Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the minimum distance from the origin to the line of intersection of the two planes $$ x+y+z=8 \text { and } 2 x-y+3 z=28 $$

Step-by-Step Solution

Answer

The minimum distance is \(2\sqrt{14}\).

Step 1: Find the Direction Vector of Line of Intersection

First, find the normal vectors of the planes: for plane 1, \(x+y+z=8\), the normal vector is \(\mathbf{n}_1 = \langle 1, 1, 1 \rangle\). For plane 2, \(2x-y+3z=28\), the normal vector is \(\mathbf{n}_2 = \langle 2, -1, 3 \rangle\). The direction vector \(\mathbf{d}\) of the line of intersection is the cross-product of these normals: \(\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2\). Calculate this to get \(\mathbf{d} = \langle 4, -1, -3 \rangle\).

Step 2: Find a Point on the Line of Intersection

To find a point on the line of intersection, set one of the variables to a convenient value. Let \(z=0\). Substitute in the equations to solve for \(x\) and \(y\). Solving \(x+y=8\) and \(2x-y=28\) gives \(x=12\) and \(y=-4\). The point \(P = (12, -4, 0)\) lies on both planes and hence is on the line of intersection.

Step 3: Calculate the Vector from Origin to the Point on the Line

The vector from the origin \((0, 0, 0)\) to the point \(P = (12, -4, 0)\) is \(\mathbf{OP} = \langle 12, -4, 0 \rangle\).

Step 4: Compute the Projection of Vector on the Direction Vector

Compute the projection \(\mathbf{proj}_{\mathbf{d}}(\mathbf{OP})\) of \(\mathbf{OP}\) onto the direction vector \(\mathbf{d}\). The formula is: \[ \mathbf{proj}_{\mathbf{d}}(\mathbf{OP}) = \frac{\mathbf{OP} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d} \]. Calculate \(\mathbf{OP} \cdot \mathbf{d} = 52\) and \(\mathbf{d} \cdot \mathbf{d} = 26\), thus, \[ \mathbf{proj}_{\mathbf{d}}(\mathbf{OP}) = \frac{52}{26} \langle 4, -1, -3 \rangle = 2 \langle 4, -1, -3 \rangle = \langle 8, -2, -6 \rangle \].

Step 5: Find the Perpendicular Vector

The vector perpendicular from the origin to the line is \(\mathbf{OP} - \mathbf{proj}_{\mathbf{d}}(\mathbf{OP})\). Thus, subtract \(\langle 8, -2, -6 \rangle\) from \(\langle 12, -4, 0 \rangle\) to get \(\langle 4, -2, 6 \rangle\).

Step 6: Calculate the Minimum Distance

Find the magnitude of the perpendicular vector \(\langle 4, -2, 6 \rangle\) using the formula \(\sqrt{x^2 + y^2 + z^2}\). Calculate \(\sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56}\). Hence, the minimum distance is \(\sqrt{56}\), which simplifies to \(2\sqrt{14}\).

Key Concepts

Line of Intersection

When dealing with two planes in space, the line where these planes intersect is called the line of intersection. This line is essentially the common boundary both planes share. By understanding the intersection, you can solve complex spatial problems. For example, consider two planes defined by the equations \(x+y+z=8\) and \(2x-y+3z=28\). To identify the line of intersection, you need to determine a direction vector along this line. You do this by taking the cross product of their normal vectors. Finding a point on this line also involves substituting values strategically to solve the system of equations, resulting in an accurate depiction of where the planes meet. This intersection forms a line, which can be further analyzed for related problem-solving, such as calculating the minimum distance to a point.

Normal Vectors

For any plane in three-dimensional space, mathematicians define a normal vector. This vector is perpendicular to the plane and essential for calculations involving intersections and projections. For the planes given in our exercise, first observe their equations:

• Plane 1: \(x+y+z=8\) has a normal vector of \(\mathbf{n}_1 = \langle 1, 1, 1 \rangle\).
• Plane 2: \(2x-y+3z=28\) has a normal vector of \(\mathbf{n}_2 = \langle 2, -1, 3 \rangle\).

These vectors are pivotal as they help compute the line of intersection by locating a direction vector. Simply compute the cross product of the normal vectors from each plane. Not only does this identify the line's direction, but it also helps visualize the three-dimensional structure of the planes.

Projection of Vectors

Vector projection is a technique used to find a component of one vector along the direction of another vector. This mathematical operation simplifies complex vector relationships by effectively aligning them. When computing the projection of vector \(\mathbf{OP}\) to the direction vector \(\mathbf{d}\), you use the formula: \[ \mathbf{proj}_{\mathbf{d}}(\mathbf{OP}) = \frac{\mathbf{OP} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d} \] First, calculate the dot product of \(\mathbf{OP}\) and \(\mathbf{d}\), then divide by the dot product of \(\mathbf{d}\) with itself. In this exercise, this results in the vector \(\langle 8, -2, -6 \rangle\). Notably, projections are incredibly useful for finding perpendicular or parallel components in physics and engineering.

Minimum Distance

The concept of minimum distance in vector calculus often involves finding the shortest length between a point and a line or plane. In our scenario, it's between the origin and the line of intersection.To find this distance, identify a vector perpendicular from the origin to the line. This vector is \(\mathbf{OP} - \text{proj}_{\mathbf{d}}(\mathbf{OP})\). The calculation gives \(\langle 4, -2, 6 \rangle\).Once you have this, measure its length using the magnitude formula: \[ \sqrt{x^2 + y^2 + z^2} = \sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{56} = 2\sqrt{14} \]This result, \(2\sqrt{14}\), is the shortest distance from the origin to the line, providing an essential solution to geometrical problems in physics and spatial analysis.