Problem 72 - Calculus Early Transcendentals

Question

Find an equation of the plane that passes through the point \(P_{0}\) and contains the line \(\ell\) a. \(P_{0}(1,-2,3) ; \ell: \mathbf{r}=\langle t,-t, 2 t\rangle\) b. \(P_{0}(-4,1,2) ; \ell: \mathbf{r}=\langle 2 t,-2 t,-4 t\rangle\)

Step-by-Step Solution

Answer

Based on the given information, we have found the equations of the planes as: a) The plane containing the point \(P_{0}(1, -2, 3)\) and the line with parametric equation \(\mathbf{r} =\langle t, -t, 2t\rangle\) has the equation: \(3x - 5y - z = 4\). b) The plane containing the point \(P_{0}'(-4, 1, 2)\) and the line with parametric equation \(\mathbf{r}=\langle 2t, -2t, -4t\rangle\) has the equation: \(-16y + 10z = -58\).

Step 1: Cartesian form of the line \(\ell\) #

The parametric equation of line \(\ell\) is given as \(\mathbf{r}=\langle t,-t, 2 t\rangle\). So, the line has the Cartesian form: $$ x = 1t = t \\ y = -1t = -t \\ z = 2t = 2t$$ The direction vector of the line is \(\mathbf{d} = \begin{bmatrix}1\\-1\\2\end{bmatrix}\).#tag_end# ##Step 2: Find two non-parallel points on the line##

Step 2: Points \(A\) and \(B\) on the line #

Let's choose \(t=0\) and \(t=1\) to find two distinct points on the line: $$ A = (x_A, y_A, z_A) = (0, 0, 0) \\ B = (x_B, y_B, z_B) = (1, -1, 2)$$#tag_end# ##Step 3: Calculate two vectors using points \(A, B\) and \(P_{0}\)##

Step 3: Vectors \(\mathbf{AP_{0}}\) and \(\mathbf{BP_{0}}\) #

Now, we determine the vectors from points \(A\) and \(B\) to \(P_{0}(1,-2,3)\): $$ \mathbf{AP_{0}} = \overrightarrow{AAP_{0}} = \begin{bmatrix}1-0\\-2-0\\3-0\end{bmatrix} = \begin{bmatrix}1\\-2\\3\end{bmatrix} \\ \mathbf{BP_{0}} = \overrightarrow{BBP_{0}} = \begin{bmatrix}1-1\\-2-(-1)\\3-2\end{bmatrix} = \begin{bmatrix}0\\1\\1\end{bmatrix}$$#tag_end# ##Step 4: Compute the cross product of vectors \(\mathbf{AP_{0}}\) and \(\mathbf{BP_{0}}\)##

Step 4: Normal vector \(\mathbf{n}\) #

To find the normal vector \(\mathbf{n}\) of the plane, we compute the cross product of the vectors \(\mathbf{AP_{0}}\) and \(\mathbf{BP_{0}}\): $$\mathbf{n}=\mathbf{AP_{0}}\times\mathbf{BP_{0}}=\begin{bmatrix}1 \times (1) - (-2) \times (1) \\ -2 \times (1) - 3 \times (1) \\ 3 \times (0) - 1 \times (1)\end{bmatrix}=\begin{bmatrix}3\\-5\\-1\end{bmatrix}$$#tag_end# ##Step 5: Use the point-normal form of the plane equation##

Step 5: Plane equation #

Our point-normal equation for the plane is: \(A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0\), where \(\mathbf{n}=\begin{bmatrix}A\\B\\C\end{bmatrix}\) and \(P_{0}(x_{0},y_{0},z_{0})\). Plugging in the values, we get: $$3(x-1) - 5(y+2) - (z-3) = 0$$ Which can be written as: $$3x - 5y - z = 4$$ Hence, the equation of the plane is \(3x - 5y - z = 4\). #-} #b# ##Step 1: Write the line in Cartesian form##

Step 6: Cartesian form of the line \(\ell'\) #

The parametric equation of line \(\ell'\) is given as \(\mathbf{r}=\langle 2t,-2t, -4t\rangle\). So, the line has the Cartesian form: $$ x = 2t = 2t \\ y = -2t = -2t \\ z = -4t = -4t$$ The direction vector of the line is \(\mathbf{d'} = \begin{bmatrix}2\\-2\\-4\end{bmatrix}\).#tag_end# ##Step 2: Find two non-parallel points on the line##

Step 7: Points \(A'\) and \(B'\) on the line #

Let's choose \(t=0\) and \(t=1\) to find two distinct points on the line: $$ A' = (x_{A'}, y_{A'}, z_{A'}) = (0, 0, 0) \\ B' = (x_{B'}, y_{B'}, z_{B'}) = (2, -2, -4)$$#tag_end# ##Step 3: Calculate two vectors using points \(A'\), \(B'\) and \(P_{0}'\)##

Step 8: Vectors \(\mathbf{A'P_{0}'}\) and \(\mathbf{B'P_{0}'}\) #

Now, we determine the vectors from points \(A'\) and \(B'\) to \(P_{0}'(-4,1,2)\): $$ \mathbf{A'P_{0}'} = \overrightarrow{A'A'P_{0}'} = \begin{bmatrix}-4-0\\1-0\\2-0\end{bmatrix} = \begin{bmatrix}-4\\1\\2\end{bmatrix} \\ \mathbf{B'P_{0}'} = \overrightarrow{B'B'P_{0}'} = \begin{bmatrix}-4-2\\1-(-2)\\2-(-4)\end{bmatrix} = \begin{bmatrix}-6\\3\\6\end{bmatrix}$$#tag_end# ##Step 4: Compute the cross product of vectors \(\mathbf{A'P_{0}'}\) and \(\mathbf{B'P_{0}'}\)##

Step 9: Normal vector \(\mathbf{n'}\) #

To find the normal vector \(\mathbf{n'}\) of the plane, we compute the cross product of the vectors \(\mathbf{A'P_{0}'}\) and \(\mathbf{B'P_{0}'}\): $$\mathbf{n'}=\mathbf{A'P_{0}'}\times\mathbf{B'P_{0}'}=\begin{bmatrix}1 \times (6) - 2 \times (3) \\ 2 \times (-6) - (-4) \times (6) \\ (-4) \times (3) - 1 \times (-6)\end{bmatrix}=\begin{bmatrix}0\\-16\\10\end{bmatrix}$$#tag_end# ##Step 5: Use the point-normal form of the plane equation##

Step 10: Plane equation #

Our point-normal equation for the plane is: \(A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0\), where \(\mathbf{n'}=\begin{bmatrix}A\\B\\C\end{bmatrix}\) and \(P_{0}'(x_{0},y_{0},z_{0})\). Plugging in the values, we get: $$0(x+4) - 16(y-1) + 10(z-2) = 0$$ Which can be written as: $$-16y + 10z = -58$$ Hence, the equation of the plane is \(-16y + 10z = -58\).

Key Concepts

Parametric Equation of a Line

To understand how a line is represented in three-dimensional space, we use the parametric equation. This allows us to express the coordinates of every point on the line as a function of a single parameter, often denoted by \( t \). For example, if we have a line \( \ell \) represented by \( \mathbf{r} = \langle t, -t, 2t \rangle \), each component of the vector is scaled by \( t \):

• \( x = t \)
• \( y = -t \)
• \( z = 2t \)

This form is useful because it clearly shows the direction in which the line extends in space. The coefficients of \( t \) define the direction vector \( \mathbf{d} = \begin{bmatrix}1\-1\2\end{bmatrix} \). The parametric equation highlights how the line's position changes as \( t \) varies.

Cross Product of Vectors

The cross product is an operation on two vectors in three-dimensional space that results in another vector that is perpendicular to the plane containing the original vectors. This operation is crucial when finding the normal vector to a plane. Given two vectors \( \mathbf{A} = \begin{bmatrix} x_1 \ y_1 \ z_1 \end{bmatrix} \) and \( \mathbf{B} = \begin{bmatrix} x_2 \ y_2 \ z_2 \end{bmatrix} \), the cross product \( \mathbf{A} \times \mathbf{B} \) is calculated as follows: \[ \mathbf{A} \times \mathbf{B} = \begin{bmatrix} (y_1 z_2 - z_1 y_2) \ (z_1 x_2 - x_1 z_2) \ (x_1 y_2 - y_1 x_2) \end{bmatrix} \] This new vector is vital in defining the plane because it acts as a normal vector that is orthogonal to the direction vectors derived from the line and the point it intersects.

Point-Normal Form

The point-normal form of a plane is an equation used to define a plane using a given point on the plane and a normal vector to the plane. This form helps us understand how the plane is oriented in three-dimensional space. The equation is generally expressed as: \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \] where \( (x_0, y_0, z_0) \) is a point on the plane, and \( \mathbf{n} = \begin{bmatrix} A \ B \ C \end{bmatrix} \) is the normal vector. Each \( A, B, \) and \( C \) defines the plane's interaction with the x, y, and z axes, respectively. This point-normal form is pivotal when solving for an equation of the plane since it ties together the spatial geometry with algebraic expressions.

Vector Arithmetic

Vector arithmetic involves operations such as addition and subtraction of vectors, scaling them by a scalar, and more complex operations like cross and dot products. These operations are the foundational blocks of vector analysis, making it crucial for solving geometric problems. When finding the equation of a plane, vector arithmetic enables us to:

• Determine vectors between points, like \( \mathbf{AP_0} \) and \( \mathbf{BP_0} \).
• Calculate the cross product of these vectors to find a normal vector.
• Employ the normal vector in deriving the plane's equation using point-normal form.

For instance, if vectors \( \mathbf{AP_0} = \begin{bmatrix} 1 \ -2 \ 3 \end{bmatrix} \) and \( \mathbf{BP_0} = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \) are given, vector arithmetic enables us to compute their cross product, yielding a normal vector crucial for the plane's equation.