Problem 63 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The intersection point is \(\left(\frac{7}{4}, -\frac{5}{4}, \frac{15}{4}\right)\).

Step 1: Equation of the Plane

To find the equation of the plane, we need a point (\(1, -1, 2\)) and a normal vector, which is given as \(\begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix}\). The plane equation format is \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \), where \((x_0, y_0, z_0)\) is a point on the plane.Plugging in the values:\[ 1(x - 1) + 2(y + 1) + 1(z - 2) = 0 \]Simplifying, we obtain:\[ x + 2y + z = 3 \].

Step 2: Parametric Equation of the Line

The line is determined by points \((0, -3, 2)\) and \((1, -2, 3)\). The direction vector is found by subtracting the coordinates:\[ \begin{bmatrix} 1 - 0 \ -2 + 3 \ 3 - 2 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \]The parametric equations are then:\[ x = 0 + t, \quad y = -3 + t, \quad z = 2 + t \] where \(t\) is a parameter.

Step 3: Substituting Line Into Plane Equation

Substitute \(x = t\), \(y = -3 + t\), and \(z = 2 + t\) from the line's parametric equations into the plane equation:\[ t + 2(-3 + t) + (2 + t) = 3 \].

Step 4: Solving for Parameter \( t \)

Simplifying the substituted equation:\[ t + 2(-3 + t) + 2 + t = 3 \]\[ t - 6 + 2t + 2 + t = 3 \]\[ 4t - 4 = 3 \]Adding 4 to both sides:\[ 4t = 7 \]Dividing both sides by 4:\[ t = \frac{7}{4} \].

Step 5: Finding Intersection Point

Substitute \(t = \frac{7}{4}\) back into the parametric equations:\[ x = \frac{7}{4}, \quad y = -3 + \frac{7}{4} = -\frac{5}{4}, \quad z = 2 + \frac{7}{4} = \frac{15}{4} \]Thus, the intersection point is \(\left(\frac{7}{4}, -\frac{5}{4}, \frac{15}{4}\right)\).

Key Concepts

Equation of a Plane

A plane in three-dimensional space can be defined by a point and a normal vector. The point is a location on the plane, and the normal vector is perpendicular to every line that lies on the plane itself. This is crucial for describing the orientation of the plane in space.

The general form of a plane's equation is:

• \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \)

Where \((x_0, y_0, z_0)\) is a point on the plane, and \(a, b, c\) are the components of the normal vector. These values affect the tilt and direction of the plane.

For example, in our problem, the point on the plane is \(1, -1, 2\) and the normal vector is \(1, 2, 1\). By replacing the variables, we find the specific equation \( x + 2y + z = 3 \). This equation captures all points \((x, y, z)\) on the plane.

Parametric Equation of a Line

In 3D space, a line can be represented parametrically, which means describing each coordinate as a function of a single variable, typically denoted as \(t\). A specific point on the line anchors it, while a direction vector indicates its trajectory.

For parametric equations, the form is:

• \( x = x_0 + ta \)
• \( y = y_0 + tb \)
• \( z = z_0 + tc \)

Where \((x_0, y_0, z_0)\) is a point on the line, and \( (a, b, c) \) are the direction vector's components.

In our case, the line passes through points \((0, -3, 2)\) and \((1, -2, 3)\), yielding a direction vector \((1, 1, 1)\). Thus, the parametric equations become \( x = t \), \( y = -3 + t \), and \( z = 2 + t \). These equations enable us to generate all possible points along the line by varying \(t\).

Vector Operations

Vector operations are essential for dealing with geometric problems, especially when it involves lines and planes. Basic operations include addition, subtraction, and finding a vector magnitude.

One often starts with **finding a direction vector** between two points by subtraction:

• For two points, \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), the direction vector \((x_2-x_1, y_2-y_1, z_2-z_1)\). This gives a vector pointing from one point to the other.

Another crucial concept is **dot product**, which assesses the angle between vectors:

• For vectors \( \vec{u} = (a, b, c) \) and \( \vec{v} = (x, y, z) \), their dot product is \( ax + by + cz \).

When this product is zero, vectors are perpendicular.

Involving vectors in equations helps solve complex spatial problems. For example, using a normal vector to define a plane or finding a direction vector for a line, as done in this exercise, is an application of these operations.