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For the following exercises, point \(P\) and vector \(\mathbf{n}\) are given. Find the scalar equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n}\). Find the general form of the equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n}\). \(P(1,2,3), \mathbf{n}=\langle 1,2,3\rangle\)

Short Answer

Expert verified
The general equation of the plane is \( x + 2y + 3z = 14 \).

Step by step solution

01

Understand the problem

We need to find the scalar equation of a plane. The plane passes through a given point \( P(1,2,3) \) and has a normal vector \( \mathbf{n}=\langle 1,2,3 \rangle \). This normal vector is perpendicular to every vector in the plane.
02

Recall the plane equation

The general scalar equation of a plane with a normal vector \( \langle a, b, c \rangle \) and passing through a point \( (x_0, y_0, z_0) \) is given by: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
03

Substitute values into the plane equation

Given that \( \mathbf{n}=\langle 1,2,3 \rangle \) and point \( P(1,2,3) \), substitute \( a = 1, b = 2, c = 3, x_0 = 1, y_0 = 2, z_0 = 3 \) into the plane equation: \[ 1(x - 1) + 2(y - 2) + 3(z - 3) = 0 \]
04

Simplify the equation

Distribute and combine like terms: \[ x - 1 + 2y - 4 + 3z - 9 = 0 \] Combine: \[ x + 2y + 3z - 14 = 0 \] This is the scalar equation of the plane.
05

Write the general form of the equation

The equation of the plane in general form is \( x + 2y + 3z = 14 \).

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Key Concepts

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

Normal Vector
In linear algebra, the concept of a normal vector is fundamental when dealing with planes. A normal vector is a vector that is perpendicular to a plane's surface. For any plane in a three-dimensional space, there is always a normal vector that is perpendicular to every vector lying on the plane. This perpendicularity makes the normal vector unique and crucial in defining the orientation of the plane.

When a plane is described with a normal vector, it represents a direction that defines how the plane is "tilted" or oriented in space. In the context of the exercise, the normal vector is given as \( \mathbf{n} = \langle 1,2,3 \rangle \), which tells us that this plane is not aligned with any of the axes alone, but rather it is tilted.

Normal vectors are vital because they allow us to write the equation of a plane in a concise manner as the dot product of the normal vector and a position vector, providing a systematic approach to finding this plane equation.
Plane Equation
The plane equation plays a central role in linear algebra as it allows us to describe a flat, two-dimensional surface extending infinitely in three dimensions. To find the plane's scalar equation, we need a point through which it passes and a normal vector.

As per the original solution, the general scalar equation of a plane with a normal vector \( \langle a, b, c \rangle \) passing through a point \((x_0, y_0, z_0)\) is expressed as:
  • \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
This equation leverages the normal vector components \( a, b, c \) to ensure the perpendicularity condition necessary for defining the plane.

The exercise further shows how the equation becomes \( x + 2y + 3z - 14 = 0 \) after substituting and simplifying. This represents a particular plane in its general form and reflects how we position the plane in space related to a specific point and orientation.
Linear Algebra
Linear algebra is a powerful tool that helps us address problems involving planes and vectors in multidimensional space efficiently. It provides the language of vectors and matrix operations, enabling us to perform complex calculations with ease.

Understanding linear algebraic concepts like vectors, dot products, and matrix equations enables students to not only write equations of planes but also solve more complicated geometric problems. It underpins the methods used in the exercises by offering a systematic approach to solving problems like finding the scalar equation of a plane.

In the context of the exercise, linear algebra allows us to take the normal vector and point and smoothly transition to the equation of the plane. With tools such as matrix operations, these methods can be applied to much larger systems, highlighting the versatility and power of linear algebra in both theoretical and applied mathematics.

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