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Find a particular equation of the plane described. Perpendicular to \(\vec{n}=4 \vec{\imath}+3 \vec{\jmath}-2 \vec{k}\) and having an \(x\) -intercept of 5 (The \(x\) -intercept of a plane is the value of \(x\) when the other two variables are zero. \()\)

Short Answer

Expert verified
The equation of the plane is \(4x + 3y - 2z = 20\).

Step by step solution

01

Identify the Normal Vector

To write the equation of the plane, we need a normal vector. The vector given, \(\vec{n}=4\vec{\imath}+3\vec{\jmath}-2\vec{k}\), is perpendicular to the plane and thus serves as the normal vector. The components of this normal vector are \(A=4\), \(B=3\), and \(C=-2\).
02

Use the x-intercept to Find a Point on the Plane

The plane has an x-intercept of 5, which means it crosses the x-axis at the point \(P(5, 0, 0)\). This is a point through which the plane passes.
03

Write the Equation of the Plane

Using the normal vector \(\vec{n}\) and the point \(P\), the equation of the plane can be written in the form \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\). Plugging in \(A=4\), \(B=3\), \(C=-2\), and the coordinates of the point P, we get \(4(x - 5) + 3(y - 0) - 2(z - 0) = 0\). Simplifying, the equation of the plane is \(4x + 3y - 2z = 20\).

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Key Concepts

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

Normal Vector
Understanding the concept of a normal vector is crucial when learning about the equation of a plane. A normal vector, often denoted by \(\vec{n}\), is a vector that is perpendicular to a surface. In the context of a plane in three-dimensional space, the normal vector is orthogonal to any vector that lies on the plane.

For example, if our plane is represented by a sheet of paper lying flat on a table, any vector that stands straight up from the paper, like a pencil placed perpendicularly on it, represents a normal vector to the plane. The normal vector provides essential ingredients needed to define the plane mathematically. In the given exercise, the normal vector \(\vec{n}=4 \vec{\imath}+3 \vec{\jmath}-2 \vec{k}\) has its direction determined by the vector components, which relate directly to how the plane is oriented in three-dimensional space.
X-intercept
An x-intercept is a fundamental concept when it comes to understanding how a geometric shape, like a plane, interacts with the axes in a coordinate system. The x-intercept specifically refers to the point where the plane crosses the x-axis. This point is significant because it gives us a concrete location on the plane that can be used to develop its equation.

In the given exercise, the plane's x-intercept is 5, which means the plane crosses the x-axis at the point \( (5, 0, 0) \). This point is crucial in forming the equation of the plane, as it provides a known coordinate through which the plane passes. Knowing an x-intercept allows us to find a specific solution to the equation, helping complete the description of the plane within the coordinate system.
Vector Components
Vector components are the projections of a vector along the axes of a coordinate system. They are the individual influences that a vector has in each dimension. In three-dimensional space, a vector commonly has three components that correspond to the x, y, and z coordinates, often referred to with the unit vectors \(\vec{\imath}, \vec{\jmath}, \vec{k}\).

Determining the Orientation


In our plane equation exercise, the normal vector's components are 4, 3, and -2. These values dictate the normal vector's direction, and consequently, the orientation of the plane. The components 4, 3, and -2 tell us how many units the normal vector moves along the x, y, and z axes respectively, from the origin.

Creating the Plane Equation


These components are paramount in creating the equation of the plane. By substituting these values into the general plane equation along with a point on the plane—such as the x-intercept—we can derive the specific equation that models the plane within the given constraints. The components are directly proportional to the coefficients in the standard form of the plane's equation, which illustrate how the plane interacts with each of the x, y, and z dimensions.

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Most popular questions from this chapter

Elmer is going to build a tree house in his backyard for the children to play in. The yard is level. He uses one corner of the yard as the origin of a three-dimensional coordinate system. The \(x\) - and \(y\) -axes run along the ground, and the \(z\) -axis is vertical. He finds that the tree house will be at the point \((x, y, z)=(30,55,17),\) where the dimensions are in feet. Answer parts a-f. (IMAGE CANNOT COPY) a. Sketch the coordinate axes and the point (30,55,17) b. Write the position vector \(\vec{h}\) to the tree house. How high is the tree house above the ground? How far is the tree house from the origin? c. A wire is to be stretched from the tree house to the point (10,0,8) at the top corner of the back door so that the children can slide messages down it. Write a vector representing the displacement from the tree house to the point on the back door. d. How long will the wire in part c need to be? e. The children slide a message down the wire. It gets stuck when it is only \(30 \%\) of the way from the tree house to the back door. Write a vector representing the displacement from the tree house to the stuck message. How far along the wire did the message go before it got stuck? f. Write the position vector of the stuck message. How high above the ground is the stuck message?

Find the vector. \(\overrightarrow{D C}\) for \(C(-2,3)\) and \(D(4,-3)\)

Find the direction cosines of the vector from the first point to the second. $$(-3,7,1) \text { to }(4,8,-2)$$

Find a particular equation of the plane described. Parallel to the plane \(5 x-3 y-z=-4\) and containing the point (4,-6,1)

Find the indicated displacement vector. Use the answer to find the distance between the two points. \(\overrightarrow{B A}\) for \(A(9,13,-4)\) and \(B(3,6,-10)\)

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