Question

In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.

Short answer

Question: Find the point where the medians of the given triangle PQR with coordinates P(2, 4, 0), Q(4, 1, 0), and R(6, 3, 4) intersect. Answer: Follow the steps outlined in the solution to calculate the intersection point of the medians of triangle PQR. After completing these steps, you will find that the medians intersect at point G, which has the coordinates (4, 2.67, 1.33).

Step-by-step solution

Find the midpoint M1 of the side PQ

To find the midpoint M1 of the side connecting points P(x1, y1, 0) and Q(x2, y2, 0), we will use the midpoint formula: M1 = ((x1 + x2)/2, (y1 + y2)/2, 0).

Find the position vector R to M1

To find the vector from R(x3, y3, z3) to M1, we perform R - M1, that is, the coordinate-wise subtraction of R and M1.

Find the vector from the origin to Z1

The vector from the origin O to Z1 is along the vector RM1 multiplied by 2/3, where 1/3 is the length of the vector OZ1 with respect to RM1.

Repeat steps 1, 2, and 3 for sides QR and RP

Do the same calculations for the QR and RP sides and find the resultant vectors OZ2 and OZ3, respectively.

Check the intersection of the medians

Compare the OZ1, OZ2, and OZ3 vectors for their components. If the coordinates are the same, then the medians intersect at a common point.

Calculate the coordinates of the intersection point

Once it is confirmed that medians intersect, calculate the coordinates of this intersection point by 4/3 multiplication of R, M2, and M3 vectors and sum them up. Divide the results by three to get centroid coordinates.

Apply the given coordinates of points P, Q, and R

Now, insert the given coordinates of P(2, 4, 0), Q(4, 1, 0) and R(6, 3, 4) into the earlier derived centroid formula to find the point where the medians of the triangle PQR intersect.

Key concepts

Coordinate Geometry

Coordinate geometry is a branch of mathematics that deals with points, lines, and shapes in a coordinate plane. In this context, we often use the Cartesian coordinate system, which represents points by their position along the x, y, and z axes.

In problems involving triangles, each vertex can be assigned coordinates, like P(x₁, y₁, z₁), Q(x₂, y₂, z₂), and R(x₃, y₃, z₃). This method allows for algebraic solutions, involving vector operations and calculations, instead of relying solely on geometric intuition.

Coordinate geometry is especially useful when proving concepts such as the intersection of medians in a triangle, as it combines algebraic and geometric properties to provide a clear and systematic approach.

Midpoint Formula

The midpoint formula is a simple, yet powerful tool in coordinate geometry. It helps us find the halfway point between two points in the coordinate plane. For any points P(x₁, y₁) and Q(x₂, y₂):

\[ M(x_m, y_m) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

In a 3D space, it expands to include the z-coordinates:

\[ M(x_m, y_m, z_m) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]

This formula is essential for determining the midpoints of sides in a triangle, which are key in finding medians and further solving complex geometric properties.

Medians of a Triangle

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Every triangle has three medians, and they intersect at a point known as the centroid.

The centroid can be considered the triangle's center of gravity, and it divides each median into a 2:1 ratio. This means the section of the median closest to the vertex is twice as long as the section near the midpoint.

To find the centroid using coordinate geometry, you calculate the intersection of these medians. The formula involving the vertices of the triangle P(x₁, y₁, z₁), Q(x₂, y₂, z₂), and R(x₃, y₃, z₃) is:

\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]

This shows the centroid's coordinates are the average of the vertices' coordinates.

Vector Subtraction

Vector subtraction is the process of subtracting one vector from another. This operation allows us to determine the directional difference between two points in space.

Given two vectors, A represented by A(x₁, y₁, z₁) and B represented by B(x₂, y₂, z₂), the subtraction of vector B from vector A is:

\[ A - B = (x_1 - x_2, y_1 - y_2, z_1 - z_2) \]

This results in a new vector, representing the direction and distance from point B to point A.

Vector subtraction is crucial when finding a vector between two points, such as in the steps of determining medians and their intersections, by considering these new vectors in your overall calculation strategy.