Question

. The plane \(2 x+3 y+6 z=6\) intersects the coordinate axes at points \(P, Q, R\) forming a triangle, Find the vectors \(P Q\) and \(\overrightarrow{P R}\). Write a vector formula for the area of the triangle \(P Q R\), and find the area.

Short answer

The area of the triangle PQR is 3.5 square units.

Step-by-step solution

Find Coordinates of Intersections

First, find the points where the plane intersects the x, y, and z-axes.

Calculate Point P

Set y = 0 and z = 0 in the plane equation to find where it intersects the x-axis. \[2x = 6 \rightarrow x = 3 So, P = (3, 0, 0)\]

Calculate Point Q

Set x = 0 and z = 0 in the plane equation to find where it intersects the y-axis. \[3y = 6 \rightarrow y = 2 So, Q = (0, 2, 0)\]

Calculate Point R

Set x = 0 and y = 0 in the plane equation to find where it intersects the z-axis. \[6z = 6 \rightarrow z = 1 So, R = (0, 0, 1)\]

Find Vector PQ

The vector \(PQ\) is obtained by subtracting the coordinates of P from Q: \[PQ = Q - P = (0, 2, 0) - (3, 0, 0) = (-3, 2, 0)\]

Find Vector PR

The vector \(PR\) is obtained by subtracting the coordinates of P from R: \[PR = R - P = (0, 0, 1) - (3, 0, 0) = (-3, 0, 1)\]

Calculate Cross Product

The area of the triangle can be found using the cross product of vectors \(PQ\) and \(PR\). Compute the cross product: \[PQ \times PR = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 2 & 0 \ -3 & 0 & 1 \end{vmatrix}\] Which results in: \[\vec{PQ} \times \vec{PR} = (2, 3, 6)\]

Find Magnitude of Cross Product

Calculate the magnitude of the cross product to get the area: \[\|2, 3, 6\| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Calculate Area

The area of triangle \(PQR\) is half the magnitude of the cross product: \[ \text{Area} = \frac{1}{2} \times 7 = 3.5 \]

Key concepts

vectors

A vector is a quantity that has both magnitude and direction. In geometry and physics, vectors are used to represent things like force, velocity, and position. Vectors are typically denoted by letters with arrows on top, like \(\rightarrow{v}\). They can be expressed in different dimensions, based on their components. For example, a 3D vector is written as \(\rightarrow{v} = (v_x, v_y, v_z)\).

Operations such as addition, subtraction, and scalar multiplication can be performed on vectors. The result of subtracting one vector from another gives a new vector called the displacement between two points.

In our exercise, we found vectors \(\rightarrow{PQ}\) and \(\rightarrow{PR}\) by subtracting the coordinates of point P from Q and R, respectively. The results were:
- \(\rightarrow{PQ} = (-3, 2, 0)\)
- \(\rightarrow{PR} = (-3, 0, 1)\)

cross product

The cross product is a way to multiply two vectors in three-dimensional space. It results in a third vector that is perpendicular to the plane containing the original vectors. Symbolically, the cross product of vectors \(\rightarrow{A}\) and \(\rightarrow{B}\) is written as \(\rightarrow{A} \times \rightarrow{B}\).

The formula for calculating the cross product involves a determinant of a matrix. For vectors \(\rightarrow{A} = (a_1, a_2, a_3)\) and \(\rightarrow{B} = (b_1, b_2, b_3)\), the cross product is determined as follows:

\(\rightarrow{A} \times \rightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\).

In our exercise, we computed the cross product of vectors \(\rightarrow{PQ}\) and \(\rightarrow{PR}\) as:
\(\rightarrow{PQ} \times \rightarrow{PR} = (2, 3, 6)\).

This result is essential for calculating the area of the triangle.

coordinate geometry

Coordinate geometry (or analytic geometry) involves placing geometric figures in a coordinate plane and using algebraic equations to represent them. This approach allows for solving geometric problems using coordinates and equations.

In a 3D coordinate system, points are represented by their coordinates \(x, y, z\), and planes can be expressed using linear equations. The plane's equation can be written in different forms, such as the general form \(Ax + By + Cz = D\).

In our exercise, we start with the plane equation \(\text{2x + 3y + 6z = 6}\). By setting two variables to zero, we can find the points of intersection on the x, y, and z-axes, which gave us P = (3, 0, 0), Q = (0, 2, 0), and R = (0, 0, 1).

These intersections form the vertices of a triangle in the 3D coordinate system.

area calculation

One practical application of vectors and the cross product in geometry is calculating the area of a triangle formed by three points in space.

To find the area of triangle \(PQR\), we follow these steps:

• Find the vectors \(\rightarrow{PQ}\) and \(\rightarrow{PR}\).
• Compute the cross product \(\rightarrow{PQ} \times \rightarrow{PR}\).
• Find the magnitude of the resulting vector, which gives twice the area of the triangle.
• Divide this magnitude by 2 to obtain the area.

In our specific case, we calculated:

• The cross product: \(\rightarrow{PQ} \times \rightarrow{PR} = (2, 3, 6)\).
• The magnitude: \(\sqrt{2^2 + 3^2 + 6^2} = 7\).
• The area: \(\frac{7}{2} = 3.5\).

Understanding these steps allows you to determine the area of any triangle in 3D space given its vertices.