Math Problems: PMI, Sphere Equation, And Vectors

Dec 26, 2025 by Andrew McMorgan 49 views

Hey math enthusiasts! Today, we're diving into some classic problems that are super common in math classes. Whether you're prepping for an exam or just love flexing those brain muscles, we've got you covered. We'll tackle proofs using the Principle of Mathematical Induction (PMI), find the equation of a sphere given its diameter, and explore vector properties. So grab your calculators, notebooks, and let's get this mathematical party started!

1. Proving a Summation Formula Using the Principle of Mathematical Induction (PMI)

Alright guys, let's kick things off with a proof using the Principle of Mathematical Induction (PMI). This is a fundamental technique for proving statements about all natural numbers. Our goal here is to prove the following summation formula: $1 cdot 2+2 cdot 3+ cdots+n(n+1)= rac{n(n+1)(n+2)}{3}$ . This formula essentially tells us how to sum up a series where each term is the product of two consecutive integers. We'll be using the two-step process inherent in PMI: the base case and the inductive step. It's like building a domino chain – you push the first one (base case), and then you show that if any domino falls, the next one will too (inductive step). This guarantees that all the dominos will eventually fall, proving our formula for all natural numbers $n$ . The structure of the formula itself, involving products of consecutive integers and a cubic term in the result, hints at the combinatorial nature of the problem, which PMI elegantly unravels. The left side of the equation represents the sum of products of consecutive integers, often appearing in areas like combinatorics and discrete mathematics. The right side, a cubic polynomial in $n$ , provides a compact and elegant closed-form solution. Proving this relationship using PMI requires careful algebraic manipulation, ensuring that our assumptions in the inductive step hold true for the general case. Understanding PMI is crucial because it forms the bedrock for proving many algorithms and properties in computer science and advanced mathematics. It’s a logical framework that allows us to generalize specific observations to universal truths, provided we can establish the initial condition and the recursive relationship.

The Base Case: $n=1$

First things first, we need to establish the base case. This means showing that the formula holds true for the smallest natural number, which is $n=1$ . Let's plug $n=1$ into both sides of the equation:

Left-hand side (LHS): For $n=1$ , the sum has only one term: $1 cdot (1+1) = 1 cdot 2 = 2$ .
Right-hand side (RHS): Plugging $n=1$ into the formula $rac{n(n+1)(n+2)}{3}$ , we get $rac{1(1+1)(1+2)}{3} = rac{1 cdot 2 cdot 3}{3} = rac{6}{3} = 2$ .

Since LHS = RHS, the formula is true for $n=1$ . Hooray! We've completed the first crucial step.

The Inductive Step

Now, for the more involved part: the inductive step. Here, we make an assumption – called the inductive hypothesis – and use it to prove the statement for the next case. Assume that the formula is true for some arbitrary natural number $k$ . That is, we assume:

$1 cdot 2+2 cdot 3+ cdots+k(k+1)= rac{k(k+1)(k+2)}{3}$ (Inductive Hypothesis)

Our mission now is to show that if this is true for $k$ , it must also be true for the next natural number, $k+1$ . So, we need to prove:

$1 cdot 2+2 cdot 3+ cdots+k(k+1)+(k+1)((k+1)+1)= rac{(k+1)((k+1)+1)((k+1)+2)}{3}$

Let's simplify the target equation for $n=k+1$ :

$rac{(k+1)(k+2)(k+3)}{3}$

Now, let's start with the left-hand side of the equation for $n=k+1$ and use our inductive hypothesis. The sum up to the $(k+1)^{th}$ term is the sum up to the $k^{th}$ term plus the $(k+1)^{th}$ term:

$(1 cdot 2+2 cdot 3+ cdots+k(k+1)) + (k+1)((k+1)+1)$

Using our inductive hypothesis, we can replace the sum of the first $k$ terms:

$rac{k(k+1)(k+2)}{3} + (k+1)(k+2)$

Now, we need to manipulate this expression algebraically to arrive at our target RHS for $n=k+1$ . Let's find a common denominator:

$rac{k(k+1)(k+2)}{3} + rac{3(k+1)(k+2)}{3}$

Combine the terms:

$rac{k(k+1)(k+2) + 3(k+1)(k+2)}{3}$

We can see a common factor of $(k+1)(k+2)$ in the numerator. Let's factor it out:

$rac{(k+1)(k+2)(k + 3)}{3}$

And voilà! This is exactly the right-hand side of the formula for $n=k+1$ . So, we've shown that if the formula holds for $k$ , it also holds for $k+1$ . By the Principle of Mathematical Induction, the formula $1 cdot 2+2 cdot 3+ cdots+n(n+1)= rac{n(n+1)(n+2)}{3}$ is true for all natural numbers $n$ . This rigorous process confirms the validity of the summation, showcasing the power of PMI in establishing general truths from specific cases.

2. Finding the Equation of a Sphere

Next up, we're jumping into 3D geometry to find the equation of a sphere. The key piece of information here is that the join of two points, $(1,-1,1)$ and $(-3,4,5)$ , serves as the diameter of the sphere. Awesome! The general equation of a sphere with center $(h, k, l)$ and radius $r$ is $(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2$ . So, our task boils down to finding the center and the radius of the sphere. The center of the sphere will be the midpoint of the diameter, and the radius will be half the length of the diameter. It's a pretty straightforward process once you recall these basic geometric principles. Finding the center is like finding the exact middle point between two given points, and the radius is simply the distance from that center to either endpoint of the diameter. This problem beautifully blends coordinate geometry and basic geometric definitions. The points themselves define the extent of the sphere in 3D space, and their relationship as the endpoints of a diameter gives us direct clues to its central location and size. This kind of problem is fundamental for understanding spheres and their representation in Cartesian coordinates, forming a basis for more complex applications in fields ranging from physics to computer graphics.

Finding the Center

The center of the sphere is the midpoint of the line segment connecting $(1,-1,1)$ and $(-3,4,5)$ . The midpoint formula for two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D is $ig( rac{x_1+x_2}{2}, rac{y_1+y_2}{2}, rac{z_1+z_2}{2}ig)$ .

Let $(x_1, y_1, z_1) = (1,-1,1)$ and $(x_2, y_2, z_2) = (-3,4,5)$ .

x-coordinate of the center: $rac{1 + (-3)}{2} = rac{-2}{2} = -1$
y-coordinate of the center: $rac{-1 + 4}{2} = rac{3}{2}$
z-coordinate of the center: $rac{1 + 5}{2} = rac{6}{2} = 3$

So, the center of the sphere $(h, k, l)$ is $ig(-1, rac{3}{2}, 3ig)$ .

Finding the Radius

The radius ( $r$ ) is half the length of the diameter. The length of the diameter is the distance between the two given points. We use the distance formula in 3D: $d = sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ .

Diameter length: $d = sqrt{(-3-1)^2 + (4-(-1))^2 + (5-1)^2}$ $d = sqrt{(-4)^2 + (5)^2 + (4)^2}$ $d = sqrt{16 + 25 + 16}$ $d = sqrt{57}$

The radius is half of this length:

$r = rac{ sqrt{57}}{2}$

Now, we need $r^2$ for the sphere equation:

$r^2 = igg( rac{ sqrt{57}}{2}igg)^2 = rac{57}{4}$

The Equation of the Sphere

Now we plug the center $(h, k, l) = ig(-1, rac{3}{2}, 3ig)$ and $r^2 = rac{57}{4}$ into the general sphere equation $(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2$ :

$(x - (-1))^2 + igg(y - rac{3}{2}igg)^2 + (z - 3)^2 = rac{57}{4}$

$(x+1)^2 + igg(y - rac{3}{2}igg)^2 + (z-3)^2 = rac{57}{4}$

And there you have it! That's the equation of the sphere with the given diameter. Pretty neat how we can define a whole 3D shape with just a few numbers and an equation. This equation defines every single point on the surface of the sphere, satisfying the geometric constraint that it passes through the specified points as a diameter. Understanding how to derive this equation is key to working with spherical geometry in various scientific and engineering applications, from calculating volumes to simulating physical phenomena.

3. Vector Properties: Orthogonality and Magnitude

Finally, let's get our hands dirty with some vectors. We're given two vectors, $vec{a}$ and $vec{b}$ , with some specific properties: they are orthogonal ( $vec{a} ot vec{b}$ ), and we know their magnitudes: $| vec{a}|=3$ and $| vec{b}|=1$ . This scenario is super common when dealing with vector operations, especially in physics and linear algebra. Orthogonality means the vectors are perpendicular, forming a 90-degree angle. The magnitude just tells us their length. From these two pieces of information, we can deduce a lot, particularly about operations involving both vectors, like their dot product or the magnitude of their sum/difference. The dot product of two orthogonal vectors is always zero, which is a direct consequence of the definition of orthogonality and the dot product formula. The magnitudes provide the scale for each vector, allowing us to quantify results when combining them. This understanding is fundamental for decomposing forces, analyzing motion, or performing transformations in higher dimensions. Vectors are the language of direction and magnitude, and problems like this help us master that language. The implications of orthogonality are profound, simplifying many complex calculations and forming the basis for orthogonal bases in vector spaces, which are essential for advanced mathematical concepts and their applications. The given magnitudes allow us to anchor these abstract properties to concrete scales, making the vector relationships tangible.

Key Properties and Implications

Orthogonality ( $vec{a} ot vec{b}$ ): The defining characteristic of orthogonal vectors is that their dot product is zero. Recall that the dot product of two vectors $vec{a}$ and $vec{b}$ is given by $vec{a} cdot vec{b} = | vec{a}| | vec{b}| ext{cos}( heta)$ , where $ heta$ is the angle between them. If $vec{a} ot vec{b}$ , then $ heta = 90^ circ$ (or $rac{ pi}{2}$ radians), and $ext{cos}(90^ circ) = 0$ . Therefore, $vec{a} ot vec{b}$ implies $vec{a} ot vec{b} = 0$ . This is a fundamental property that simplifies many vector calculations.
Magnitudes: We are given $| vec{a}|=3$ and $| vec{b}|=1$ . These are the lengths of the vectors.

What Can We Deduce?

Given these properties, we can make several deductions, for example:

Dot Product: As mentioned, $vec{a} ot vec{b} = | vec{a}| | vec{b}| ext{cos}(90^ circ) = 3 cdot 1 cdot 0 = 0$ .
Magnitude of the Sum of Vectors: Consider the magnitude of the sum of the two vectors, $| vec{a}+ vec{b}|$ . We know that $| vec{a}+ vec{b}|^2 = ( vec{a}+ vec{b}) ot ( vec{a}+ vec{b})$ . Expanding this using the distributive property of the dot product: $| vec{a}+ vec{b}|^2 = vec{a} ot vec{a} + vec{a} ot vec{b} + vec{b} ot vec{a} + vec{b} ot vec{b}$ Since $vec{a} ot vec{a} = | vec{a}|^2$ , $vec{b} ot vec{b} = | vec{b}|^2$ , and $vec{a} ot vec{b} = vec{b} ot vec{a} = 0$ (due to orthogonality), we get: $| vec{a}+ vec{b}|^2 = | vec{a}|^2 + 0 + 0 + | vec{b}|^2$ $| vec{a}+ vec{b}|^2 = | vec{a}|^2 + | vec{b}|^2$ Plugging in the given magnitudes: $| vec{a}+ vec{b}|^2 = 3^2 + 1^2 = 9 + 1 = 10$ Therefore, $| vec{a}+ vec{b}| = sqrt{10}$ . This result is a direct consequence of the Pythagorean theorem in vector space, where orthogonal vectors form the legs of a right triangle.
Magnitude of the Difference of Vectors: Similarly, for the magnitude of the difference, $| vec{a}- vec{b}|^2$ : $| vec{a}- vec{b}|^2 = ( vec{a}- vec{b}) ot ( vec{a}- vec{b})$ $| vec{a}- vec{b}|^2 = vec{a} ot vec{a} - vec{a} ot vec{b} - vec{b} ot vec{a} + vec{b} ot vec{b}$ $| vec{a}- vec{b}|^2 = | vec{a}|^2 - 0 - 0 + | vec{b}|^2$ $| vec{a}- vec{b}|^2 = | vec{a}|^2 + | vec{b}|^2$ $| vec{a}- vec{b}|^2 = 3^2 + 1^2 = 9 + 1 = 10$ Therefore, $| vec{a}- vec{b}| = sqrt{10}$ . In this specific case, because the vectors are orthogonal, the magnitude of their sum and difference are the same. This is a unique property arising from their perpendicular relationship.

These deductions highlight how fundamental properties like orthogonality and magnitude drastically simplify vector algebra and geometry. Understanding these relationships is crucial for a wide array of applications, from quantum mechanics to computer graphics, where vectors are used to represent physical quantities and transformations.

And that's a wrap on these math problems, guys! We've covered induction, 3D geometry, and vector properties. Keep practicing, and don't hesitate to explore these topics further. Math is a journey, and every solved problem is a step forward!