Problem 30 - Thomas Calculus

Question

In Exercises 29 and \(30,\) find the plane containing the intersecting lines. $$L 1 : x=t y=3-3 t z=-2-t -\infty < t < \infty$$ $$L 2 : x=1+s y=4+s z=-1+s -\infty < s < \infty$$

Step-by-Step Solution

Answer

The plane containing the intersecting lines is given by the equation: \(x + y - 2z = 6\).

Step 1: Determine Direction Vectors of the Lines

For line \( L1 \), the direction vector \( \mathbf{d}_1 \) is derived from the coefficients of \( t \), which gives us \( \mathbf{d}_1 = \langle 1, -3, -1 \rangle \). For line \( L2 \), the direction vector \( \mathbf{d}_2 \) is derived from the coefficients of \( s \), which results in \( \mathbf{d}_2 = \langle 1, 1, 1 \rangle \).

Step 2: Find a Common Point on Both Lines

Set \( x \) equations equal: \( t = 1 + s \). Set \( y \) equations equal: \( 3 - 3t = 4 + s \). Set \( z \) equations equal: \(-2 - t = -1 + s\). Solve these equations simultaneously to find the common point. After solving, we find that both lines intersect at the point \( (1, 1, -3) \).

Step 3: Calculate the Cross Product for the Normal Vector

With intersecting direction vectors \( \mathbf{d}_1 = \langle 1, -3, -1 \rangle \) and \( \mathbf{d}_2 = \langle 1, 1, 1 \rangle \), calculate the cross product to find the plane's normal vector: \( \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \langle -2, -2, 4 \rangle \).

Step 4: Formulate the Equation of the Plane

With the normal vector \( \mathbf{n} = \langle -2, -2, 4 \rangle \) and the point of intersection \( (1, 1, -3) \), use the plane formula \( ax + by + cz = d \). Substitute \((a, b, c) = (-2, -2, 4)\) and point \((1, 1, -3)\) to find \(-2(x - 1) - 2(y - 1) + 4(z + 3) = 0\). Simplifying gives the plane equation: \(-2x - 2y + 4z = -12\).

Step 5: Simplify the Plane Equation

Divide the entire equation by \(-2\) to simplify it further, which results in \(x + y - 2z = 6\).

Key Concepts

Direction Vectors

Understanding direction vectors is crucial when dealing with lines in space. A direction vector indicates the direction in which a line extends. In the example given, we have two lines: \(L1\) and \(L2\). To find the direction vector of these lines, we look at the coefficients of the parameters \(t\) and \(s\) in the parametric equations of the lines.

For \(L1: x = t, y = 3 - 3t, z = -2 - t\), the parameter is \(t\), and the corresponding direction vector \(\mathbf{d}_1\) is \(\langle 1, -3, -1 \rangle\). Similarly, in line \(L2: x = 1 + s, y = 4 + s, z = -1 + s\), \(s\) is the parameter, and its direction vector \(\mathbf{d}_2\) is \(\langle 1, 1, 1 \rangle\).

Direction vectors are fundamental because they help us determine how lines relate to each other, such as being parallel or intersecting. In this case, both lines intersect at a single point, which we can further explore using these vectors.

Cross Product

The cross product is an operation on two vectors in three-dimensional space that results in another vector, which is perpendicular (or normal) to the plane formed by the original two vectors. Here, we have direction vectors \( \mathbf{d}_1 = \langle 1, -3, -1 \rangle \) and \( \mathbf{d}_2 = \langle 1, 1, 1 \rangle \).

To find the plane that contains both lines \(L1\) and \(L2\), we calculate the cross product \( \mathbf{d}_1 \times \mathbf{d}_2 \), resulting in the vector \( \mathbf{n} = \langle -2, -2, 4 \rangle \).

The resulting vector \(\mathbf{n}\) is normal to the plane formed by the intersecting lines. Calculating the cross product is a key step in finding the equation of a plane, as the normal vector is used in its formula.

Plane Equation

The equation of a plane in three-dimensional space can be expressed in the standard form: \( ax + by + cz = d \), where \( \langle a, b, c \rangle \) is the normal vector to the plane and \((x, y, z)\) are the coordinates. Once we have the normal vector and a point on the plane, we can formulate the equation of the plane.

In this exercise, the normal vector \( \mathbf{n} = \langle -2, -2, 4 \rangle \) is obtained from the cross product, and the point \((1, 1, -3)\) is where the lines intersect. By substituting these into the formula, we establish the initial plane equation: \(-2(x - 1) - 2(y - 1) + 4(z + 3) = 0\).

Simplifying gives us \(-2x - 2y + 4z = -12\). For ease, this is further simplified by dividing by \(-2\), resulting in \(x + y - 2z = 6\). This final equation represents the plane encompassing the intersecting lines, hinging on the critical understanding of both the normal vector and its relationship to the plane's form.