Finding A Vector Orthogonal To Given Vectors
Hey guys! Let's dive into a cool problem in geometry: finding a vector that's not only perpendicular to two other vectors but also makes a wide, obtuse angle with another specified vector, all while having a particular magnitude. This is a classic problem that combines concepts of orthogonality, angles between vectors, and vector magnitudes. It might sound a bit complex, but we'll break it down step by step so it's super clear. We'll explore the conditions given – orthogonality, obtuse angle, and magnitude – and use these to build equations. These equations will then help us solve for the unknown components of our target vector. Think of it like a puzzle where each piece of information is a clue, and putting them together reveals the solution. So, grab your thinking caps, and let's get started!
Understanding the Problem
Before we jump into calculations, let's make sure we understand what the problem is asking. We're looking for a vector, let's call it x, that satisfies three conditions:
- Orthogonality: Vector x must be orthogonal (perpendicular) to two given vectors, a(2, 3, -1) and b(1, 1, -3). Remember, two vectors are orthogonal if their dot product is zero.
- Obtuse Angle: Vector x must form an obtuse angle with the vector i (which is the unit vector along the x-axis, or (1, 0, 0)). An obtuse angle means the angle is greater than 90 degrees, and this condition will help us determine the sign of one of the components of x.
- Magnitude: The magnitude (or length) of vector x is given as |x| = 138. This provides a crucial piece of information for finding the exact components of x.
By understanding these three conditions, we can translate the geometric requirements into algebraic equations. This is a key step in solving any problem involving vectors and their properties. Visualizing these conditions can also be helpful. Imagine the vectors a and b defining a plane, and x being perpendicular to this plane. The obtuse angle condition then restricts x to a specific direction within that perpendicular line. The magnitude finally fixes the length of x, giving us a unique solution.
Setting Up the Equations
Now that we understand the conditions, let's translate them into mathematical equations. This is where the real problem-solving begins! We'll start by expressing vector x in terms of its components: x = (x₁, x₂, x₃). Our goal is to find these components.
Orthogonality Conditions
Since x is orthogonal to a and b, their dot products must be zero:
- x ⋅ a = 0 which translates to 2x₁ + 3x₂ - x₃ = 0
- x ⋅ b = 0 which translates to x₁ + x₂ - 3x₃ = 0
These two equations give us a system of linear equations. Solving this system will provide a relationship between the components x₁, x₂, and x₃. It's like finding the common ground between the two orthogonality constraints. Each equation represents a plane in 3D space, and their intersection will be a line. The vector x must lie along this line.
Obtuse Angle Condition
The obtuse angle condition tells us that the angle between x and i is greater than 90 degrees. We can use the dot product formula to express this condition:
x ⋅ i = |x| |i| cos θ, where θ is the angle between x and i.
Since θ is obtuse, cos θ is negative. Also, |i| = 1. So, we have:
x ⋅ i = x₁ = |x| cos θ < 0
This tells us that the first component, x₁, must be negative. This is a crucial piece of information because it restricts the possible solutions we find from the orthogonality conditions. It's like adding a direction constraint – x not only has to be perpendicular to the plane defined by a and b, but it also has to point in a general direction away from the positive x-axis.
Magnitude Condition
The magnitude of x is given as |x| = 138. This gives us another equation:
√(x₁² + x₂² + x₃²) = 138
Squaring both sides, we get:
x₁² + x₂² + x₃² = 138² = 19044
This equation represents a sphere in 3D space, centered at the origin, with a radius of 138. The solution vector x must lie on the surface of this sphere. This condition ensures that the length of x matches the given magnitude. It's like adding a distance constraint – x has to be a certain distance away from the origin.
So, now we have a system of three equations (two from orthogonality and one from magnitude) and an inequality (from the obtuse angle condition). This is a well-defined system that we can solve to find the components of x. Let's move on to the next step: solving these equations.
Solving the Equations
Okay, guys, let's get our hands dirty with the math! We have a system of equations and an inequality to solve. Here's a recap of what we're working with:
- 2x₁ + 3x₂ - x₃ = 0
- x₁ + x₂ - 3x₃ = 0
- x₁² + x₂² + x₃² = 19044
- x₁ < 0
Solving the System of Linear Equations
First, let's focus on the two equations from the orthogonality conditions (1 and 2). We can solve this system for two variables in terms of the third. A common technique here is to use substitution or elimination. Let's use elimination. Multiply equation (2) by 2:
2x₁ + 2x₂ - 6x₃ = 0
Now, subtract this new equation from equation (1):
(2x₁ + 3x₂ - x₃) - (2x₁ + 2x₂ - 6x₃) = 0
This simplifies to:
x₂ + 5x₃ = 0
So, we have:
x₂ = -5x₃
Now, substitute this back into equation (2):
x₁ + (-5x₃) - 3x₃ = 0
This simplifies to:
x₁ = 8x₃
Great! Now we have x₁ and x₂ expressed in terms of x₃. This means our solution vector x can be written as x = (8x₃, -5x₃, x₃). It's like reducing the problem from finding three independent variables to finding just one – x₃. The other components are now tied to it.
Using the Magnitude Condition
Next, we'll use the magnitude condition (equation 3) to find the value of x₃. Substitute the expressions for x₁ and x₂ into the magnitude equation:
(8x₃)² + (-5x₃)² + (x₃)² = 19044
This simplifies to:
64x₃² + 25x₃² + x₃² = 19044
Which further simplifies to:
90x₃² = 19044
Now, divide both sides by 90:
x₃² = 19044 / 90 = 211.6
Take the square root of both sides:
x₃ = ±√211.6 ≈ ±14.55
So, we have two possible values for x₃. But we're not done yet! We need to consider the obtuse angle condition.
Applying the Obtuse Angle Condition
Remember, the obtuse angle condition (x₁ < 0) tells us that the first component of x must be negative. Since x₁ = 8x₃, this means 8x₃ < 0, which implies x₃ < 0. Therefore, we choose the negative value for x₃:
x₃ ≈ -14.55
Now we can find x₁ and x₂:
x₁ = 8x₃ ≈ 8 * (-14.55) ≈ -116.4
x₂ = -5x₃ ≈ -5 * (-14.55) ≈ 72.75
So, we have our vector x! It's like we've navigated through the maze of equations and found our treasure. But before we declare victory, let's check our solution.
Checking the Solution
Alright, we've found a potential solution for vector x, but let's not pop the champagne just yet! It's always good practice to check if our solution actually satisfies all the conditions of the problem. This step is crucial to avoid silly mistakes and ensure we've got the right answer. So, let's verify our solution: x ≈ (-116.4, 72.75, -14.55).
Checking Orthogonality
First, let's check if x is orthogonal to a and b. We'll do this by calculating the dot products and making sure they are close to zero (allowing for some rounding errors):
- x ⋅ a = (-116.4)(2) + (72.75)(3) + (-14.55)(-1) ≈ -232.8 + 218.25 + 14.55 ≈ 0
- x ⋅ b = (-116.4)(1) + (72.75)(1) + (-14.55)(-3) ≈ -116.4 + 72.75 + 43.65 ≈ 0
Great! The dot products are indeed close to zero, so x is orthogonal to both a and b. It's like our key fits the lock – the vector x satisfies the perpendicularity condition.
Checking the Obtuse Angle
Next, let's check if x forms an obtuse angle with i. We know that x₁ should be negative, and it is! x₁ ≈ -116.4, which is clearly negative. This confirms that the angle between x and i is greater than 90 degrees. It's like our compass points in the right direction – the vector x satisfies the obtuse angle condition.
Checking the Magnitude
Finally, let's check the magnitude of x:
|x| = √((-116.4)² + (72.75)² + (-14.55)²) ≈ √(13548.96 + 5292.5625 + 211.6025) ≈ √19053.125 ≈ 138.03
The magnitude is approximately 138, which is what we were given. The small difference is due to rounding errors in our calculations. It's like our measuring tape matches the blueprint – the vector x satisfies the magnitude condition.
Conclusion
We've checked all the conditions, and our solution x ≈ (-116.4, 72.75, -14.55) satisfies them all! So, we can confidently say that this is the vector we were looking for. High five! You've successfully navigated a complex problem involving vectors, orthogonality, angles, and magnitudes. This is a valuable skill in geometry and physics.
Final Answer
Therefore, the coordinates of vector x are approximately (-116.4, 72.75, -14.55). Remember, guys, the key to solving problems like this is to break them down into smaller, manageable steps. Understand the conditions, translate them into equations, solve the equations, and always, always check your solution! Geometry can be fun, especially when you conquer a challenging problem. Keep practicing, and you'll become a vector whiz in no time! Keep exploring the world of math and geometry, and who knows what fascinating challenges you'll conquer next!