Problem 6 - Active Calculus Multivariable

Question

(A) If the positive z-axis points upward, an equation for a horizontal plane through the point (-4,1,5) is ___________ (B) An equation for the plane perpendicular to the \(x\) -axis and passing through the point (-4,1,5) is ___________ (C) An equation for the plane parallel to the \(\mathrm{xz}\) -plane and passing through the point (-4,1,5) is ____________

Step-by-Step Solution

Answer

(A) \(z=5\) (B) \(x=-4\) (C) \(y=1\)

Step 1: (A) Horizontal plane through the point (-4,1,5)

To find the equation of a horizontal plane, we need to recognize that the direction of the normal vector is aligned with the positive z-axis. As such, the normal vector is given by \(\mathbf{n}=(0,0,1)\). Now, the equation of a plane can be written in the form: \[\mathbf{n}\cdot(\mathbf{x}-\mathbf{p}) = 0\] where \(\mathbf{n}\) is the normal vector, \(\mathbf{x}\) is a generic point on the plane, and \(\mathbf{p}\) is a given point on the plane. In this case, \(\mathbf{n}=(0,0,1)\) and \(\mathbf{p}=(-4,1,5)\). Plugging in the given values, we get: \[(0,0,1)\cdot((x,y,z)-(-4,1,5)) = 0\] Upon simplifying, we get: \[z=5\]

Step 2: (B) Plane perpendicular to the x-axis and passing through (-4,1,5)

For a plane to be perpendicular to the x-axis, the normal vector of the plane should point in the direction of the x-axis. So, the normal vector is given by \(\mathbf{n}=(1,0,0)\). Applying the same formula for the equation of a plane with \(\mathbf{n}=(1,0,0)\) and \(\mathbf{p}=(-4,1,5)\), we get: \[(1,0,0)\cdot((x,y,z)-(-4,1,5))=0\] Simplifying this, we get: \[x=-4\]

Step 3: (C) Plane parallel to the xz-plane and passing through (-4,1,5)

To find a plane parallel to the xz-plane, it must have the same normal vector as the xz-plane. The normal vector for the xz-plane is \(\mathbf{n}=(0,1,0)\) since it is perpendicular to the y-axis. Using the plane equation formula with \(\mathbf{n}=(0,1,0)\) and \(\mathbf{p}=(-4,1,5)\), we get: \[(0,1,0)\cdot((x,y,z)-(-4,1,5)) = 0\] Simplifying, we find: \[y=1\] So the answers are: (A) z=5 (B) x=-4 (C) y=1

Key Concepts

Normal Vector

When we talk about planes in geometry, the idea of a "normal vector" is very important. A normal vector is a straight arrow that pokes directly out from a plane. It helps to understand the plane's orientation in space. Here’s how it works:

• The normal vector is perpendicular to every line lying directly on the plane.
• In 3D space, if you have a normal vector given as \( \mathbf{n} = (a, b, c) \), then it describes a plane where \( ax + by + cz = d \).
• This vector influences the entire plane’s position and direction in space.

Understanding the normal vector allows you to write the equation for a plane quickly. For instance, if you know the normal vector \((0, 0, 1)\), it suggests the plane is horizontal, as it points in the z-axis direction. This means wherever the z-coordinate stays fixed, you have a horizontal plane.

Horizontal Plane

A horizontal plane is a flat, level surface, much like the calm waters of a lake. If we imagine the z-axis as pointing upwards, a horizontal plane holds constant at one z-value across all its points. This means:

• The normal vector for a horizontal plane is \((0, 0, 1)\), showing no change in the x or y directions.
• An example equation for a horizontal plane is \(z = c\), where \(c\) is a constant.
• All points on this plane will share the same z-coordinate, making it easy to visualize!

For example, through the point \((-4, 1, 5)\), a horizontal plane can be written as \(z = 5\). Every point on this plane will have z-coordinate 5, making it level in the direction of the horizon.

Perpendicular Plane

The term "perpendicular plane" refers to a plane that forms a right angle, or 90 degrees, with a particular line or another plane. To understand a perpendicular plane:

• The normal vector is key—it's directed along the line with which the plane is perpendicular.
• For a plane perpendicular to the x-axis, its normal vector is \(\mathbf{n} = (1, 0, 0)\).
• Such a plane would have an equation like \(x = a\), where all points share the x-coordinate \(a\).

When a plane passes through \((-4, 1, 5)\) and is perpendicular to the x-axis, it holds \(x = -4\) for every point on the plane. This indicates that anywhere on this plane, the x-component remains constant.

Parallel Plane

Parallel planes are fascinating because they never intersect, much like railroad tracks stretching into the distance. Understanding parallel planes involves thinking about planes that have the same orientation but at different positions. Here’s how:

• A plane parallel to another shares the same normal vector.
• If we consider a plane parallel to the xz-plane, its normal vector is \(\mathbf{n} = (0, 1, 0)\).
• The equation would be \(y = b\), meaning all points on such a plane have the same y-coordinate, standing parallel to the xz-plane.

So, a plane parallel to the xz-plane, passing through the point \((-4, 1, 5)\), keeps \(y = 1\) constant for every point on it. This shows that while the x and z values can vary, the y component remains fixed.