Problem 41
Question
Find the equation of the plane through \((0,0,0)\) and perpendicular to \([1,0,0]\) '.
Step-by-Step Solution
Verified Answer
The equation of the plane is x = 0.
1Step 1: Understanding the Problem
We need to find the equation of a plane. The plane should pass through the point
(0,0,0)
and be perpendicular to a direction vector
[1,0,0]
. These details will help us derive the necessary components of the plane's equation.
2Step 2: Identify the Normal Vector
The plane is perpendicular to the vector
[1,0,0]
. Therefore, this vector is the normal vector of the plane. The normal vector is crucial because it defines the orientation of the plane in space.
3Step 3: Equation of a Plane
The general formula for the equation of a plane with a normal vector
(a, b, c)
that passes 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.
4Step 4: Substitute Known Values
We substitute the components of the normal vector
(a, b, c) = (1, 0, 0)
and the point
(x_0, y_0, z_0) = (0, 0, 0)
into the general equation of the plane:
1(x - 0) + 0(y - 0) + 0(z - 0) = 0.
5Step 5: Simplify the Equation
The substitution from the previous step simplifies to:
x = 0.
This is the equation of the plane. It denotes all points that have their x-coordinate as 0, effectively being parallel to both the yz-plane and perpendicular to the x-axis.
Key Concepts
Normal VectorPerpendicular VectorPlane in 3D Space
Normal Vector
In geometry, especially when dealing with planes in 3D space, the concept of a normal vector is essential. A normal vector is a vector that is perpendicular to a plane or a surface.
For a plane, the normal vector uniquely determines its orientation. If you know the normal vector, you essentially know how the plane is "tilted" in space.
When finding the equation of a plane, having the normal vector allows you to apply the necessary formula to define the plane completely:
\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]
where \((a, b, c)\) is the normal vector, and \((x_0, y_0, z_0)\) is a point through which the plane passes.
This vector is pivotal in calculations because it provides directionality to the mathematical expression, ensuring that the plane aligns correctly with the given vectors in the space. In the problem at hand, the normal vector is \([1, 0, 0]\), meaning the plane is perpendicular to the x-direction.
For a plane, the normal vector uniquely determines its orientation. If you know the normal vector, you essentially know how the plane is "tilted" in space.
When finding the equation of a plane, having the normal vector allows you to apply the necessary formula to define the plane completely:
\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]
where \((a, b, c)\) is the normal vector, and \((x_0, y_0, z_0)\) is a point through which the plane passes.
This vector is pivotal in calculations because it provides directionality to the mathematical expression, ensuring that the plane aligns correctly with the given vectors in the space. In the problem at hand, the normal vector is \([1, 0, 0]\), meaning the plane is perpendicular to the x-direction.
Perpendicular Vector
When a vector is perpendicular to another object in space, it forms a right angle with it. In the context of 3D geometry, if we say a plane is perpendicular to a given vector, it means the plane is at a right angle to that vector.
This perpendicular relationship simplifies determining equations of planes. Whenever a plane is described as being perpendicular to a vector, that vector automatically becomes the normal vector of the plane.
As seen in the original exercise, the vector \([1, 0, 0]\) is perpendicular to the plane, meaning it acts as the plane's normal vector. This gives us the components needed to plug into the general plane equation:
This perpendicular relationship simplifies determining equations of planes. Whenever a plane is described as being perpendicular to a vector, that vector automatically becomes the normal vector of the plane.
As seen in the original exercise, the vector \([1, 0, 0]\) is perpendicular to the plane, meaning it acts as the plane's normal vector. This gives us the components needed to plug into the general plane equation:
- The vector \([1, 0, 0]\) indicates the plane is perpendicular to the x-axis.
- The resulting equation is \(x = 0\), showing that all points on the plane have the same x-value.
Plane in 3D Space
A plane in 3D space is a flat, two-dimensional surface extending infinitely in three-dimensional surroundings. It is defined using either:
\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]
This formula ensures that any point \((x, y, z)\) on the plane will satisfy the equation.
The simplicity of visualizing a plane comes when aligning it with coordinate planes. In our example, the plane equation \(x = 0\) signifies that the plane is parallel to the yz-plane and covers all coordinates where the x-coordinate remains zero.
This basic understanding helps students see how planes function in a 3D environment and how they relate spatially to axes and other geometric entities.
- A point through which it passes and a normal vector.
- Or more commonly, by specifying an equation.
\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]
This formula ensures that any point \((x, y, z)\) on the plane will satisfy the equation.
The simplicity of visualizing a plane comes when aligning it with coordinate planes. In our example, the plane equation \(x = 0\) signifies that the plane is parallel to the yz-plane and covers all coordinates where the x-coordinate remains zero.
This basic understanding helps students see how planes function in a 3D environment and how they relate spatially to axes and other geometric entities.
Other exercises in this chapter
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View solution