Problem 21 - APEX Calculus

Question

In Exercises 21-22, give the equation of the line that is the intersection of the given planes. \(p 1: 3(x-2)+(y-1)+4 z=0,\) and \(p 2: 2(x-1)-2(y+3)+6(z-1)=0\)

Step-by-Step Solution

Answer

Line: \(x = 3.5 + 14t, \; y = -3.5 - 10t, \; z = -8t\).

Step 1: Rewrite the Plane Equations in Standard Form

The given equations need to be expanded and rewritten in standard form. For plane \(p_1\), expand to get \(3x - 6 + y - 1 + 4z = 0\). Simplify this to \(3x + y + 4z = 7\). For plane \(p_2\), expand to obtain \(2x - 2 - 2y - 6 + 6z - 6 = 0\). Simplify to \(2x - 2y + 6z = 14\). Now, we have the equations \(3x + y + 4z = 7\) and \(2x - 2y + 6z = 14\).

Step 2: Find the Normal Vectors of the Planes

For plane \(p_1: 3x + y + 4z = 7\), the normal vector is \(\mathbf{n_1} = (3, 1, 4)\). For plane \(p_2: 2x - 2y + 6z = 14\), the normal vector is \(\mathbf{n_2} = (2, -2, 6)\).

Step 3: Calculate the Direction Vector of the Line of Intersection

The direction vector \(\mathbf{d}\) of the line of intersection can be found by taking the cross product of the normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\): \[\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 1 & 4 \ 2 & -2 & 6 \end{vmatrix} \]\[= \mathbf{i}(1\cdot6 - 4\cdot(-2)) - \mathbf{j}(3\cdot6 - 4\cdot2) + \mathbf{k}(3\cdot(-2) - 1\cdot2)\]\[= \mathbf{i}(6 + 8) - \mathbf{j}(18 - 8) + \mathbf{k}(-6 - 2)\]\[= 14\mathbf{i} - 10\mathbf{j} - 8\mathbf{k}\]Thus, the direction vector of the line is \((14, -10, -8)\).

Step 4: Find a Point on the Line of Intersection

To find a point that lies on both planes, you can choose values for two variables and solve for the third. Set \(z = 0\), substitute into both equations to find \((x, y)\) as follows. Substitute into \(3x + y + 4(0) = 7\) to get \(3x + y = 7\). From \(2x - 2y + 6(0) = 14\), simplify to get \(x - y = 7\). Solve the system:\[3x + y = 7\]\[x - y = 7\]Adding the equations, we get:\[4x = 14\]Thus, \(x = 3.5\). Substitute back to find \(y\): \(3.5 - y = 7\Rightarrow y = -3.5\). The point is \((3.5, -3.5, 0)\).

Step 5: Write the Parametric Equations of the Line

Using the direction vector \((14, -10, -8)\) and the point \((3.5, -3.5, 0)\), the parametric equations of the line of intersection are:\[x = 3.5 + 14t\]\[y = -3.5 - 10t\]\[z = -8t\]Here, \(t\) is the parameter.

Key Concepts

Equation of Line

When working with planes, finding the equation of the line that intersects two planes is a common task. This line is important because it is the only line that both planes have in common. To find this, it is essential to first manipulate the equations of the planes to a friendlier form. Start by expanding and arranging the plane equations into standard form, such as \(ax + by + cz = d\). This makes it easier to see the relationships between the coefficients, which are used to determine normal vectors.

Parametric Equations

Once you identify a direction vector and a point on the line of intersection, you can write the parametric equations of the line. Parametric equations describe a line by expressing each coordinate (\(x\), \(y\), \(z\)) as a function of a parameter, usually \(t\).
For example:

• \(x = x_0 + at\)
• \(y = y_0 + bt\)
• \(z = z_0 + ct\)

Where \((x_0, y_0, z_0)\) is a point on the line, and \((a, b, c)\) is the direction vector. Each equation increases or decreases the corresponding coordinate as \(t\) changes, effectively tracing out the line in three-dimensional space. This approach makes it very easy to find any point along the line or to understand the line's trajectory in space.

Normal Vectors

A crucial step in finding the intersection of two planes is identifying their normal vectors. The normal vector of a plane is a vector that is perpendicular to the plane.
This vector is derived from the coefficients of \(x\), \(y\), and \(z\) in the plane's standard form. For example, in the equation \(ax + by + cz = d\), the normal vector is given by \((a, b, c)\).
Normal vectors are important because they help find the direction vector of the line of intersection. By taking the cross product of the normal vectors of the two planes, you can derive the direction vector of their line of intersection. Thus, understanding and calculating normal vectors is vital in operations involving planes.

Cross Product

The cross product is a vector operation that can find a vector perpendicular to two given vectors in three-dimensional space. When you have the normal vectors of two planes, the cross product tells you the direction of the line that lies at the intersection of these planes.
You calculate the cross product of two vectors, \(\mathbf{n_1}\) and \(\mathbf{n_2}\), using this determinant format:\[\mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ n_{1x} & n_{1y} & n_{1z} \ n_{2x} & n_{2y} & n_{2z} \end{vmatrix} \]This results in a new vector \((\mathbf{d})\) that is perpendicular to both, representing the direction of the intersection line.
This method is powerful as it finds the direction without solving complicated linear equations. The result vector serves as a backbone for forming the parametric equations of the line. Understanding the cross product can thus simplify solving many geometry problems involving planes.