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Find the two points trisecting the segment between \(P(2,3,5)\) and \(Q(8,-6,2)\).

Short Answer

Expert verified
The two points trisecting the segment are \( A = (4, 0, 4) \) and \( B = (6, -3, 3) \).

Step by step solution

01

Understand the Problem

We need to find two points on the line segment between the points \( P(2,3,5) \) and \( Q(8,-6,2) \) such that the segment is divided into three equal parts. This means we need to find two points that trisect the segment.
02

Find the Direction Vector

Calculate the direction vector \( \overrightarrow{PQ} \) by subtracting the coordinates of \( P \) from \( Q \):\[ \overrightarrow{PQ} = (8 - 2, -6 - 3, 2 - 5) = (6, -9, -3). \]
03

Determine the Trisection Points

To trisect the segment, we divide the direction vector into three equal parts. Trisecting means finding points at \( \frac{1}{3} \) and \( \frac{2}{3} \) along this vector from \( P \). Let’s find the new points \( A \) and \( B \).
04

Calculate the First Trisection Point (Point A)

Point \( A \) is at \( \frac{1}{3} \) of \( \overrightarrow{PQ} \) from \( P \). We use the formula for point division: \( A = P + \frac{1}{3}\overrightarrow{PQ} \).Substitute \( \overrightarrow{PQ} \):\[ A = (2, 3, 5) + \frac{1}{3}(6, -9, -3). \]Calculate each coordinate:\[ A_x = 2 + \frac{1}{3} \times 6 = 2 + 2 = 4, \]\[ A_y = 3 + \frac{1}{3} \times (-9) = 3 - 3 = 0, \]\[ A_z = 5 + \frac{1}{3} \times (-3) = 5 - 1 = 4. \]Thus, \( A = (4, 0, 4). \)
05

Calculate the Second Trisection Point (Point B)

Point \( B \) is at \( \frac{2}{3} \) of \( \overrightarrow{PQ} \) from \( P \). Use the formula: \( B = P + \frac{2}{3}\overrightarrow{PQ} \).Substitute \( \overrightarrow{PQ} \):\[ B = (2, 3, 5) + \frac{2}{3}(6, -9, -3). \]Calculate each coordinate:\[ B_x = 2 + \frac{2}{3} \times 6 = 2 + 4 = 6, \]\[ B_y = 3 + \frac{2}{3} \times (-9) = 3 - 6 = -3, \]\[ B_z = 5 + \frac{2}{3} \times (-3) = 5 - 2 = 3. \]Thus, \( B = (6, -3, 3). \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Coordinate Geometry
Coordinate geometry, often referred to as analytic geometry, is a branch of mathematics that allows for the representation of geometric shapes in a numerical manner. This approach helps connect algebra with geometry, linking equations and geometric shapes.

Specifically, for a 3-dimensional space, a point is represented by three coordinates \(x, y, z\), which signify its location relative to a defined origin. For instance, in this exercise, point \(P(2,3,5)\) denotes a location where the x-coordinate is 2, the y-coordinate is 3, and the z-coordinate is 5. By using these coordinates, we can specify positions precisely in 3D space, allowing the solving of geometric problems through numerical methods.

In the context of trisection, coordinate geometry enables the solving process by using vector operations to divide a line segment into desired proportions, such as into thirds.
The Role of Vector Operations
Vector operations are fundamental in solving problems related to points and lines in space, like dividing line segments into specific ratios. A vector is represented as \(\overrightarrow{AB} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)\), outlining the direction and magnitude from point A to point B.

Here, the direction vector \(\overrightarrow{PQ}\) results from subtracting the coordinates of point P from point Q, giving us \(6, -9, -3\). By performing operations such as scalar multiplication, vectors can be scaled, allowing us to step proportionally on the line segment. For trisection, we find intermediate points by splitting this vector into thirds, thus easily identifying points \(\frac{1}{3}\) and \(\frac{2}{3}\) along the segment.

This method provides a solid mathematical approach, ensuring accuracy in computations, which is essential for diverse applications in physics, engineering, and computer graphics.
Exploring Point Division
Point division in coordinate geometry refers to finding specific points along a line segment that divide it into a specified ratio. This is a common task in geometric problems where proportional division is required.

The formula used for this is based on vector operations and is expressed as \(A = P + r \cdot \overrightarrow{PQ}\), where r is the ratio of division. In our trisection problem, the ratio for the first point \(A\) is \(\frac{1}{3}\), dividing \(\overrightarrow{PQ}\) accordingly, leading us to \(A = (4, 0, 4)\).

Similarly, for the second trisection point \(B\), a ratio of \(\frac{2}{3}\) is used. Applying this method ensures that calculations remain consistent and that the points align perfectly with the desired segments of the line. Understanding point division allows students to perceive mathematics beyond numbers, connecting it with spatial reasoning and problem-solving in real-world contexts.

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Most popular questions from this chapter

Let \(\alpha, \beta,\) and \(\gamma\) be the angles a vector \(\mathbf{v} \neq \mathbf{0}\) makes with the positive \(x, y,\) and \(z\) axes, respectively. Then \(\cos \alpha, \cos \beta,\) and \(\cos \gamma\) are called the direction cosines of the vector \(\mathbf{v}\). a. If \(\mathbf{v}=\left[\begin{array}{l}a \\ b \\ c\end{array}\right],\) show that \(\cos \alpha=\frac{a}{\|\mathbf{v}\|}, \cos \beta=\frac{b}{\|\mathbf{v}\|}\), and \(\cos \gamma=\frac{c}{\|\mathbf{v}\|}\). b. Show that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\).

Compute \(\mathbf{u} \times \mathbf{v}\) where: a. \(\mathbf{u}=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right], \mathbf{v}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-6 \\ 2 \\ 0\end{array}\right]\) c. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}1 \\ 1 \\ -1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}2 \\ 0 \\ -1\end{array}\right], \mathbf{v}=\left[\begin{array}{l}1 \\ 4 \\ 7\end{array}\right]\)

Let \(\mathbf{u}=\left[\begin{array}{r}2 \\ 0 \\ -4\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{r}2 \\ 1 \\ -2\end{array}\right] .\) In each case find \(\mathbf{x}\) : a. \(2 \mathbf{u}-\|\mathbf{v}\| \mathbf{v}=\frac{3}{2}(\mathbf{u}-2 \mathbf{x})\) b. \(3 \mathbf{u}+7 \mathbf{v}=\|\mathbf{u}\|^{2}(2 \mathbf{x}+\mathbf{v})\)

a. Does the line through \(P(1,2,-3)\) with direction vector \(\mathbf{d}=\left[\begin{array}{r}1 \\ 2 \\ -3\end{array}\right]\) lie in the plane \(2 x-y-z=3 ?\) Explain. b. Does the plane through \(P(4,0,5), Q(2,2,1),\) and \(R(1,-1,2)\) pass through the origin? Explain.

Show that the triangle with vertices \(A(4,-7,9), B(6,4,4),\) and \(C(7,10,-6)\) is not a rightangled triangle.

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