Acute Angle Between Vectors: Finding The Value Of 'a'

Hey guys! Today, we're diving into a super interesting problem from algebra: figuring out when two vectors form an acute angle. Specifically, we want to find the value (or values) of 'a' that make the vectors a(-1; 4; a) and b(5; -1; a) create an acute angle. Buckle up, because we're about to get our vector math on!

Understanding Acute Angles and Vectors

So, before we jump into the nitty-gritty calculations, let's make sure we're all on the same page about what an acute angle is and how it relates to vectors. An acute angle, simply put, is an angle that's less than 90 degrees. Think of it as a nice, sharp angle, not a wide or obtuse one. Now, how does this connect to vectors? Well, the angle between two vectors tells us something about their relationship in space. If the angle is acute, it means the vectors are pointing in roughly the same direction. If it's obtuse (greater than 90 degrees), they're pointing more in opposite directions. And if it's exactly 90 degrees, they're perpendicular, meaning they're at right angles to each other.

To determine the angle between two vectors, we use a handy tool called the dot product. The dot product of two vectors a and b is defined as:

a · b = |a| |b| cos θ

Where |a| and |b| represent the magnitudes (lengths) of the vectors, and θ (theta) is the angle between them. The key here is the cos θ term. The cosine function behaves in a specific way: it's positive for angles between 0 and 90 degrees (acute angles), zero for 90 degrees (right angles), and negative for angles between 90 and 180 degrees (obtuse angles). Therefore, if we want an acute angle, we need cos θ to be positive, which means the dot product a · b must also be positive (since the magnitudes |a| and |b| are always positive).

Breaking Down the Dot Product and Magnitudes

Let's break down how to calculate the dot product and magnitudes for our specific vectors. Given a(-1; 4; a) and b(5; -1; a), the dot product is calculated as:

a · b = (-1)(5) + (4)(-1) + (a)(a) = -5 - 4 + a² = a² - 9

So, we know that for an acute angle, a² - 9 > 0. This is a crucial inequality we'll use to find the values of 'a'.

Next, let's think about the magnitudes of the vectors. The magnitude of a vector is its length, and it's calculated using the Pythagorean theorem in three dimensions. For vector a, the magnitude is:

|a| = √((-1)² + 4² + a²) = √(1 + 16 + a²) = √(17 + a²)

And for vector b, the magnitude is:

|b| = √(5² + (-1)² + a²) = √(25 + 1 + a²) = √(26 + a²)

While the magnitudes themselves are important for calculating the exact angle, in our case, we primarily care about the sign of the dot product. The magnitudes will always be positive (or zero), so they don't affect whether the angle is acute or obtuse. Our focus remains on the condition a² - 9 > 0.

Solving the Inequality: a² - 9 > 0

Okay, guys, we've arrived at the heart of the problem! To find the values of 'a' that make the angle between our vectors acute, we need to solve the inequality a² - 9 > 0. This is a quadratic inequality, and there are a couple of ways we can tackle it.

Method 1: Factoring and Critical Points

The first method involves factoring the quadratic expression. We can rewrite a² - 9 as (a - 3)(a + 3). So, our inequality becomes:

(a - 3)(a + 3) > 0

To solve this, we find the critical points, which are the values of 'a' that make the expression equal to zero. These are a = 3 and a = -3. These critical points divide the number line into three intervals: (-∞, -3), (-3, 3), and (3, ∞).

Now, we need to test a value from each interval to see where the inequality holds true. Let's pick a = -4 (from the first interval), a = 0 (from the second interval), and a = 4 (from the third interval).

• For a = -4: (-4 - 3)(-4 + 3) = (-7)(-1) = 7 > 0 (True)
• For a = 0: (0 - 3)(0 + 3) = (-3)(3) = -9 < 0 (False)
• For a = 4: (4 - 3)(4 + 3) = (1)(7) = 7 > 0 (True)

So, the inequality holds true for the intervals (-∞, -3) and (3, ∞). This means that if 'a' is less than -3 or greater than 3, the angle between the vectors will be acute.

Method 2: Graphing the Parabola

Another way to visualize this is by thinking about the graph of the quadratic function y = a² - 9. This is a parabola that opens upwards (since the coefficient of a² is positive) and has roots at a = -3 and a = 3. The inequality a² - 9 > 0 asks us to find the values of 'a' where the parabola is above the x-axis (i.e., where y is positive). Looking at the graph, we can see that this occurs when a < -3 or a > 3, which confirms our solution from the previous method.

The Solution: Putting it All Together

Alright, guys, we've done the hard work! We've figured out that the values of 'a' that make the vectors a(-1; 4; a) and b(5; -1; a) form an acute angle are those in the intervals (-∞, -3) and (3, ∞). This means that 'a' can be any number less than -3 or any number greater than 3, but it cannot be between -3 and 3 (inclusive). We found this by understanding the relationship between the dot product and the angle between vectors, setting up an inequality based on the condition for an acute angle, and then solving that inequality using either factoring and critical points or by visualizing the graph of the quadratic function.

So, there you have it! We've successfully navigated the world of vectors and angles to find the solution to our problem. This kind of problem is a great example of how different concepts in math—like vectors, dot products, inequalities, and quadratic functions—can all come together to solve something interesting. Keep practicing, and you'll be a vector whiz in no time!