Problem 28 - University Calculus: Early Transcendentals

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 lines intersect at (0, 2, 1), and the plane is defined by 6x + 3y - 3z = -3.

Step 1: Equate the parameters for x-coordinates

Given the two lines with parametric equations, equate the expressions involving the parameters for the x-coordinates. The equations to solve are:1. For the first line: \(x = t\)2. For the second line: \(x = 2s + 2\)Set \(t = 2s + 2\).

Step 2: Solve for t and s from x-coordinate equation

From the equation \(t = 2s + 2\), we can express \(t\) in terms of \(s\) or vice versa:If we solve for \(s\), we get:\(s = \frac{t - 2}{2}\).

Step 3: Use y-coordinate equations to find t and s

Substitute \(s = \frac{t-2}{2}\) into the y-coordinate equations for both lines:1. First line: \(y = -t + 2\)2. Second line: \(y = s + 3\)Set the y-values equal: \(-t + 2 = \frac{t-2}{2} + 3\) and solve for \(t\).

Step 4: Solve for t

Rearrange and solve the equation from Step 3:\(-t + 2 = \frac{t-2}{2} + 3\)Multiplying through by 2 to clear the fraction:\(-2t + 4 = t - 2 + 6\)Simplifying, we get:\(-2t + 4 = t + 4\)Subtract \(4\) from both sides and solve:\(-2t = t\)\(t = 0\).

Step 5: Find corresponding s and z-value

With \(t = 0\), substitute back into \(s = \frac{t - 2}{2}\):\(s = \frac{0 - 2}{2} = -1\)Now substitute \(t = 0\) into \(z = t + 1 = 0 + 1 = 1\), and \(s = -1\) into \(z = 5s + 6 = 5(-1) + 6 = 1\). Both lines intersect at \((x, y, z) = (0, 2, 1)\).

Step 6: Find direction vectors for the lines

The direction vector for the first line is \((1, -1, 1)\) and for the second line is \((2, 1, 5)\). Using these vectors, we can find the plane.

Step 7: Find the normal vector of the plane

The normal vector of the plane is the cross product of the direction vectors of the lines:Direction vectors:\(\mathbf{v_1} = \langle 1, -1, 1 \rangle\)\(\mathbf{v_2} = \langle 2, 1, 5 \rangle\)Cross product:\(\mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \langle -6, -3, 3 \rangle\).

Step 8: Find the equation of the plane

Using the normal vector \((-6, -3, 3)\) and the point of intersection \((0, 2, 1)\), the equation of the plane is:\(-6(x - 0) - 3(y - 2) + 3(z - 1) = 0\)Which simplifies to:\(-6x - 3y + 3z = 3\)or:\(6x + 3y - 3z = -3\).

Key Concepts

Parametric Equations

A parametric equation expresses the coordinates of the points making up a geometric object, such as a line, in terms of one or more parameters. In our exercise, the lines are given in parametric form:

• First Line: \( x = t, \; y = -t + 2, \; z = t + 1 \)
• Second Line: \( x = 2s + 2, \; y = s + 3, \; z = 5s + 6 \)

Here, \(t\) and \(s\) are parameters that vary along the lines. The equations tell us how each coordinate depends on this parameter value.
The main advantage of using parametric equations is they offer a simple way to describe graphical objects, and they are especially powerful in 3D for describing lines, as they allow tracing out points as parameters change.

Direction Vectors

Direction vectors play a crucial role in understanding parametric equations of lines. A direction vector provides the direction along which the points on the line extend. For each line in our exercise:

• First Line's Direction Vector: \( \langle 1, -1, 1 \rangle \)
• Second Line's Direction Vector: \( \langle 2, 1, 5 \rangle \)

These vectors guide how you trace the lines in space. Imagine them as arrows pointing in the direction your line will extend when the parameter changes.
By examining these vectors, one can determine if two lines are parallel or if they intersect. In our exercise, the intersection is found at a common point by equating the parametric equations.

Cross Product

The cross product is a vector operation that takes two vectors and returns a third vector that is perpendicular to the plane containing the first two vectors. In 3D geometry, this is particularly useful for finding a normal vector to a plane.

• Given Vectors: \( \mathbf{v_1} = \langle 1, -1, 1 \rangle \) and \( \mathbf{v_2} = \langle 2, 1, 5 \rangle \)

To find the normal vector, \[ \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} \] Results in: \( \mathbf{n} = \langle -6, -3, 3 \rangle \)This perpendicular vector is key to defining the plane that contains the intersecting lines. The cross product not only gives direction but also helps in constructing the equation of the plane.

Normal Vector

The normal vector is an important concept in 3D geometry, especially when dealing with planes. A normal vector is perpendicular to the plane, and it serves as a key element in the plane's equation.
In the given exercise, we derived the normal vector by taking the cross product of direction vectors from both lines:

• Normal Vector: \( \langle -6, -3, 3 \rangle \)

This vector is vital because it allows us to form the plane's equation using the point of intersection. The plane's equation has the form: \[ ax + by + cz = d \]where \((a, b, c)\) are the components of our normal vector.
This simplifies the process of representing the plane's orientation and position in space, translating our understanding of the problem into a mathematical language highly useful for solving geometry problems.