Question

Give the equation of the line that is the intersection of the given planes. \(p 1: 5(x-5)+2(y+2)+4(z-1)=0,\) and \(p 2: 3 x-4(y-1)+2(z-1)=0\)

Short answer

The line's parametric equations are \( x = \frac{48}{13} + 20t \), \( y = \frac{85}{26} + 2t \), and \( z = -26t \).

Step-by-step solution

Expand Plane Equations

First, we'll expand the equations of the planes to put them in the standard form:For Plane 1, expand: \[5(x-5) + 2(y+2) + 4(z-1) = 0\]\[5x - 25 + 2y + 4 + 4z - 4 = 0\]\[5x + 2y + 4z - 25 = 0\]For Plane 2, expand: \[3x - 4(y-1) + 2(z-1) = 0\]\[3x - 4y + 4 + 2z - 2 = 0\]\[3x - 4y + 2z + 2 = 0\]

Identify Normal Vectors

Next, identify the normal vectors of both planes:- For Plane 1, the normal vector is \( \mathbf{n_1} = (5, 2, 4) \).- For Plane 2, the normal vector is \( \mathbf{n_2} = (3, -4, 2) \).

Find the Direction Vector of the Line

The direction vector of the line of intersection is the cross product of the normal vectors of the two planes:\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]Compute the cross product:\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 5 & 2 & 4 \ 3 & -4 & 2 \end{vmatrix} \]Solving the determinant, we get:\[ \mathbf{d} = (2 \cdot 2 - 4 \cdot (-4))\mathbf{i} - (5 \cdot 2 - 4 \cdot 3)\mathbf{j} + (5 \cdot (-4) - 2 \cdot 3)\mathbf{k} \]\[ \mathbf{d} = (4 + 16)\mathbf{i} - (10 - 12)\mathbf{j} + (-20 - 6)\mathbf{k} \]\[ \mathbf{d} = 20\mathbf{i} + 2\mathbf{j} - 26\mathbf{k} \]So, the direction vector is \( (20, 2, -26) \).

Solve the System for a Point on the Line

To find a point on the line, solve the system formed by the equations of the two planes by setting one of the variables (usually the easiest) as a parameter:1. From Plane 1: \[5x + 2y + 4z = 25\]2. From Plane 2: \[3x - 4y + 2z = -2\]Choose \(z = 0\) (as it simplifies the arithmetic):- Substitute \(z = 0\) into both plane equations:1. \(5x + 2y = 25\)2. \(3x - 4y = -2\)- Solve this system. Start by solving for \(y\) in the second equation:\[3x = 4y - 2\]\[y = \frac{3x + 2}{4}\]- Substitute \(y = \frac{3x + 2}{4}\) in the first equation:\[5x + 2\left(\frac{3x + 2}{4}\right) = 25\]\[5x + \left(\frac{6x + 4}{4}\right) = 25\]\[20x + 6x + 4 = 100\]\[26x = 96\]\[x = \frac{96}{26} = \frac{48}{13}\]- Substitute \(x = \frac{48}{13}\) back to find \(y\):\[y = \frac{3(\frac{48}{13}) + 2}{4} = \frac{144/13 + 26/13}{4} = \frac{170}{52} = \frac{85}{26}\]Thus, point on the line is \( \left(\frac{48}{13}, \frac{85}{26}, 0 \right) \).

Write the Parametric Equations of the Line

Using the point found and the direction vector, write the parametric equations of the line:\[ x = \frac{48}{13} + 20t \]\[ y = \frac{85}{26} + 2t \]\[ z = 0 - 26t \]

Key concepts

Normal Vector

In geometry and linear algebra, a normal vector is essential for understanding planes. It is a vector that is perpendicular (orthogonal) to the surface of a plane. For any given plane expressed as\[ ax + by + cz = d \]the coefficients \( a, b, \text{and} c \) form the components of the normal vector.

• A normal vector is crucial in identifying the orientation of a plane in three-dimensional space.
• It helps in defining the geometrical properties of the plane.

In our exercise, the normal vectors for the planes \( p_1 \) and \( p_2 \) were identified as \( \mathbf{n_1} = (5, 2, 4) \) and \( \mathbf{n_2} = (3, -4, 2) \). These vectors are derived directly from the coefficients of the variables in the plane equations.

Cross Product

The cross product is a mathematical operation applicable between two vectors in three-dimensional space, resulting in a third vector. This third vector is orthogonal to the plane generated by the initial vectors. It's vital for finding the direction vector of a line that becomes the intersection of two planes.

• It provides a way to calculate a perpendicular direction from two directions.
• It's only defined in three-dimensional space and is represented as a determinant.

To compute the cross product \( \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \), set up a 3x3 matrix determinant. The result of this computation, \( (20, 2, -26) \), is the direction vector for our line.

Direction Vector

The direction vector is fundamental in determining the line's direction in space. This vector shows the alignment and orientation of the line with respect to the planes. It emerges as the cross product of the normal vectors from the intersecting planes.
Recognizing the direction vector is important because it:

• Defines the line's trajectory.
• Helps write parametric equations that fully express the line geometrically.

In our example, after calculating the cross product, the resulting direction vector was \( (20, 2, -26) \), indicating the specific path the line follows as it crosses the planes.

Parametric Equations

Parametric equations are powerful tools for representing lines mathematically in Cartesian coordinates. They express the coordinates \((x, y, z)\) of any point on the line as a function of a parameter, often \( t \).
This is particularly useful as it:

• Provides a systematic way to capture all possible points on the line through the parameter's variation.
• Makes it easy to visualize and compute specific points on the line.

For the intersecting line in our scenario, the parametric equations are obtained by using a known point \( \left(\frac{48}{13}, \frac{85}{26}, 0\right) \) and the direction vector \( (20, 2, -26) \), formulated as:
\[ x = \frac{48}{13} + 20t \] \[ y = \frac{85}{26} + 2t \] \[ z = 0 - 26t \] These equations describe every point on the line formed by the intersection.

System of Equations

A system of equations consists of multiple equations that are solved simultaneously, often involving common variables. In the context of finding the intersection of planes, it's used to identify a specific point that lies on both planes.
Solving such systems is invaluable because they allow:

• Isolation and determination of shared variables' values.
• Identification of intersection points which are pivotal in line equations.

When solving our exercise, the system comprising the expanded plane equations was solved by parameterizing a variable, leading to solutions for \(x\) and \(y\). Setting \(z = 0\) simplified the solving process, yielding \( x = \frac{48}{13} \) and \( y = \frac{85}{26} \), helping us find a specific point on the line of intersection.