Question

The equation of the line of the intersection of the planes \(x+2 y-3 z=6\) and \(2 x-y+z=7\) is (A) \([(\mathrm{x}-4) / 1]=[(\mathrm{y}-1) / 7]=(\mathrm{z} / 5)\) (B) \([(\mathrm{x}+4) / 1]=[(\mathrm{y}-1) / 7)=(\mathrm{z} / 5)\) (C) \([(\mathrm{x}+1) / 1]=[(\mathrm{y}+1) / 7]=(\mathrm{z} / 5)\) (D) \([(\mathrm{x}-1) /(-1)]=[(\mathrm{y}-1) /(-7)]=(\mathrm{z} / 5)\)

Short answer

The short answer is: The equation of the line of intersection of the planes \(x + 2y - 3z = 6\) and \(2x - y + z = 7\) is \(\frac{x-4}{1}=\frac{y-1}{7}=\frac{z}{5}\), which corresponds to choice (A).

Step-by-step solution

Find a point on the line of intersection

To find a point on the line of intersection, we need to solve the system of equations formed by the two given plane equations: \(x + 2y - 3z = 6\) \(2x - y + z = 7\) We can eliminate the variable y by multiplying the first equation by 2 and adding it to the second equation: \[(2)(x + 2y - 3z) + (2x - y + z) = (2)(6) + 7\] \[5x - 5z = 19\] Now, let us solve for x in terms of z: \[x = \frac{19 + 5z}{5}\] Choosing z=0 (and since x, y, and z will be the coordinates of the point on the line), we get: \[x = \frac{19}{5}, y = \frac{6-(19/5)+3(0)}{2}= \frac{11}{5}\] So, the point on the line is \((\frac{19}{5}, \frac{11}{5}, 0)\).

Find the direction vector of the line

To find the direction vector for the line, we need to take the cross product of the normal vectors of the two planes. The normal vector of the first plane is \(A =\begin{bmatrix}1\\2\\-3\end{bmatrix}\) and the normal vector of the second plane is \(B =\begin{bmatrix}2\\-1\\1\end{bmatrix}\). The cross product AxB is given by: \[ \begin{bmatrix} (2(1) - (-3)(-1)) \\ (1(1) - (-3)(2)) \\ (1(-1) - 2(2)) \end{bmatrix} = \begin{bmatrix} -1 \\ -7 \\ -5 \end{bmatrix}\] Thus, the direction vector is \([-1, -7, -5]\).

Write the parametric equation of the line

Now that we have a point on the line and the direction vector, we can write the parametric equation of the line: \(\frac{x - (\frac{19}{5})}{(-1)} = \frac{y - (\frac{11}{5})}{(-7)} = \frac{z - 0}{(-5)}\) To make the fractions in the form like the given choices, multiply each side by the denominators 5, 7,5. We get: \[(\frac{x - (\frac{19}{5})}{(-1)})\times5\times7=(\frac{y - (\frac{11}{5})}{-7})\times5\times5=(\frac{z - 0}{-5})\times5\times7\] Simplifying gives us the final parametric equation: \[(\frac{x - 4}{1})=(\frac{y - 1}{7})=(\frac{z}{5})\] This corresponds to choice (A) from the given options.

Key concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. In problems involving the intersection of planes, these equations often represent each plane in a 3D space. For example, the given planes in the exercise can be written as:

• Equation 1: \(x + 2y - 3z = 6\)
• Equation 2: \(2x - y + z = 7\).

The goal is to find a common solution for these equations, which represents the line of intersection. To achieve this, we eliminate one variable, leaving a solvable set of equations for the other two variables.
By manipulating and combining the equations, we can isolate variables to find specific values that fit both planes simultaneously. This process is crucial in determining the specific point that lies on the line of intersection.

Cross Product

The cross product is a vital mathematical operation used in vector algebra. It helps determine a vector perpendicular to two given vectors. When dealing with planes, finding the cross product of their normal vectors gives us the direction vector of their intersection line.
For the exercise, the normal vectors of the given planes are:

• Normal vector of Plane 1: \( \begin{bmatrix}1\2\-3\end{bmatrix} \)
• Normal vector of Plane 2: \( \begin{bmatrix}2\-1\1\end{bmatrix} \)

Using the cross product formula, \[\begin{bmatrix}a_2b_3 - a_3b_2\a_3b_1 - a_1b_3\a_1b_2 - a_2b_1\end{bmatrix},\]we calculate their cross product to find the direction vector of the intersection line as:\[\begin{bmatrix}-1\-7\-5\end{bmatrix}.\]This vector forms the backbone of the line, dictating its orientation in 3D space.

Normal Vector

A normal vector is a vector that is perpendicular to a surface or a plane. It uniquely defines the orientation of a plane in space.
For a plane given by an equation like \(ax + by + cz = d\), the normal vector is \(\begin{bmatrix} a \ b \ c \end{bmatrix}\). It serves as an essential component when calculating the line of intersection between two planes.

• The normal vector indicates all the directions perpendicular to the surface, making it a crucial tool for understanding geometric shapes and forms in higher dimensions.

In the context of the exercise, identifying and using the normal vectors of the planes was key to finding the direction vector for the line. Thus, normal vectors are central to linking the algebraic representation of planes to their geometric intersections.

Parametric Equation

A parametric equation expresses a set of quantities as explicit functions of one or more parameters. In the context of lines and planes, such equations describe the position of any point along a line using a single variable, often denoted as \(t\).
The parametric equation for a line is given by:

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

Here \((x_0, y_0, z_0)\) is a point on the line, and \((a, b, c)\) is the direction vector of the line. This equation simplifies finding all points along the line, as we need only vary \(t\). For our exercise, after determining a point on the line and its direction vector, we can write the parametric equation, \[\frac{x - 4}{1} = \frac{y - 1}{7} = \frac{z}{5},\]expressing the coordinates of any point on the line as a function of parameters, providing a comprehensive description of the line's geometry in space.