Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane containing the line \(x=1+2 t, y=-1+3 t, z=4+t\) and the point \((1,-1,5) .\)

Step-by-Step Solution

Answer

The equation of the plane is \(-3x + 2y + 1 = 0\).

Step 1: Identify Direction Vector of the Line

The given line is in parametric form: \(x=1+2t\), \(y=-1+3t\), \(z=4+t\). From this, we can identify the direction vector of the line as \( \mathbf{d} = \langle 2, 3, 1 \rangle \).

Step 2: Use the Given Point on the Plane

We are given a point \((1, -1, 5)\) on the plane. We also notice this is another point on the line when \(t=1\). Thus, this point will be used to construct two vectors that lie on the plane.

Step 3: Find Two Vectors on the Plane

The point on the line when \(t=0\) is \((1, -1, 4)\). We use two points \((1, -1, 5)\) and \((1, -1, 4)\) on the line to create two vectors lying on the plane: \(\vec{v_1} = (0, 0, 1)\) (from \((1, -1, 5) - (1, -1, 4)\)) and \(\vec{v_2} = (2, 3, 1)\) (as given directly by the line's direction vector).

Step 4: Find the Normal Vector of the Plane

To find the normal vector of the plane, compute the cross product \( \vec{n} = \vec{v_1} \times \vec{v_2} \). Calculate the cross product: \( (0, 0, 1) \times (2, 3, 1) = (-3, 2, 0) \). Thus, the normal vector is \((-3, 2, 0)\).

Step 5: Write the Equation of the Plane

Using the point \((1, -1, 5)\) and the normal vector \((-3, 2, 0)\), substitute these into the plane equation formula: \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \). This gives: \(-3(x-1) + 2(y+1) + 0(z-5) = 0\). Simplify to \(-3x + 2y + 1 = 0\).

Key Concepts

Parametric equations

Parametric equations are a way of expressing the coordinates of points on a line using a parameter, typically denoted as \(t\). They are particularly useful in analytical geometry for describing lines and curves in space.

For instance, in the parametric equations given:
\[x = 1 + 2t, \quad y = -1 + 3t, \quad z = 4 + t\]
each point on the line is represented as a function of \(t\). By changing the value of \(t\), we generate different points along the line.

• \(x = 1 + 2t\) indicates that as \(t\) increases or decreases, the x-coordinate of points along the line shifts linearly with this expression.
• \(y = -1 + 3t\) shows a similar linear change for the y-coordinate.
• \(z = 4 + t\) does the same for the z-coordinate.

These equations help identify a line's direction and are foundational in forming other geometric constructs such as planes.

Direction vector

Direction vectors are a key concept when working with parametric equations. They provide a straight-line path's orientation in space and are derived from the coefficients of the parameter \(t\) in the parametric equations.

For the line described by the equations above:
The direction vector is given by \( \mathbf{d} = \langle 2, 3, 1 \rangle \). This vector indicates the direction in which the line extends.

• The first component, \(2\), corresponds to how much the x-coordinate changes as \(t\) changes.
• The second component, \(3\), corresponds to the y-coordinate change.
• The third component, \(1\), aligns with the change in z-coordinate.

Understanding direction vectors is crucial as they are utilized to define and find perpendicular planes in space.

Normal vector

A normal vector to a plane is a vector that is perpendicular to every vector lying on that plane.

In this exercise, the normal vector is found by computing the cross product of two vectors lying on the plane.
These two vectors are:

• \( \vec{v_1} = (0, 0, 1) \)
• \( \vec{v_2} = (2, 3, 1) \)

Calculating their cross product gives us the normal vector \( \vec{n} = (-3, 2, 0) \).

This vector is perpendicular to the plane containing the given line
and directed by \( \vec{v_2} \), while also passing through the point \((1, -1, 5)\).

Normal vectors are essential for determining the equation of a plane,
as they provide the necessary orientation information.

Cross product

The cross product is a mathematical operation used to find a vector perpendicular to two given vectors.

To find a normal vector to a plane, we take the cross product of two vectors on the plane.
Here's how the calculation is executed in this context:

• The vectors are \((0, 0, 1)\) and \((2, 3, 1)\).
• The cross product formula for these vectors is:
  \[(0, 0, 1) \times (2, 3, 1) = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \0 & 0 & 1 \2 & 3 & 1 \\end{array} \right| \]
• Expanding this determinant gives \((-3, 2, 0)\).

The result is a vector that is orthogonal, or perpendicular, to both original vectors.

In geometric problems, finding a cross product is vital for determining directions of normals
which, in turn, help solve equations of planes and other spatial phenomena.