Problem 24 - A Graphical Approach to Precalculus with Limits

Question

Given vectors u and v, find (a) \(2 u\) (b) \(2 u+3 v\) (c) \(v-3 u\) Do not use a calculator. $$\mathbf{u}=-\mathbf{i}+2 \mathbf{j}, \mathbf{v}=\mathbf{i}-\mathbf{j}$$

Step-by-Step Solution

Answer

(a) \(-2\mathbf{i} + 4\mathbf{j}\), (b) \(\mathbf{i} + \mathbf{j}\), (c) \(4\mathbf{i} - 7\mathbf{j}\)."}

Step 1: Find the Expression for 2u

The vector \(\mathbf{u}\) is given as \(-\mathbf{i} + 2\mathbf{j}\). To find \(2\mathbf{u}\), we multiply each component of \(\mathbf{u}\) by 2:\[2\mathbf{u} = 2(-\mathbf{i}) + 2(2\mathbf{j}) = -2\mathbf{i} + 4\mathbf{j}\]

Step 2: Calculate the Sum 2u + 3v

First, find \(3\mathbf{v}\) by multiplying each component of \(\mathbf{v}\) by 3. Given \(\mathbf{v} = \mathbf{i} - \mathbf{j}\), we have:\[3\mathbf{v} = 3(\mathbf{i} - \mathbf{j}) = 3\mathbf{i} - 3\mathbf{j}\]Next, add \(2\mathbf{u}\) from Step 1 to \(3\mathbf{v}\) to get \(2\mathbf{u} + 3\mathbf{v}\):\[2\mathbf{u} + 3\mathbf{v} = (-2\mathbf{i} + 4\mathbf{j}) + (3\mathbf{i} - 3\mathbf{j}) = (-2\mathbf{i} + 3\mathbf{i}) + (4\mathbf{j} - 3\mathbf{j}) = \mathbf{i} + \mathbf{j}\]

Step 3: Solve the Expression v - 3u

First, compute \(3\mathbf{u}\) by multiplying each component of \(\mathbf{u}\) by 3. Given \(\mathbf{u} = -\mathbf{i} + 2\mathbf{j}\), we get:\[3\mathbf{u} = 3(-\mathbf{i} + 2\mathbf{j}) = -3\mathbf{i} + 6\mathbf{j}\]Now, subtract \(3\mathbf{u}\) from \(\mathbf{v}\) to find \(\mathbf{v} - 3\mathbf{u}\):\[\mathbf{v} - 3\mathbf{u} = (\mathbf{i} - \mathbf{j}) - (-3\mathbf{i} + 6\mathbf{j}) = (\mathbf{i} + 3\mathbf{i}) + (-\mathbf{j} - 6\mathbf{j}) = 4\mathbf{i} - 7\mathbf{j}\]

Key Concepts

Vector Multiplication

Vector multiplication, particularly scalar multiplication, involves multiplying each component of a vector by a scalar (a single number). This operation stretches or shrinks the vector while retaining its direction, unless multiplied by a negative, which reverses it.
Consider the vector \( \mathbf{u} = -\mathbf{i} + 2\mathbf{j} \). To find \( 2\mathbf{u} \), each term in the vector is multiplied by 2:

• Multiply the \( \mathbf{i} \)-component: \( 2(-\mathbf{i}) = -2\mathbf{i} \).
• Multiply the \( \mathbf{j} \)-component: \( 2(2\mathbf{j}) = 4\mathbf{j} \).

After performing these operations, the result is \( 2\mathbf{u} = -2\mathbf{i} + 4\mathbf{j} \). This demonstrates how scalar multiplication can modify a vector's magnitude while maintaining its inherent properties.

Vector Addition

Vector addition combines two vectors to produce a third vector, called the resultant. Each corresponding component of the vectors is added together. This operation can be visualized as placing the tail of the second vector at the head of the first and drawing a vector from the tail of the first to the head of the second.
For example, with vectors \( 2\mathbf{u} = -2\mathbf{i} + 4\mathbf{j} \) and \( 3\mathbf{v} = 3\mathbf{i} - 3\mathbf{j} \), the addition is performed on the components:

• Add the \( \mathbf{i} \)-components: \( -2\mathbf{i} + 3\mathbf{i} = \mathbf{i} \).
• Add the \( \mathbf{j} \)-components: \( 4\mathbf{j} - 3\mathbf{j} = \mathbf{j} \).

Hence, \( 2\mathbf{u} + 3\mathbf{v} = \mathbf{i} + \mathbf{j} \), creating a new vector with components derived from the sums of respective parts of \( \mathbf{u} \) and \( \mathbf{v} \).

Vector Subtraction

Vector subtraction finds the difference between two vectors by reversing the direction of the vector being subtracted and then performing vector addition. Seen another way, it involves subtracting each component of the second vector from the corresponding component of the first.
To illustrate this, consider \( \mathbf{v} - 3\mathbf{u} \). Start with \( \mathbf{v} = \mathbf{i} - \mathbf{j} \) and \( 3\mathbf{u} = -3\mathbf{i} + 6\mathbf{j} \). The subtraction goes as follows:

• Subtract the \( \mathbf{i} \)-components: \( \mathbf{i} - (-3\mathbf{i}) = \mathbf{i} + 3\mathbf{i} = 4\mathbf{i} \).
• Subtract the \( \mathbf{j} \)-components: \( -\mathbf{j} - 6\mathbf{j} = -7\mathbf{j} \).

After calculation, we arrive at \( 4\mathbf{i} - 7\mathbf{j} \). This operation effectively determines the resultant vector by offsetting the influence of one vector on another.