Problem 57 - Advanced Engineering Mathematics

Question

In Problems, find an equation of the plane that satisfies the given conditions. Contains the parallel lines \(x=1+t, y=1+2 t, z=3+t ;\) \(x=3+s, y=2 s, z=-2+s\)

Step-by-Step Solution

Answer

The equation of the plane is \(3x + 7y - 5z = -11\).

Step 1: Identify Points on the Lines

The first step is to find points on each of the given lines. For the line given by \(x=1+t, y=1+2t, z=3+t\), let's choose \(t=0\), giving point \((1, 1, 3)\). For the other line, \(x=3+s, y=2s, z=-2+s\), choose \(s=0\), giving point \((3, 0, -2)\).

Step 2: Find Direction Vectors of the Lines

The direction vector for the first line is \(\vec{v_1} = (1, 2, 1)\), which is the coefficients of \(t\) in the parametric equations. The direction vector for the second line is \(\vec{v_2} = (1, 2, 1)\) too, as it also matches the coefficients of \(s\) in its parametric equations.

Step 3: Calculate Cross Product for the Normal Vector

Since both lines have the same direction vectors, they are parallel. We need a vector orthogonal to this direction for the plane's normal, so we use the cross product of \(\vec{v_1}\) with a vector formed from the two points. The vector between the points we found in Step 1 is \((3 - 1, 0 - 1, -2 - 3) = (2, -1, -5)\). Compute \(\vec{n} = \vec{v_1} \times (2, -1, -5)\). The cross product is \(\vec{n} = (3, 7, -5)\).

Step 4: Form Equation of the Plane

Using one of the points, say \((1, 1, 3)\), and the normal vector \(\vec{n} = (3, 7, -5)\), the equation is formed as: \[ 3(x - 1) + 7(y - 1) - 5(z - 3) = 0 \]. Simplifying this, the equation becomes \(3x + 7y - 5z = -11\).

Key Concepts

Parallel Lines

In the world of geometry, parallel lines are lines that never meet. These lines run in the same direction, staying a constant distance apart from each other. It's like train tracks; no matter how far they extend, they never intersect.

For any two lines to be parallel, they must have the same direction vector. In our exercise, both lines gave us the same direction vector of \((1, 2, 1)\). This means that in three-dimensional space, the lines move in exactly the same direction.

• Parallel lines have identical or proportional direction vectors.
• They maintain constant separation and will never intersect, no matter how far extended.

Understanding these concepts is crucial for geometric problem solving, especially when determining planes that connect such lines.

Direction Vectors

Direction vectors are essential in describing how a line moves through space. They represent the path in which a line travels and can be visualized as arrows that point in the same direction as the line.

For the first line in our exercise, the direction vector is derived from \(x=1+t, y=1+2t, z=3+t\), resulting in \(\vec{v_1} = (1, 2, 1)\). The second line is \(x=3+s, y=2s, z=-2+s\), which also produces \(\vec{v_2} = (1, 2, 1)\), indicating both lines are parallel. These vectors show the lines' directional movement.

• The direction vector is calculated based on the coefficients of the parameter in line equations.
• They are fundamental in determining whether lines are parallel or intersecting.

Knowing the direction vectors allows us to understand how lines navigate through a given space.

Normal Vector

The normal vector is a key element in defining a plane in three-dimensional space. It is always perpendicular to the plane and any two vectors that lie within the plane.

To determine the plane containing our parallel lines, we need a normal vector perpendicular to the direction vectors. In this exercise, we used the cross product to find a vector perpendicular to both provided direction vectors and an additional vector between two arbitrary points on each line.

• A normal vector is orthogonal to any vector within the plane it defines.
• It helps in forming the equation of the plane.
• A plane is uniquely defined by a point in space and a normal vector.

By calculating the normal vector, we establish a unique plane on which the parallel lines reside.

Cross Product

The cross product is a mathematical operation used to determine a vector that is perpendicular to two given vectors. It's crucial when working with three-dimensional space as it helps us find direction and orientation.

In our exercise, the cross product was used to find the normal vector of the plane. We took the direction vector \(\vec{v_1} = (1, 2, 1)\) and a vector between two points on our lines \((2, -1, -5)\), calculating \(\vec{n} = (3, 7, -5)\) as the normal using the formula: \[ \vec{a} \times \vec{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1) \]

• The result of a cross product is always orthogonal to the original vectors.
• It is instrumental in finding normal vectors to planes.
• This operation applies specifically to three-dimensional space.

Using the cross product efficiently allowed us to deduce the correct normal vector for our plane equation.