Problem 28 - Thomas Calculus

Question

Find the point of intersection of the lines \(x=t, y=\) $$-t+2, z=t+1,\( and \)x=2 s+2, y=s+3, z=5 s+6$$ and then find the plane determined by these lines.

Step-by-Step Solution

Answer

The intersection point is (0, 2, 1). The plane is 6x + 3y - 3z = 3.

Step 1: Identify the parametric equations

The first line is given by the parametric equations: \[x = t, \ y = -t + 2, \ z = t + 1\]. The second line is given by the parametric equations: \[x = 2s + 2, \ y = s + 3, \ z = 5s + 6\].

Step 2: Equate the parameters to find intersection

For the lines to intersect, we need the coordinates to be equal. Therefore, \[t = 2s + 2,\]\[-t + 2 = s + 3,\]\[t + 1 = 5s + 6\]. We solve these equations simultaneously to find the values of \(t\) and \(s\).

Step 3: Solve the system of equations

Using the first equation: \[t = 2s + 2\].Substitute \(t\) in the second equation: \[-(2s + 2) + 2 = s + 3\],\[-2s - 2 + 2 = s + 3\],which simplifies to \(-2s = s + 3\), leading to \(3s = -3\) and \(s = -1\). Substitute \(s = -1\) into \(t = 2s + 2\) to find \(t = 0\).

Step 4: Verify intersection point in parametric equations

Substitute \(t = 0\) into first line equations: \[x = 0, \ y = 2, \ z = 1\]. Substitute \(s = -1\) into second line equations: \[x = 0, \ y = 2, \ z = 1\]. Both sets match, confirming the intersection point is \((0, 2, 1)\).

Step 5: Find direction vectors of lines

The direction vector of the first line, derived from its parametric equations, is \((1, -1, 1)\). The direction vector of the second line is \((2, 1, 5)\).

Step 6: Find the normal vector to the plane

The normal vector to the plane can be found by taking the cross product of the direction vectors: \[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -1 & 1 \2 & 1 & 5 \end{vmatrix}\],which simplifies to \((-6, -3, 3)\).

Step 7: Formulate the equation of the plane

The equation of the plane is formed using the normal vector \((-6, -3, 3)\) and the point \((0, 2, 1)\) as \[-6(x - 0) - 3(y - 2) + 3(z - 1) = 0\],which simplifies to \[-6x - 3y + 3z = -3\].This can be further simplified to \[6x + 3y - 3z = 3\].

Key Concepts

Parametric Equations

Parametric equations are a way to express the coordinates of the points that make up geometric objects, like lines or curves, using one or more parameters. For example, the line equations \(x = t, \ y = -t + 2, \ z = t + 1\) and \(x = 2s + 2, \ y = s + 3, \ z = 5s + 6\) utilize parameters \(t\) and \(s\) to denote the position of any point on the lines as these parameters vary.

• The parameter \(t\) may represent time, moving along the line according to your choice.
• The set of equations describe a line in three-dimensional space.

The purpose of using these is to simplify complex geometric constructs into simpler algebraic formulas, allowing for straightforward computation to analyze points and paths. They are especially handy in determining points of intersection by equalizing the parametric expressions.

System of Equations

A system of equations is essentially a collection of two or more equations with a common set of unknowns. For lines to meet, their coordinates have to be equal at some parameter values. Here, we need the values of \(t\) and \(s\) that solve the system:

• \[ t = 2s + 2 \]
• \[ -t + 2 = s + 3 \]
• \[ t + 1 = 5s + 6 \]

Solving such systems usually involves substitution or elimination methods. We begin with one equation to express one parameter in terms of another, like substituting \(t = 2s + 2\) into the others, then solving for one parameter. Through these steps, we found \(s = -1\) and \(t = 0\).
This solution process is essential in determining the specific point where two lines intersect, which is not always obvious when lines are represented parametrically.

Cross Product

The cross product is a binary operation on two vectors in three-dimensional space which results in another vector, sometimes called the normal vector. This is particularly useful in finding a vector perpendicular to a plane formed by two direction vectors of lines.

• The cross product formula involves a determinant of a matrix consisting of unit vectors and the components of the given directional vectors.
• For example, consider the matrix \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -1 & 1 \ 2 & 1 & 5 \end{vmatrix} \]

The result of calculating this is \((-6, -3, 3)\), which is the normal vector to the plane. This process is fundamental to establish a plane's orientation in 3D space and is particularly used in applications like computer graphics and physics.

Plane Equation

A plane in three-dimensional space can be described using an equation that relates the coordinates of any point on the plane to a fixed point and a normal vector. Through the example, we use the intersection point \((0, 2, 1)\) and the normal vector \((-6, -3, 3)\) we earlier derived.
The general form used for the plane equation is \[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]where \((x_0, y_0, z_0)\) is a point on the plane, and \((n_1, n_2, n_3)\) is the normal vector. By substituting our values:

• We get \[-6(x - 0) - 3(y - 2) + 3(z - 1) = 0\]
• Which simplifies to \[6x + 3y - 3z = 3\]

This equation tells us everything we need to know about the plane's orientation and position in space. Understanding plane equations is vital for mastering spatial reasoning and solving geometry problems in three dimensions.