Problem 33 - Precalculus Mathematics for Calculus

Question

Find \(2 u,-3 v, u+v,\) and \(3 u-4 v\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) $$\mathbf{u}=\langle 0,-1\rangle, \quad \mathbf{v}=\langle- 2,0\rangle$$

Step-by-Step Solution

Answer

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

Step 1: Determine 2u

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

Step 2: Determine -3v

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

Step 3: Calculate u+v

To find \( \mathbf{u} + \mathbf{v} \), add the corresponding components of \( \mathbf{u} = \langle 0, -1 \rangle \) and \( \mathbf{v} = \langle -2, 0 \rangle \).\[\mathbf{u} + \mathbf{v} = \langle 0 + (-2), -1 + 0 \rangle = \langle -2, -1 \rangle\]

Step 4: Calculate 3u - 4v

First find \( 3\mathbf{u} \) by multiplying each component of \( \mathbf{u} = \langle 0, -1 \rangle \) by 3, and \( 4\mathbf{v} \) by multiplying each component of \( \mathbf{v} = \langle -2, 0 \rangle \) by 4, then subtract \( 4\mathbf{v} \) from \( 3\mathbf{u} \).\[3\mathbf{u} = 3 \langle 0, -1 \rangle = \langle 0, -3 \rangle\]\[4\mathbf{v} = 4 \langle -2, 0 \rangle = \langle -8, 0 \rangle\]Subtract the vectors:\[3\mathbf{u} - 4\mathbf{v} = \langle 0, -3 \rangle - \langle -8, 0 \rangle = \langle 0 - (-8), -3 - 0 \rangle = \langle 8, -3 \rangle\]

Key Concepts

Scalar Multiplication

Scalar multiplication is an operation that involves multiplying each component of a vector by the same scalar (a real number). This operation helps in scaling the vector either up or down.
For instance, if we have a vector \( \mathbf{u} = \langle a, b \rangle \) and we want to multiply it by a scalar \( k \), the result is \( k\mathbf{u} = \langle k \, a, k \, b \rangle \). Each component of the vector is multiplied by the scalar.
To illustrate this more concretely, consider our original vector \( \mathbf{u} = \langle 0, -1 \rangle \). If we multiply by \( 2 \) to get \( 2\mathbf{u} \), we compute:

• Multiply the first component: \( 2 \times 0 = 0 \)
• Multiply the second component: \( 2 \times (-1) = -2 \)

So, \( 2\mathbf{u} = \langle 0, -2 \rangle \). Scalar multiplication provides a quick way to stretch or shrink a vector.

Vector Addition

Vector addition is the process of adding two vectors together to obtain a new vector. This operation involves adding the respective components of the vectors. It is one of the simplest yet profoundly useful vector operations.
Given vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), the sum \( \mathbf{u} + \mathbf{v} = \langle u_1 + v_1, u_2 + v_2 \rangle \).
For example, with \( \mathbf{u} = \langle 0, -1 \rangle \) and \( \mathbf{v} = \langle -2, 0 \rangle \), calculate \( \mathbf{u} + \mathbf{v} \):

• Add the first components: \( 0 + (-2) = -2 \)
• Add the second components: \( -1 + 0 = -1 \)

Thus \( \mathbf{u} + \mathbf{v} = \langle -2, -1 \rangle \).
Vector addition constructs a new vector that is geometrically the diagonal of the parallelogram created by the two original vectors.

Linear Combination of Vectors

A linear combination of vectors involves creating a new vector by adding together scalar multiples of two or more vectors. This operation is foundational in vector spaces and is widely used in solving vector equations.
Consider two vectors \( \mathbf{u} \) and \( \mathbf{v} \) with scalars \( a \) and \( b \). The linear combination is expressed as \( a\mathbf{u} + b\mathbf{v} \).
In the context of our exercise, we explore the combination \( 3\mathbf{u} - 4\mathbf{v} \).

• First compute \( 3\mathbf{u} = 3 \langle 0, -1 \rangle = \langle 0, -3 \rangle \)
• Then compute \( -4\mathbf{v} = -4 \langle -2, 0 \rangle = \langle 8, 0 \rangle \)
• Combine these results: \( 3\mathbf{u} + (-4\mathbf{v}) = \langle 0 + 8, -3 + 0 \rangle = \langle 8, -3 \rangle \)

Linear combinations are key in expressing complex equations and movements, such as in physics and engineering, through simpler vector components.