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For the following exercises, the equation of a plane is given. a. Find normal vector \(\mathbf{n}\) to the plane. Express \(\mathbf{n}\) using standard unit vectors. b. Find the intersections of the plane with the axes of coordinates. C. Sketch the plane. \(x+z=0\)

Short Answer

Expert verified
The normal vector is \(\mathbf{i} + \mathbf{k}\); the plane intersects the axes at \((0,0,0)\).

Step by step solution

01

Identify the Normal Vector

Given the plane equation is \(x + z = 0\). The general form of a plane equation is \(Ax + By + Cz = D\). Here, \(A = 1\), \(B = 0\), and \(C = 1\). The normal vector \(\mathbf{n}\) is \(\langle A, B, C \rangle = \langle 1, 0, 1 \rangle\). Expressing it using standard unit vectors, \(\mathbf{n} = \mathbf{i} + \mathbf{k}\).
02

Find Intersections with the Axes

To find the intersection with the x-axis, set \(y = 0\) and \(z = 0\). Solving \(x + 0 = 0\) gives \(x = 0\). The intersection with the x-axis is at the origin \((0,0,0)\).For the intersection with the y-axis, set \(x = 0\) and \(z = 0\). The plane equation \(0 + z = 0\) does not involve \(y\), so there is no specific intersection point with the y-axis other than the whole axis itself.For the intersection with the z-axis, set \(x = 0\) and \(y = 0\). Solving \(0 + z = 0\) gives \(z = 0\). The intersection with the z-axis is also at the origin \((0,0,0)\).
03

Sketch the Plane

The equation \(x + z = 0\) represents a plane that passes through the origin and is a diagonal plane in the x-z plane. It forms a 45-degree angle with both the negative x-axis and the positive z-axis. Since \(y\) does not appear, it extends infinitely parallel to the y-axis.

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Key Concepts

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

Understanding the Normal Vector
When we talk about the normal vector to a plane in 3D geometry, we mean a vector that is perpendicular to every line within that plane. This vector is like a direction signpost, defining the orientation of the plane in space. In our exercise, the plane equation given is \(x + z = 0\). We can compare this with the general form of a plane equation \(Ax + By + Cz = D\) to identify the coefficients that will build our normal vector.
  • The coefficient of \(x\) is \(A = 1\).
  • The coefficient of \(y\) is \(B = 0\).
  • The coefficient of \(z\) is \(C = 1\).
Using these, the normal vector \(\mathbf{n}\) is \(\langle 1, 0, 1 \rangle\). This simplifies to using the standard unit vectors as \(\mathbf{i} + \mathbf{k}\), making the direction of the normal vector clear and easy to understand. The absence of the \(\mathbf{j}\) component tells us that there is no change in the y-direction for the orientation of this plane.
Coordinate Axes Intersection Points
Finding where a plane intersects the coordinate axes can tell us a lot about its orientation in space. Each intersection gives us a point that can help visualize the position and orientation of the plane.
  • X-axis Intersection: To find where the plane intersects the x-axis, set \(y = 0\) and \(z = 0\). Solving the equation \(x + 0 = 0\) gives \(x = 0\). Therefore, the intersection point is at the origin, \((0, 0, 0)\).
  • Y-axis Intersection: Setting \(x = 0\) and \(z = 0\), we realize that the equation \(0 + z = 0\) doesn't involve \(y\). This means the plane extends along the entire y-axis, so there isn't a distinct intersection other than overlapping with the whole axis.
  • Z-axis Intersection: Set \(x = 0\) and \(y = 0\), which results in \(0 + z = 0\). Solving gives \(z = 0\), placing the intersection again at the origin \((0,0,0)\).
This exercise underlines that the plane passes through the origin and is fixed within the xz-plane, cutting the axes at a common point.
Visualize with 3D Sketching
Sketching planes in three-dimensional space can be a challenging but rewarding exercise. It helps us translate mathematical equations into real-world geometry. For the plane \(x + z = 0\), there are certain characteristics to note:- The plane is situated symmetrically between the negative x-axis and the positive z-axis. This results in a diagonal plane structure relative to these axes.- As the plane is described by \(x + z = 0\) and lacks a \(y\) term, it implies that it extends infinitely along the y-axis direction without any tilt or slope.To visualize this plane:- Imagine a sheet of paper standing upright on the floor. Along the xz-plane, it will form a 45-degree angle with both the negative x and positive z directions.- Anywhere you look down the y-axis, you'll see this fundamental angle remaining the same.By creating mental or drawn sketches of planes, we develop a deeper understanding of their spatial orientation and relationships.

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