Problem 35 - Algebra and Trigonometry

Question

\(31-36\) Find \(2 \mathbf{u},-3 \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(3 \mathbf{u}-4 \mathbf{v}\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v} .\) $$ \mathbf{u}=2 \mathbf{i}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j} $$

Step-by-Step Solution

Answer

\(2 \mathbf{u} = 4 \mathbf{i}, -3 \mathbf{v} = -9 \mathbf{i} + 6 \mathbf{j}, \mathbf{u} + \mathbf{v} = 5 \mathbf{i} - 2 \mathbf{j}, 3 \mathbf{u} - 4 \mathbf{v} = 18 \mathbf{i} - 8 \mathbf{j}.\)

Step 1: Calculate \(2 \mathbf{u}\)

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

Step 2: Calculate \(-3 \mathbf{v}\)

The vector \(\mathbf{v}\) is given as \(3 \mathbf{i} - 2 \mathbf{j}\). To find \(-3 \mathbf{v}\), multiply each component of \(\mathbf{v}\) by -3:\[-3 \mathbf{v} = -3 \times (3 \mathbf{i} - 2 \mathbf{j}) = -9 \mathbf{i} + 6 \mathbf{j}. \] Thus, \(-3 \mathbf{v} = -9 \mathbf{i} + 6 \mathbf{j}\).

Step 3: Determine \(\mathbf{u} + \mathbf{v}\)

Add the vectors \(\mathbf{u} = 2 \mathbf{i}\) and \(\mathbf{v} = 3 \mathbf{i} - 2 \mathbf{j}\):\[\mathbf{u} + \mathbf{v} = (2 \mathbf{i}) + (3 \mathbf{i} - 2 \mathbf{j}) = (2 \mathbf{i} + 3 \mathbf{i}) - 2 \mathbf{j} = 5 \mathbf{i} - 2 \mathbf{j}. \] Thus, \(\mathbf{u} + \mathbf{v} = 5 \mathbf{i} - 2 \mathbf{j}\).

Step 4: Calculate \(3 \mathbf{u} - 4 \mathbf{v}\)

First, compute \(3 \mathbf{u}\):\[3 \mathbf{u} = 3 \times (2 \mathbf{i}) = 6 \mathbf{i}.\] Next, compute \(-4 \mathbf{v}\):\[-4 \mathbf{v} = -4 \times (3 \mathbf{i} - 2 \mathbf{j}) = -12 \mathbf{i} + 8 \mathbf{j}.\]Now, subtract \(-4 \mathbf{v}\) from \(3 \mathbf{u}\):\[3 \mathbf{u} - 4 \mathbf{v} = 6 \mathbf{i} - (-12 \mathbf{i} + 8 \mathbf{j}) = 6 \mathbf{i} + 12 \mathbf{i} - 8 \mathbf{j} = 18 \mathbf{i} - 8 \mathbf{j}.\] Thus, \(3 \mathbf{u} - 4 \mathbf{v} = 18 \mathbf{i} - 8 \mathbf{j}\).

Key Concepts

Scalar Multiplication

Scalar multiplication is one of the fundamental operations in vector algebra. It involves multiplying a vector by a scalar (a real number), which alters the magnitude of the vector without changing its direction. This operation affects each component of the vector.
For instance, consider vector \( \mathbf{u} = 2\mathbf{i} \). If you multiply it by a scalar 2, you obtain \( 2 \mathbf{u} = 4 \mathbf{i} \). Here, each component of \( \mathbf{u} \) is doubled.

• Direction: The vector remains in the same direction unless multiplied by a negative scalar, which reverses the direction.
• Magnitude: The length of the vector is scaled by the absolute value of the scalar.

Understanding scalar multiplication helps in scaling vectors to desired lengths, which is particularly useful in physics and engineering applications.

Vector Addition

Vector addition is a method of combining two vectors to yield a resultant vector. This operation is performed by adding the respective components of the vectors together.
For example, given vectors \( \mathbf{u} = 2\mathbf{i} \) and \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \), their sum is calculated as follows:

• Add the \( \mathbf{i} \) components: \( 2\mathbf{i} + 3\mathbf{i} = 5\mathbf{i} \).
• Add the \( \mathbf{j} \) components (bearing in mind the sign): \( 0\mathbf{j} - 2\mathbf{j} = -2\mathbf{j} \).

Thus, the resulting vector from \( \mathbf{u} + \mathbf{v} \) is \( 5\mathbf{i} - 2\mathbf{j} \).
This operation is important in fields like physics, where it is used to determine resultant forces or velocities.

Vector Subtraction

Vector subtraction follows principles similar to vector addition, but involves subtracting the components of one vector from another. This operation allows us to find the vector between two points or opposing a given vector.
To subtract vector \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \) from vector \( \mathbf{u} = 2\mathbf{i} \), perform the following calculations:

• Subtract \( \mathbf{i} \) components: \( 2\mathbf{i} - 3\mathbf{i} = -1\mathbf{i} \).
• Subtract \( \mathbf{j} \) components (not present in \( \mathbf{u} \)): \( 0\mathbf{j} - (-2\mathbf{j}) = 2\mathbf{j} \).

Subtraction can also be visualized by reversing the direction of the subtracting vector and then performing vector addition.

Linear Combinations

A linear combination involves multiplying vectors by scalars and then adding the results. This concept is crucial for solving many algebraic and geometric problems.
Consider finding \( 3\mathbf{u} - 4\mathbf{v} \), which is a linear combination of \( \mathbf{u} = 2\mathbf{i} \) and \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \):

• First, compute \( 3\mathbf{u} = 3 \times 2\mathbf{i} = 6\mathbf{i} \).
• Next, compute \(-4\mathbf{v} = -4(3\mathbf{i} - 2\mathbf{j}) = -12\mathbf{i} + 8\mathbf{j} \).
• Finally, add these results: \( 3\mathbf{u} - 4\mathbf{v} = 6\mathbf{i} - 12\mathbf{i} + 8\mathbf{j} = 18\mathbf{i} - 8\mathbf{j} \).

Linear combinations are foundational for determining vectors that lie within a plane or defining equations of lines using vectors.