Problem 19 - Algebra and Trigonometry

Question

Find \(a\) so that the vectors \(\mathbf{v}=\mathbf{i}-a \mathbf{j}\) and \(\mathbf{w}=2 \mathbf{i}+3 \mathbf{j}\) are orthogonal.

Step-by-Step Solution

Answer

a = \frac{2}{3}

Step 1 - Understand Orthogonality Condition

Two vectors are orthogonal if their dot product is zero. The dot product of two vectors \(\textbf{v} = a\textbf{i} + b\textbf{j}\) and \(\textbf{w} = c\textbf{i} + d\textbf{j}\) is given by \(a \times c + b \times d = 0\).

Step 2 - Identify Components of the Vectors

The given vectors are \(\textbf{v} = \textbf{i} - a\textbf{j}\) and \(\textbf{w} = 2\textbf{i} + 3\textbf{j}\). Therefore, the components are: \( \textbf{v}: (1, -a)\), and \( \textbf{w}: (2, 3)\).

Step 3 - Set Up the Dot Product Equation

Form the dot product of the vectors: \(1 \times 2 + (-a) \times 3 = 0\).

Step 4 - Solve for a

Simplify the equation from Step 3: \(2 - 3a = 0\). Now, solve for \(a\) by isolating the variable: \(2 = 3a\), so \(a = \frac{2}{3}\).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar (a single number). The dot product of two vectors \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) and \(\mathbf{w} = c\mathbf{i} + d\mathbf{j}\) is computed as \[a \times c + b \times d\]. This operation combines the respective components of the vectors. If the result is zero, the vectors are orthogonal, meaning they are perpendicular to each other. Remember this key point: If \(\mathbf{v} \cdot \mathbf{w} = 0\), then \(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal.
In our exercise, we verified orthogonality by setting the dot product to zero and solving the resulting equation.

Vector Components

Vectors can be broken down into components. For example, the vector \(\mathbf{v} = \mathbf{i} - a \mathbf{j}\) can be written as \(\mathbf{v} = (1, -a)\). Here, the vector has a component of 1 in the direction of \(\mathbf{i}\) (the x-axis) and \(-a\) in the direction of \(\mathbf{j}\) (the y-axis).
Understanding vector components helps in calculating the dot product. By identifying the components of \(\mathbf{v}\) and \(\mathbf{w}\), we were able to set up the equation \[1 \times 2 + (-a) \times 3 = 0\]. Breaking vectors into components is an essential step when dealing with vector operations.

Solving Equations

To solve for a variable in an equation, follow these steps:
1. Isolate the variable on one side of the equation.
2. Perform algebraic operations to simplify the expression.
3. Check your solution by plugging it back into the original equation.
In our exercise, we set up the equation \[2 - 3a = 0\] from the dot product condition. By isolating \(a\), we rearranged the equation to \[2 = 3a\]. Then, by dividing both sides by 3, we found \[a = \frac{2}{3}\]. This step-by-step process is useful for solving any linear equation involving a single variable.