Problem 34 - Essential Calculus Early Transcendentals

Question

Where does the line through \((1,0,1)\) and \((4,-2,2)\) intersect the plane \(x+y+z=6 ?\)

Step-by-Step Solution

Answer

The line intersects the plane at the point (7, -4, 3).

Step 1: Parametrize the line

To find where the line intersects the plane, first we need to write the parametric equation of the line passing through the points \(1, 0, 1\) and \(4, -2, 2\). The direction vector of the line can be obtained by computing the difference between these points:\[\vec{v} = (4-1, -2-0, 2-1) = (3, -2, 1)\].Let the point \(P = (1, 0, 1)\) be a point on the line. Therefore, the parametric equation for the line can be given by:\[(x, y, z) = (1, 0, 1) + t(3, -2, 1) = (1+3t, -2t, 1+t)\] where \(t\) is a parameter.

Step 2: Substitute into the Plane Equation

The equation of the plane is given by \ x + y + z = 6 \. Substitute the parametric equations from the line into the plane:\[(1+3t) + (-2t) + (1+t) = 6\].This substitution will help us find the value of \(t\) at which the line intersects the plane.

Step 3: Solve the Equation for t

Simplify and solve the equation:\[(1 + 3t - 2t + 1 + t) = 6\]\[2 + 2t = 6\]Subtract 2 from both sides:\[2t = 4\]Divide by 2:\[t = 2\].This tells us that when \(t = 2\), the line intersects the plane.

Step 4: Find the Intersection Point

Substitute \(t = 2\) back into the parametric equations of the line to find the intersection point:- \ x = 1 + 3(2) = 7 \ - \ y = -2(2) = -4 \ - \ z = 1 + 2 = 3 \ Therefore, the line intersects the plane at the point \( (7, -4, 3) \).

Key Concepts

Parametric Equation

A parametric equation is a way to represent a line or curve using a parameter, often denoted as \(t\). In the context of our problem, we have a line passing through two points: \((1, 0, 1)\) and \((4, -2, 2)\). By utilizing these points, we can derive a parametric equation.

The first step is to calculate the direction vector, which we'll do in the next section. Here, let's focus on writing the equation once we have our direction vector \(\vec{v}\). This vector helps us express the line's movement through space.

For the points given, we selected \((1, 0, 1)\) as our starting point. The equation is formulated as:

• \((x, y, z) = (1, 0, 1) + t(3, -2, 1) = (1+3t, -2t, 1+t)\).

This expression gives us a set of equations for each coordinate \(x, y, z\) in terms of the parameter \(t\). By varying \(t\), we can find any point along the line. This is how the line’s position is described in parametric form.

Direction Vector

The direction vector is essential in defining the direction of a line in space. It shows us how the line progresses from one point to another, and it's derived from the differences in coordinates between two points on the line. For our line through \((1, 0, 1)\) and \((4, -2, 2)\), the direction vector \(\vec{v}\) is calculated as follows:

- Subtract the coordinates of the starting point from the ending point:

• \(\vec{v} = (4-1, -2-0, 2-1) = (3, -2, 1)\).

This vector \(\vec{v} = (3, -2, 1)\) indicates the line’s change in \(x\), \(y\), and \(z\) for any change in \(t\).

When you include this vector in the parametric equation, it essentially captures how the line extends along the direction defined by the vector. The line moves 3 units in the \(x\)-direction, decreases 2 units in the \(y\)-direction, and increases 1 unit in the \(z\)-direction, per unit of \(t\). This idea is visualized with the parametric equation derived above.

Intersection Point

Finding an intersection point where a line meets a plane involves combining the parametric equation of the line and the equation of the plane. In our case, the plane is defined by the equation \(x + y + z = 6\).

To find the exact intersection point:

1. Substitute the parametric expressions of \(x\), \(y\), and \(z\) from the line into the plane’s equation:

• \((1+3t) + (-2t) + (1+t) = 6\).

2. Solve for \(t\) to determine when the plane and line intersect:

• Combine and simplify: \(2 + 2t = 6\).
• Solving gives \(t = 2\).

3. Find the coordinates of the intersection point by substituting \(t = 2\) back into the parametric equations:

• \(x = 1 + 3(2) = 7\)
• \(y = -2(2) = -4\)
• \(z = 1 + 2 = 3\)

Therefore, the intersection occurs at \((7, -4, 3)\). This point is where the line crosses the plane, showcased by substituting the direction vector into the line's equation and setting it equal to the plane equation.