Question

Find the equation of the plane through the given points. $$ (1,3,2),(0,3,0), \text { and }(2,4,3) $$

Short answer

The equation of the plane is \(2x - y - z = -3\).

Step-by-step solution

Determine Two Vectors in the Plane

To find the equation of a plane, we first need two vectors that lie within the plane. With three points given as \((1,3,2)\), \((0,3,0)\), and \((2,4,3)\), we can form two vectors. The vector \(\mathbf{v_1}\) from \((1,3,2)\) to \((0,3,0)\) is found by subtracting the coordinates of the first point from the second: \[ \mathbf{v_1} = (0,3,0) - (1,3,2) = (-1,0,-2) \] Similarly, the vector \(\mathbf{v_2}\) from \((1,3,2)\) to \((2,4,3)\) is: \[ \mathbf{v_2} = (2,4,3) - (1,3,2) = (1,1,1) \]

Find the Normal Vector

The normal vector \(\mathbf{n}\) to the plane can be found using the cross product of \(\mathbf{v_1}\) and \(\mathbf{v_2}\). Calculate the cross product: \[ \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = (-1,0,-2) \times (1,1,1) \] Calculate each component of \(\mathbf{n}\): - \(n_x = (0)(1) - (-2)(1) = 2\)- \(n_y = (-2)(1) - (-1)(1) = -1\)- \(n_z = (-1)(1) - (0)(1) = -1\) Thus, the normal vector \(\mathbf{n}\) is \((2,-1,-1)\).

Write the Plane Equation

Given the normal vector \((2,-1,-1)\) and a point on the plane, say \((1,3,2)\), we can write the plane's equation using the formula \[ ax + by + cz = d \] Where \((a,b,c)\) is the normal vector and \(d = ax_0 + by_0 + cz_0\) with \((x_0,y_0,z_0)\) being a point on the plane. In our case:\[ d = (2)(1) + (-1)(3) + (-1)(2) = 2 - 3 - 2 = -3 \] Hence, the equation of the plane is: \[ 2x - y - z = -3 \]

Verify the Plane Equation

To ensure our plane equation is correct, substitute the other points into the plane equation and verify. For point \((0,3,0)\): \[ 2(0) - 3 - 0 = -3 \] For point \((2,4,3)\): \[ 2(2) - 4 - 3 = -3 \] Both equations hold true, confirming the plane equation \(2x - y - z = -3\) is correct.

Key concepts

Vector Calculus

Vector calculus is a key tool in mathematics, especially when dealing with physical spaces and dimensions. It combines calculus and vector algebra to study fields which vary in direction and magnitude. This is applicable in fields such as physics and engineering.

Essentially, vector calculus helps us analyze vector fields, which are functions assigning vectors to points in three-dimensional space. For example, in our problem, the vectors formed from the given points are within a 3D space. We create two vectors using the given points to determine the plane's orientation.

This discipline enables us to compute basic vector operations like addition, subtraction, and importantly, the cross product. It provides the machinery to abstract, model and solve problems in multi-dimensional spaces. Understanding vector calculus is crucial for comprehending relationships within 3D geometry and fields.

Cross Product

The cross product is a method in vector calculus that helps find a vector perpendicular to two given vectors. In three-dimensional space, this is particularly useful for finding a normal vector to a plane.

The cross product of two vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\) is calculated as follows:

• First Component: \(a_2b_3 - a_3b_2\)
• Second Component: \(a_3b_1 - a_1b_3\)
• Third Component: \(a_1b_2 - a_2b_1\)

In our case, the cross product of the vectors \((-1,0,-2)\) and \( (1,1,1)\) was used to find a normal vector: \(\mathbf{n} = (2,-1,-1)\).

Thus, employing the cross product efficiently determines a normal vector, which is integral to plane equation formulation.

Normal Vector

A normal vector is a vector perpendicular to a surface or a plane. It plays a crucial role in the geometry of three-dimensional spaces. An understanding of normal vectors helps with fundamental tasks like calculating plane equations or reflecting light off surfaces.

In our scenario, the normal vector is discovered through the cross product of two vectors within the plane. Given vectors \(\mathbf{v_1} = (-1,0,-2)\) and \(\mathbf{v_2} = (1,1,1)\), the resulting normal vector \(\mathbf{n} = (2,-1,-1)\) establishes the perpendicularity needed for the plane equation.

• It provides the coefficients for the plane equation \(ax + by + cz = d\)
• Allows computation of the constant term \(d\) in the plane equation

Using the normal vector, we translate the geometric plane into an algebraic equation, crucial for solving or modeling 3D geometrical problems.

3D Geometry

Three-dimensional (3D) geometry involves the analysis of shapes and objects in a 3D space, adding depth to the understanding of spatial properties compared to two-dimensional geometry. It encompasses points, lines, planes, and angles between them.

In 3D geometry, a plane can be defined by three points, which do not lie on a single line, such as \(\{(1,3,2), (0,3,0), (2,4,3)\}\) from our task. With these points, two vectors can be drawn, providing the necessary information to find the normal vector through the cross product.

3D geometry is essential for modeling real-world scenarios where objects have height, width, and depth. It provides the framework to construct the equation of a plane: \(2x - y - z = -3\) in our example. This plane equation forms the basis for further geometric calculations or graphical representation within the space.

• Helps with visualization and solving spatial problems
• Integral for applications like computer graphics, physics simulations, and architectural modeling