Problem 34 - Precalculus Mathematics for Calculus

Question

Find \(2 u,-3 v, u+v,\) and \(3 u-4 v\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) $$\mathbf{u}=\mathbf{i}, \quad \mathbf{v}=-2 \mathbf{j}$$

Step-by-Step Solution

Answer

\(2\mathbf{u} = \langle 2, 0 \rangle\), \(-3\mathbf{v} = \langle 0, 6 \rangle\), \(\mathbf{u} + \mathbf{v} = \langle 1, -2 \rangle\), \(3\mathbf{u} - 4\mathbf{v} = \langle 3, 8 \rangle\)."

Step 1: Understand the Vector Components

The vector \( \mathbf{u} \) can be represented as \( \mathbf{u} = \langle 1, 0 \rangle \) because it's simplified to \( 1 \mathbf{i} + 0 \mathbf{j} \). Likewise, \( \mathbf{v} = \langle 0, -2 \rangle \) because it's \( 0 \mathbf{i} - 2 \mathbf{j} \).

Step 2: Determine 2u

To find \( 2\mathbf{u} \), multiply each component of \( \mathbf{u} = \langle 1, 0 \rangle \) by 2: \[ 2\mathbf{u} = 2 \times \langle 1, 0 \rangle = \langle 2 \times 1, 2 \times 0 \rangle = \langle 2, 0 \rangle \]

Step 3: Calculate -3v

To find \( -3\mathbf{v} \), multiply each component of \( \mathbf{v} = \langle 0, -2 \rangle \) by -3: \[ -3\mathbf{v} = -3 \times \langle 0, -2 \rangle = \langle -3 \times 0, -3 \times (-2) \rangle = \langle 0, 6 \rangle \]

Step 4: Sum u and v

To compute \( \mathbf{u} + \mathbf{v} \), add corresponding components: \[ \mathbf{u} + \mathbf{v} = \langle 1, 0 \rangle + \langle 0, -2 \rangle = \langle 1 + 0, 0 - 2 \rangle = \langle 1, -2 \rangle \]

Step 5: Evaluate 3u - 4v

For \( 3\mathbf{u} - 4\mathbf{v} \), compute each portion separately: First, \( 3\mathbf{u} = 3 \times \langle 1, 0 \rangle = \langle 3, 0 \rangle \).Next, \( 4\mathbf{v} = 4 \times \langle 0, -2 \rangle = \langle 0, -8 \rangle \).Now, subtract \( 4\mathbf{v} \) from \( 3\mathbf{u} \):\[ 3\mathbf{u} - 4\mathbf{v} = \langle 3, 0 \rangle - \langle 0, -8 \rangle = \langle 3 - 0, 0 - (-8) \rangle = \langle 3, 8 \rangle \]

Key Concepts

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar, which is simply a real number. This operation is straightforward and only requires us to multiply each component of the vector by the scalar.
For instance, if you have a vector \( \mathbf{u} = \langle 1, 0 \rangle \), and you want to calculate \( 2\mathbf{u} \), you multiply each component by 2. Therefore, \( 2\mathbf{u} = \langle 2 \times 1, 2 \times 0 \rangle = \langle 2, 0 \rangle \).
Similarly, scalar multiplication can involve negative numbers. When calculating \( -3\mathbf{v} \) for \( \mathbf{v} = \langle 0, -2 \rangle \), each component of the vector is multiplied by -3, resulting in \( \langle 0, 6 \rangle \).

• This operation will scale the vector, changing its magnitude while keeping the direction consistent if the scalar is positive.
• If the scalar is negative, the direction of the vector reverses.
• The procedure is a simple multiplication of vector coordinates by the scalar value.

Practice and familiarity make scalar multiplication a breeze!

Vector Addition

Vector addition combines two vectors into one by adding each of their respective components.
Imagine you have two vectors, \( \mathbf{u} = \langle 1, 0 \rangle \) and \( \mathbf{v} = \langle 0, -2 \rangle \). To add these vectors, we add the first components together and the second components together. Consequently:\[ \mathbf{u} + \mathbf{v} = \langle 1 + 0, 0 - 2 \rangle = \langle 1, -2 \rangle \]

• Vector addition is commutative, meaning \( \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} \).
• It effectively translates vectors in the coordinate plane, shifting their position without changing their properties.
• Each corresponding component is combined to form the resultant vector.

This skill is fundamental in fields like physics and engineering, where vectors represent quantities like force or velocity.

Vector Subtraction

Vector subtraction works similarly to vector addition, but instead of adding, you subtract each corresponding component of the two vectors.
For example, if we need to calculate \( 3\mathbf{u} - 4\mathbf{v} \):

• First, multiply \( \mathbf{u} = \langle 1, 0 \rangle \) by 3 to get \( \langle 3, 0 \rangle \).
• Next, multiply \( \mathbf{v} = \langle 0, -2 \rangle \) by 4 to get \( \langle 0, -8 \rangle \).
• Finally, subtract \( 4\mathbf{v} \) from \( 3\mathbf{u} \), resulting in: \[ 3\mathbf{u} - 4\mathbf{v} = \langle 3, 0 \rangle - \langle 0, -8 \rangle = \langle 3, 8 \rangle \]

Subtraction of vectors leads to a new vector that represents the directional difference between the two original vectors.

Similar to addition, subtraction shifts the vector's position in space.

Subtracting a vector reverses its direction, essentially adding its negative.

Coordinate Representation

Vectors are often expressed in a coordinate system using unit vectors, traditionally denoted as \( \mathbf{i} \) and \( \mathbf{j} \) in two-dimensional space.
In our case, vector \( \mathbf{u} = \mathbf{i} \), which implies \( \mathbf{u} = \langle 1, 0 \rangle \), representing movement along the x-axis with no movement along the y-axis.
Similarly, \( \mathbf{v} = -2 \mathbf{j} \) translates to \( \mathbf{v} = \langle 0, -2 \rangle \), indicating movement along the negative direction of the y-axis with no movement along the x-axis.

• Coordinate representation simplifies complex vector operations by standardizing their format.
• It allows for a straightforward manipulation of vectors using basic arithmetic operations.
• This representation is crucial across many disciplines that require precise and clear vector depiction.

Being comfortable with coordinate representation enables effective problem-solving in technical fields.