Problem 64 - Calculus for Biology and Medicine

Question

Where do a plane through \((2,0,-1)\) and perpendicular to \(\left[\begin{array}{r}-1 \\ 1 \\ 3\end{array}\right]\) and a line through the points \((1,0,-2)\) and \((1,-1,1)\) intersect?

Step-by-Step Solution

Answer

The plane and line intersect at the point \((1, -1, 1)\).

Step 1: Understand the Plane Equation

The plane passes through the point \((2, 0, -1)\) and is perpendicular to the vector \([-1, 1, 3]\). Thus, the equation of the plane can be established using the point-normal form: \(-1(x - 2) + 1(y - 0) + 3(z + 1) = 0\). Simplifying this gives the equation \(-x + y + 3z = 1\).

Step 2: Parametrize the Line

The line passes through the points \((1, 0, -2)\) and \((1, -1, 1)\). The direction vector of the line is calculated as the difference between these points, resulting in \([0, -1, 3]\). Therefore, the parametric equation of the line is \(x = 1, y = -t, z = -2 + 3t\).

Step 3: Set Up the System of Equations

To find where the line intersects the plane, substitute the parametric equations of the line into the plane equation. Substitute \(x = 1\), \(y = -t\), and \(z = -2 + 3t\) into \(-x + y + 3z = 1\). This results in \(-1 + (-t) + 3(-2 + 3t) = 1\). Simplifying gives \(-1 - t - 6 + 9t = 1\).

Step 4: Solve for the Parameter

Simplify the equation from Step 3: \(-1 - 6 + (-t + 9t) = 1\), which simplifies to \(8t - 7 = 1\). Solving for \(t\), we find: \(8t = 8\), so \(t = 1\).

Step 5: Find the Intersection Point

Using \(t = 1\), substitute back into the line's parametric equations: \(x = 1\), \(y = -1\times 1 = -1\), \(z = -2 + 3\times 1 = 1\). Therefore, the intersection point is \((1, -1, 1)\).

Key Concepts

Plane Equation

To find the equation of a plane, we need a point through which the plane passes and a vector that is perpendicular to it, known as the normal vector. Here, the plane passes through the point \((2, 0, -1)\) and is perpendicular to the vector \([-1, 1, 3]\). These components allow us to use the point-normal form of the plane equation, which is expressed as:\[\vec{n} \cdot (\vec{r} - \vec{r_0}) = 0\]where \(\vec{n}\) is the normal vector, \(\vec{r}\) is any point on the plane, and \(\vec{r_0}\) is the specific given point.By substituting the point and the normal vector into the equation, we get:\[-1(x - 2) + 1(y - 0) + 3(z + 1) = 0\]Simplifying this, we achieve the plane equation:\(-x + y + 3z = 1\).This formula represents all points \((x, y, z)\) that satisfy being part of the plane.

Parametric Equations

A parametric equation represents a line by expressing each coordinate as a separate equation dependent on a common parameter, often denoted as \(t\). To find the parametric equations of a line, we need two points through which the line passes. In our exercise, the points are \((1, 0, -2)\) and \((1, -1, 1)\).First, calculate the direction vector of the line by subtracting the coordinates of one point from the coordinates of the other:\[\vec{d} = (1 - 1, -1 - 0, 1 + 2) = (0, -1, 3)\]With the direction vector \((0, -1, 3)\) and a point on the line, say \((1, 0, -2)\), our parametric equations for the line are:- \(x = 1\)- \(y = -t\)- \(z = -2 + 3t\)These equations help us trace the line's path as the parameter \(t\) varies.

Intersection Point

Finding where a line intersects a plane involves substituting the parametric equations of the line into the plane equation. In our case, we use:- The plane equation: \(-x + y + 3z = 1\)- The parametric equations: - \(x = 1\) - \(y = -t\) - \(z = -2 + 3t\)Substitute these into the plane equation:\[-1(1) + (-t) + 3(-2 + 3t) = 1\]Breaking it down:- \(-1\) from \(x = 1\)- \(-t\) from \(y = -t\)- \(+ 3(-2 + 3t)\)Simplify the calculation:- Combine the terms: \(-1 - t - 6 + 9t = 1\)- Group the terms, resulting in\(8t - 7 = 1\)Solving gives us \(t = 1\).Substituting \(t = 1\) back into the parametric equations yields the intersection point:\((1, -1, 1)\).

Vector Operations

Vectors are essential in multivariable calculus for descriptions of direction and magnitude. They help define lines, planes, and their relationships in space. Here are some basic operations with vectors we used in the exercise:

• **Adding vectors** involves summing their corresponding components individually.
• **Subtracting vectors** helps find a direction vector between two points by subtracting the coordinates of one vector from another.
• **Scalar multiplication** adjusts the vector's magnitude without changing its direction, used in defining parametric equations.
• **Dot product** (or scalar product) is crucial for defining perpendicularity. The dot product of two perpendicular vectors is zero.

These operations enabled us to determine the normal vector for the plane using the dot product, calculate the direction vector for the line, and solve for their intersection point, applying both vector addition and the dot product.