Locus Of Points M: Vector Relations In The Plane

Hey guys! Today, we're diving into a fascinating problem from mathematics: determining the set of points M in a plane based on given vector relations. This involves understanding how the position vector OM (or MO) changes with the parameter m, and how these changes trace out a specific locus or path in the plane. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the solutions, let's make sure we're all on the same page. The core of this problem lies in understanding that the position vector OM (from the origin O to point M) can be expressed as a linear combination of the unit vectors i and j. The coefficients of these unit vectors depend on the parameter m, which belongs to the set of real numbers (R). As m varies, the endpoint M of the vector OM moves around the plane, tracing out a curve or a line. Our goal is to find the equation of this curve or line, which represents the locus of points M.

Why is this important? Well, these types of problems help us connect algebra and geometry. They show how algebraic equations can describe geometric shapes, and vice versa. This connection is fundamental in many areas of mathematics and physics.

Problem Breakdown

We are given three different relations for the position vector (either OM or MO) in terms of the parameter m. Let's break down each case and find the corresponding locus of points M.

a) ${\overline{OM} = (4-m)\vec{i}+(3m-2) \vec{j}}$

In this first scenario, we need to determine the set of points M in the plane given the vector equation ${\overline{OM} = (4-m)\vec{i}+(3m-2) \vec{j}}$, where m is a real number. To solve this, we'll express the coordinates of point M in terms of m and then eliminate m to find the Cartesian equation representing the locus of M.

Let M have coordinates (x, y). Then the position vector ${\overline{OM}}$ can be written as ${\overline{OM} = x\vec{i} + y\vec{j}}$. Comparing this with the given equation, we have:

• x = 4 - m
• y = 3m - 2

Our goal is to eliminate m from these two equations. From the first equation, we can express m as:

• m = 4 - x

Now, substitute this expression for m into the second equation:

• y = 3(4 - x) - 2
• y = 12 - 3x - 2
• y = -3x + 10

This equation represents a straight line in the plane. Therefore, the set of points M forms a line with the equation y = -3x + 10. This line represents all possible positions of point M as m varies over the real numbers.

Geometrically, this means that as the value of m changes, the point M moves along this specific line in the Cartesian plane. The equation we found, y = -3x + 10, defines the path or locus of all such points M. This exercise beautifully illustrates how vector equations can be used to describe geometric shapes and how algebraic manipulations can reveal these shapes.

b) ${\overline{OM} = (m-1)\vec{i}+(2m-3) \vec{j}}$

Now, let's tackle the second case. We're given the relation ${\overline{OM} = (m-1)\vec{i}+(2m-3) \vec{j}}$. Just like before, our aim is to find the locus of points M as m varies through the real numbers. This involves expressing the coordinates of M in terms of m and then eliminating m to obtain a Cartesian equation.

Let M have coordinates (x, y), so ${\overline{OM} = x\vec{i} + y\vec{j}}$. Comparing this with the given equation, we get:

• x = m - 1
• y = 2m - 3

We need to eliminate m from these equations. From the first equation, we can express m as:

• m = x + 1

Substitute this into the second equation:

• y = 2(x + 1) - 3
• y = 2x + 2 - 3
• y = 2x - 1

This equation, y = 2x - 1, represents another straight line in the plane. So, in this case, the locus of points M is a line with the equation y = 2x - 1. This means that as m changes, the point M traces a path along this particular line.

The process here mirrors the approach we used in part (a), but the resulting line is different due to the different coefficients in the original vector equation. This reinforces the idea that the coefficients of m in the vector equation directly influence the slope and intercept of the line that forms the locus of points M. Understanding this relationship is crucial for solving these types of problems efficiently.

c) ${\overline{MO}=(3m-1)(\vec{i}+\vec{j})+(1-m)(2\vec{i}+\vec{j})}$

Alright, let's move on to the third and final case. This time, we're given ${\overline{MO}=(3m-1)(\vec{i}+\vec{j})+(1-m)(2\vec{i}+\vec{j})}$. Notice that we have MO instead of OM. This means we're dealing with the vector pointing from M to O, which is the negative of the vector OM. We'll need to keep this in mind as we proceed.

Our goal remains the same: find the locus of points M. First, let's simplify the expression for MO:

${\overline{MO} = (3m - 1)\vec{i} + (3m - 1)\vec{j} + (2 - 2m)\vec{i} + (1 - m)\vec{j}}$

Combine the ${\vec{i}}$ and ${\vec{j}}$ components:

${\overline{MO} = (3m - 1 + 2 - 2m)\vec{i} + (3m - 1 + 1 - m)\vec{j}}$

${\overline{MO} = (m + 1)\vec{i} + (2m)\vec{j}}$

Now, remember that ${\overline{OM} = -\overline{MO}}$, so:

${\overline{OM} = -(m + 1)\vec{i} - (2m)\vec{j}}$

Let M have coordinates (x, y), then ${\overline{OM} = x\vec{i} + y\vec{j}}$. Comparing this with the equation above, we have:

• x = -(m + 1)
• y = -2m

Let's eliminate m. From the first equation:

• m = -x - 1

Substitute into the second equation:

• y = -2(-x - 1)
• y = 2x + 2

So, the locus of points M in this case is a straight line with the equation y = 2x + 2. This line represents all possible positions of M as the parameter m varies.

This part of the problem highlights the importance of paying attention to the direction of the vectors. The fact that we were given MO instead of OM required an extra step of negating the vector components to find the correct equation for the locus.

Key Takeaways

After working through these three cases, there are several key takeaways we can glean from this problem:

1. Expressing Position Vectors: The position vector OM is fundamental in describing the location of a point M in the plane. Representing OM as a linear combination of unit vectors ${\vec{i}}$ and ${\vec{j}}$ allows us to connect the point's coordinates (x, y) with vector equations.
2. Parametric Equations: When the coefficients of ${\vec{i}}$ and ${\vec{j}}$ in the vector equation depend on a parameter (like m in our problem), we effectively have a set of parametric equations for the coordinates of M. These equations describe how x and y change as m varies.
3. Eliminating the Parameter: To find the locus of points M, we need to eliminate the parameter m from the parametric equations. This usually involves solving one equation for m and substituting into the other equation. The resulting equation is the Cartesian equation of the locus.
4. Geometric Interpretation: The Cartesian equation we obtain represents a geometric shape in the plane. In our cases, we found straight lines. However, with different vector relations, we could encounter other shapes like circles, parabolas, or hyperbolas.
5. Vector Direction: Pay close attention to the direction of the vectors. If you're given MO instead of OM, remember that ${\overline{OM} = -\overline{MO}}$. This seemingly small detail can make a big difference in the final result.

General Strategy for Solving Locus Problems

To wrap things up, let's outline a general strategy for tackling these types of locus problems:

1. Express the Position Vector: Write the position vector (OM or MO) in terms of the coordinates of point M (x, y) and the unit vectors ${\vec{i}}$ and ${\vec{j}}$.
2. Set up Parametric Equations: Equate the components of the given vector equation with the components of the position vector you expressed in step 1. This will give you a set of equations where x and y are expressed in terms of the parameter (e.g., m).
3. Eliminate the Parameter: Solve one of the parametric equations for the parameter, and substitute that expression into the other equation. This will eliminate the parameter and give you a single equation relating x and y.
4. Identify the Locus: The equation you obtained in step 3 represents the locus of points M. Recognize the type of curve or shape described by the equation (e.g., line, circle, parabola).
5. Consider Special Cases: Sometimes, there might be special cases to consider, such as restrictions on the parameter or specific values that lead to degenerate cases (e.g., a line collapsing to a point).

By following these steps, you'll be well-equipped to tackle a wide range of locus problems involving vector relations. Remember, practice makes perfect, so don't hesitate to work through plenty of examples to solidify your understanding.

Conclusion

So, guys, we've successfully navigated through the problem of determining the locus of points M based on vector relations. We've seen how to express position vectors, set up parametric equations, eliminate parameters, and identify the resulting geometric shapes. These techniques are fundamental in connecting vector algebra and geometry, and they'll serve you well in your mathematical journey. Keep exploring, keep practicing, and you'll master these concepts in no time! Until next time, happy problem-solving! Remember, the key is understanding how algebraic equations define geometric shapes and manipulating them to find solutions. This skill is not only useful in math class, but also in many real-world applications, from computer graphics to physics simulations.