Problem 32 - Advanced Engineering Mathematics

Question

Solve for the unknown. $$ P_{1}(x, x, 1), P_{2}(0,3,5) ; d\left(P_{1}, P_{2}\right)=5 $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = 3 \).

Step 1: Understand Distance Formula

The distance between two points \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\) in three-dimensional space is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] Here, we are given that the distance \(d\) between the points \(P_1(x, x, 1)\) and \(P_2(0, 3, 5)\) is 5.

Step 2: Substitute Given Values

Using the distance formula from Step 1, substitute the given coordinates:\[d = \sqrt{(0-x)^2 + (3-x)^2 + (5-1)^2} = 5\] This equation becomes: \[ 5 = \sqrt{x^2 + (3-x)^2 + 4^2} \] Now, simplify the expression under the square root.

Step 3: Simplify the Expression

Expand \((3-x)^2\), which is \((3-x)(3-x) = 9 - 6x + x^2\). Substitute back into the equation:\[ 5 = \sqrt{x^2 + 9 - 6x + x^2 + 16} \] Combine like terms:\[ 5 = \sqrt{2x^2 - 6x + 25} \]

Step 4: Square Both Sides

To eliminate the square root, square both sides of the equation:\[ 25 = 2x^2 - 6x + 25 \] Now, simplify the equation by subtracting 25 from both sides:

Step 5: Simplify and Solve Quadratic Equation

Subtracting 25 from both sides, we have: \[ 0 = 2x^2 - 6x \] Factor out the greatest common factor:\[ 0 = 2x(x - 3) \] Set each factor equal to zero: \[ 2x = 0 \] and \[ x - 3 = 0 \]. This gives us the solutions \( x = 0 \) and \( x = 3 \).

Step 6: Verify Solutions

Verify that both solutions satisfy the original distance condition. For \(x = 0\), check:\[ d(0, 0, 1, 0, 3, 5) = \sqrt{(0-0)^2 + (3-0)^2 + (5-1)^2} = \sqrt{9 + 16} = 5 \] For \(x = 3\), check: \[ d(3, 3, 1, 0, 3, 5) = \sqrt{(0-3)^2 + (3-3)^2 + (5-1)^2} = \sqrt{9 + 16} = 5 \] Both solutions are correct.

Key Concepts

Quadratic Equations

Quadratic equations form the foundation of algebraic problem-solving. In the previous solution, we end up with the quadratic equation:\[ 0 = 2x^2 - 6x \]A quadratic equation is any equation that can be rearranged into the standard form \( ax^2 + bx + c = 0 \). Here, \( a = 2 \), \( b = -6 \), and \( c = 0 \). Quadratic equations can have two possible solutions, which we determine by either factoring, using the quadratic formula, or completing the square. In this instance, the equation was successfully solved using factoring methods:

• The equation is factored into \( 2x(x - 3) = 0 \).
• Setting each factor equal to zero gives: \( 2x = 0 \) and \( x - 3 = 0 \).
• Solving these yields solutions \( x = 0 \) and \( x = 3 \).

Understanding how to manipulate quadratic equations is crucial for solving many mathematical problems, especially in geometry and physics.

Coordinate Geometry

In coordinate geometry, the position of a point in space is defined by its coordinates, and the relationships between points are explored. The exercise involves two points \( P_1(x,x,1) \) and \( P_2(0,3,5) \). The distance formula, fundamental in coordinate geometry, is used to calculate the distance between two points in a 3D space.

• The distance between points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) in three dimensions is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
• For this problem, \( d = 5 \), which sets up an equation that helps determine the unknown \( x \).

Coordinate geometry allows us to solve real-life problems involving distance, positioning, and navigation in three-dimensional spaces.

Algebraic Manipulations

Algebraic manipulations are key to transforming and simplifying mathematical expressions, making them solvable. This exercise required the application of these skills at various stages:

• First, the square of binomial expansions helped in simplifying \((3-x)^2\), resulting in \(9 - 6x + x^2\).
• Combining like terms made the expression under the square root clearer: \(2x^2 - 6x + 25\).
• By squaring both sides of the equation, the square root was eliminated, leading to a solvable quadratic equation.
• Lastly, rearranging terms allowed the equation to be factored, yielding solutions.

Overall, algebraic manipulations involve a series of operations such as expanding, factoring, and simplifying expressions. Such manipulations are essential for transitioning from complex forms to simple, actionable equations that are easier to interpret or solve.