Problem 18 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 17-20: Determine \(m\) such that the two vectors are orthogonal. $$ 4 m \mathbf{i}+\mathbf{j}, \quad 9 m \mathbf{i}-25 \mathbf{j} $$

Step-by-Step Solution

Answer

The values of \(m\) are \(\pm \frac{5}{6}\).

Step 1: Define Orthogonal Vectors

Two vectors are orthogonal if their dot product is zero. Given two vectors \(\mathbf{a}\) and \(\mathbf{b}\), they are orthogonal if \(\mathbf{a} \cdot \mathbf{b} = 0\).

Step 2: Write Dot Product Formula

For vectors \(\mathbf{a} = 4m\mathbf{i} + \mathbf{j}\) and \(\mathbf{b} = 9m\mathbf{i} - 25\mathbf{j}\), the dot product is \((4m) \times (9m) + (1) \times (-25)\).

Step 3: Solve Dot Product Equation

Set the dot product equation to zero: \[ (4m)(9m) + (1)(-25) = 0 \] which simplifies to \[ 36m^2 - 25 = 0 \].

Step 4: Simplify and Solve Quadratic Equation

Solve the quadratic equation \(36m^2 - 25 = 0\). Add 25 to both sides to get: \(36m^2 = 25\). Divide both sides by 36 to get \(m^2 = \frac{25}{36}\).

Step 5: Solve for m

Taking the square root of both sides, we get \(m = \pm \frac{5}{6}\).

Key Concepts

Dot Product

The concept of the dot product is essential in understanding orthogonal vectors. When you multiply two vectors together using the dot product, you're essentially finding how much one vector goes in the direction of the other. If the dot product is zero, it means the vectors are perpendicular, or orthogonal.

The dot product of two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j}\) is calculated as follows:

• Multiply the coefficients of the \(\mathbf{i}\) components: \(a_1 \times b_1\).
• Multiply the coefficients of the \(\mathbf{j}\) components: \(a_2 \times b_2\).
• Add the results: \(a_1 \times b_1 + a_2 \times b_2\).

In our original problem, you calculate \(\mathbf{a} \cdot \mathbf{b}\) as \((4m)(9m) + (1)(-25)\), setting it equal to zero to satisfy the orthogonal condition.

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \(ax^2 + bx + c = 0\). In our problem, after finding the dot product, we were left with a quadratic equation \(36m^2 - 25 = 0\).

To solve a quadratic equation, you follow these general steps:

• Rearrange the equation in the form \(ax^2 + bx + c = 0\).
• If possible, simplify the equation by dividing by any common factor.
• Use methods like factoring, completing the square, or the quadratic formula to find the solutions for the variable.

In the exercise, since there is no linear term \(b\) or constant term \(c\) making the equation easier to solve by isolating the term \(36m^2 = 25\).

Solving for Unknowns

Solving for unknowns in an algebraic equation involves finding the values that satisfy the equation. Here, we needed to find values for \(m\) that make the vectors orthogonal.

Once the quadratic equation was simplified to \(36m^2 = 25\), the next step was to solve for \(m\):

• First, isolate \(m^2\) by dividing both sides by 36, resulting in \(m^2 = \frac{25}{36}\).
• To find \(m\), take the square root of both sides. Remember that taking the square root can yield two results, positive and negative. This gives us \(m = \pm \frac{5}{6}\).

These solutions for \(m\) ensure that the original condition for orthogonality is met, providing the values needed to solve the exercise.