Problem 17 - Contemporary Precalculus

Question

Find \(u+v, v-u,\) and \(2 u-3 v\). $$\mathbf{u}=8 \mathbf{i}, \mathbf{v}=2(3 \mathbf{i}-2 \mathbf{j})$$

Step-by-Step Solution

Answer

Question: Find the values of the following vector expressions: \(u+v\), \(v-u\), and \(2u-3v\), given that \(\mathbf{u} = 8\mathbf{i}\) and \(\mathbf{v} = 2(3\mathbf{i}-2\mathbf{j})\). Answer: The values of the vector expressions are: \(u+v = 14\mathbf{i} - 4\mathbf{j}\), \(v-u = -2\mathbf{i} - 4\mathbf{j}\), and \(2u-3v = -2 \mathbf{i} + 12 \mathbf{j}\).

Step 1: Vector Representation

First, represent the given vectors in component form. \(\mathbf{u} = 8\mathbf{i} = \langle 8, 0 \rangle\) \(\mathbf{v} = 2(3\mathbf{i}-2\mathbf{j}) = \langle 2(3), 2(-2) \rangle = \langle 6, -4 \rangle\) Now, we can calculate the required expressions.

Step 2: Calculate u + v

To find the sum of the vectors, add their corresponding components together. \(u+v = \langle u_x + v_x, u_y + v_y \rangle\) \(u+v = \langle 8+6, 0+(-4) \rangle = \langle 14, -4 \rangle\) So \(u+v = 14\mathbf{i} - 4\mathbf{j}\).

Step 3: Calculate v - u

To find the difference between the two vectors, subtract their corresponding components. \(v-u = \langle v_x - u_x, v_y - u_y \rangle\) \(v-u = \langle 6-8, (-4)-0 \rangle = \langle -2, -4 \rangle\) So \(v-u = -2\mathbf{i} - 4\mathbf{j}\).

Step 4: Calculate 2u - 3v

To find the linear combination of the vectors, multiply each vector by its corresponding scalar and subtract their respective components. \(2u-3v = \langle 2u_x - 3v_x, 2u_y - 3v_y \rangle\) \(2u-3v = \langle 2(8) - 3(6), 2(0) - 3(-4) \rangle = \langle 16-18, 0+12 \rangle = \langle -2, 12 \rangle\) So \(2u-3v = -2 \mathbf{i} + 12 \mathbf{j}\).

Key Concepts

vector addition

Vector addition is a fundamental operation where you combine two vectors to get a new vector. It's like taking two different journeys and finding where you end up if you add the paths together. In mathematical terms, if you have vectors \(\mathbf{u}\) and \(\mathbf{v}\), their sum \(\mathbf{u} + \mathbf{v}\) is found by adding their corresponding components.
For example, if \(\mathbf{u} = \langle 8, 0 \rangle\) and \(\mathbf{v} = \langle 6, -4 \rangle\), their sum is:

• First component: \(8 + 6 = 14\)
• Second component: \(0 + (-4) = -4\)

Thus, \(\mathbf{u} + \mathbf{v} = \langle 14, -4 \rangle\), which translates to the vector \(14\mathbf{i} - 4\mathbf{j}\). This operation is visually represented by placing the tail of vector \(\mathbf{v}\) at the head of vector \(\mathbf{u}\) and drawing a new vector from the tail of \(\mathbf{u}\) to the head of \(\mathbf{v}\). This gives you a new direction and magnitude, which is useful in navigation, physics, and graphics.

vector subtraction

Vector subtraction involves finding the difference between two vectors. It's like determining how much one journey would need to change to become another. To subtract vector \(\mathbf{u}\) from vector \(\mathbf{v}\), you compute \(\mathbf{v} - \mathbf{u}\) by subtracting the corresponding components.
Given \(\mathbf{u} = \langle 8, 0 \rangle\) and \(\mathbf{v} = \langle 6, -4 \rangle\), the subtraction is as follows:

• First component: \(6 - 8 = -2\)
• Second component: \((-4) - 0 = -4\)

This results in the vector \(\langle -2, -4 \rangle\) or \(-2\mathbf{i} - 4\mathbf{j}\). In geometric terms, subtracting vectors can be visualized by connecting their heads; this shows a vector directing from \(\mathbf{u}\) to \(\mathbf{v}\). Subtraction is especially useful in contexts like physics, where you may need to find relative velocity.

scalar multiplication

Scalar multiplication scales a vector by a constant, changing its magnitude but not its direction. Think of it as stretching or shrinking a vector on a drawing. The calculation involves multiplying each component of the vector by the scalar.
With vector \(\mathbf{u} = \langle 8, 0 \rangle\) and scalar \(2\), you calculate \(2\mathbf{u}\):

• Multiply the first component by \(2\): \(2 \times 8 = 16\)
• Multiply the second component by \(2\): \(2 \times 0 = 0\)

This gives \(2\mathbf{u} = \langle 16, 0 \rangle\). Similarly, for vector \(\mathbf{v} = \langle 6, -4 \rangle\) with a scalar \(-3\):

• First component: \(-3 \times 6 = -18\)
• Second component: \(-3 \times (-4) = 12\)

So, \(-3\mathbf{v} = \langle -18, 12 \rangle\). When combining several operations like in \(2\mathbf{u} - 3\mathbf{v}\), you first perform the scalar multiplication for each vector and then apply vector addition or subtraction, depending on the problem.