step Solution

Answer

(a) \(\langle 3, 2, 1 \rangle, \langle -1, 0, -3 \rangle, \langle \pi - 2\sqrt{2}, \pi - \sqrt{2}, -\pi - 2\sqrt{2} \rangle\); (c) \(\langle -1, 0, -3 \rangle\).

Step 1: Compute \(\vec{u} + \vec{v}\)

To add the vectors \(\vec{u}\) and \(\vec{v}\), sum their corresponding components.\[\vec{u} + \vec{v} = \langle 1 + 2, 1 + 1, -1 + 2 \rangle = \langle 3, 2, 1 \rangle\]

Step 2: Compute \(\vec{u} - \vec{v}\)

To subtract \(\vec{v}\) from \(\vec{u}\), subtract the components of \(\vec{v}\) from those of \(\vec{u}\).\[\vec{u} - \vec{v} = \langle 1 - 2, 1 - 1, -1 - 2 \rangle = \langle -1, 0, -3 \rangle\]

Step 3: Compute \(\pi \vec{u} - \sqrt{2} \vec{v}\)

Firstly, scale the vector \(\vec{u}\) by \(\pi\) and the vector \(\vec{v}\) by \(\sqrt{2}\).\[\pi \vec{u} = \langle \pi, \pi, -\pi \rangle\]\[\sqrt{2} \vec{v} = \langle 2\sqrt{2}, \sqrt{2}, 2\sqrt{2} \rangle\]Now find the resulting vector by subtracting \(\sqrt{2} \vec{v}\) from \(\pi \vec{u}\).\[\pi \vec{u} - \sqrt{2} \vec{v} = \langle \pi - 2\sqrt{2}, \pi - \sqrt{2}, -\pi - 2\sqrt{2} \rangle\]

Step 4: Find \(\vec{x}\) given \(\vec{u} + \vec{x} = \vec{v} + 2\vec{x}\)

Rearrange the equation to solve for \(\vec{x}\). Subtract \(\vec{x}\) from both sides to isolate \(\vec{x}\).\[\vec{u} + \vec{x} - \vec{x} = \vec{v} + 2\vec{x} - \vec{x}\]This simplifies to: \[\vec{u} = \vec{v} + \vec{x}\]Now, solve for \(\vec{x}\) by subtracting \(\vec{v}\) from both sides.\[\vec{u} - \vec{v} = \vec{x}\]We found \(\vec{u} - \vec{v}\) in Step 2:\[\vec{x} = \langle -1, 0, -3 \rangle\]

Key Concepts

Vector Addition

When you think of vector addition, imagine stacking arrows together, tip to tail. Each vector is an arrow, pointing in a direction, with a certain length. The addition results in a new vector starting from the tail of the first vector to the tip of the last added vector.
To add two vectors, add their corresponding components. For example, the vectors \( \vec{u} = \langle 1, 1, -1 \rangle \) and \( \vec{v} = \langle 2, 1, 2 \rangle \), when added, result in a new vector \( \langle 1+2, 1+1, -1+2 \rangle = \langle 3, 2, 1 \rangle \). This simple arithmetic gives you the position along each dimension (x, y, z).
Remember, vector addition is commutative, meaning \( \vec{u} + \vec{v} = \vec{v} + \vec{u} \). It’s also associative, so you can add multiple vectors together in any grouping. Visualize this in a plane, or better yet, try drawing it with the arrows; you'll see the path these vectors "walk".

Vector Subtraction

Vector subtraction is similar to addition but involves reversing the direction of the vector being subtracted. Picture pulling an arrow backwards to reduce the overall journey. Subtracting a vector is akin to adding a vector in the opposite direction.
Let's take the given vectors again, \( \vec{u} = \langle 1, 1, -1 \rangle \) and \( \vec{v} = \langle 2, 1, 2 \rangle \). To find \( \vec{u} - \vec{v} \), subtract each component of \( \vec{v} \) from \( \vec{u} \), giving \( \langle 1-2, 1-1, -1-2 \rangle = \langle -1, 0, -3 \rangle \).
This subtraction of vectors can be imagined as traveling to \( \vec{v} \) first and then stepping back by the route \( \vec{v} \) would take, the result shows how much further \( \vec{u} \) extends beyond \( \vec{v} \). Use vector subtraction when you need to find the distance or difference between two points represented as vectors.

Scalar Multiplication

When talking about scalar multiplication in vector algebra, think of stretching or shrinking a vector—changing its magnitude without altering its direction. If you've ever zoomed in or out on a map, you've performed a sort of scalar multiplication.
Scalar multiplication involves multiplying each component of the vector by the same scalar, or number. Consider vector \( \vec{u} = \langle 1, 1, -1 \rangle \). By multiplying it by \( \pi \), a constant, we adjust its length, resulting in \( \pi \vec{u} = \langle \pi, \pi, -\pi \rangle \). Similarly, multiplying vector \( \vec{v} \) by \( \sqrt{2} \) gives \( \sqrt{2}\vec{v} = \langle 2\sqrt{2}, \sqrt{2}, 2\sqrt{2} \rangle \).
Using scalar multiplication allows for the transformation of vectors to suit your needs, either extending them across space or squishing them down to something smaller, just by using regular multiplication. It is a fundamental operation for vector transformations, scaling the vector's size proportionately.

Vector Components

Vectors are defined by their components, each representing a portion of the vector's direction and magnitude. Think of vector components as the coordinates breaking down how far and in which directions the vector extends in space.
For instance, a vector like \( \vec{u} = \langle 1, 1, -1 \rangle \) has components 1 along the x-axis, 1 along the y-axis, and -1 along the z-axis. These components are critical in expressing vectors as they indicate precisely how much the vector moves in each dimension.
Understanding vector components is crucial for vector algebra because nearly every operation involves manipulating these pieces directly. When vectors are manipulated, whether through addition, subtraction, or scaling, it's these components that make the calculations possible. For any changes in a vector’s direction or magnitude, knowing its components provides a clear way of visualizing its impact.