Question

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A plane parallel to $x+3y-4z=7$

(b) A line orthogonal to the plane$x+3y-4z=7$

(c) A plane orthogonal to the line $x=3t-5,y=-2t+7,z=-4$

Short answer

Part (a) The plane parallel to the plane $x+3y-4z=7$is $2x+6y-8z=6$

Part (b) The equation of the line that is orthogonal to the plane $x+3y-4z=7$is $\mathbf{r}\left(t\right)=⟨3-t,1-t,-t⟩$

Part (c) The equation of the plane that is orthogonal to the line $\mathbf{r}\left(t\right)=⟨3t-5,-2t+7,-4⟩$is $2x+3y+5z=6$

Step-by-step solution

Part (a) Step 1: Given information

The plane $x+3y-4z=7$

Part (a) Step 2: Explanation

The objective is to find the plane which is parallel to the plane$x+3y-4z=7$

The normal vector of the plane $x+3y-4z=7$is $\mathbf{N}=⟨1,3,-4⟩$

Consider the plane $2x+6y-8z=6$

The normal vector of the plane $2x+6y-8z=6$is ${\mathbf{N}}_1=⟨2,6,-8⟩$

The normal vector ${\mathbf{N}}_1=⟨2,6,-8⟩$can be written as:

${\mathbf{N}}_1=2\mathbf{N}$

Therefore, the normal vector of the plane $2x+6y-8z=6$is the scalar multiple of normal vector of the plane $x+3y-4z=7$

Therefore, the plane parallel to the plane $x+3y-4z=7$is $2x+6y-8z=6$

Part (b) Step 1: Given information

The plane $x+3y-4z=7$

Part (b) Step 2: Calculation

The objective is to find the line which is orthogonal to the plane $x+3y-4z=7$The normal vector of the plane $x+3y-4z=7$is $\mathbf{N}=⟨1,3,-4⟩$

Consider the line $\mathbf{r}\left(t\right)=⟨3-t,1-t,-t⟩$

The direction vector of the line $\mathbf{r}\left(t\right)=⟨3-t,1-t,-t⟩$is $\mathbf{d}=⟨-1,-1,-1⟩$

The dot product of $\mathbf{d}=⟨-1,-1,-1⟩$and $\mathbf{N}=⟨1,3,-4⟩$is:

$$\begin{gathered} \mathbf{d}·\mathbf{N}=⟨-1,-1,-1⟩·⟨1,3,-4⟩ \\ =-1-3+4 \\ =0 \end{gathered}$$

The dot product of $\mathbf{d}=⟨-1,-1,-1⟩$ and $\mathbf{N}=⟨1,3,-4⟩$ is zero. Therefore, the line and the plane are orthogonal.

Therefore, the equation of the line that is orthogonal to the plane $x+3y-4z=7$ is $\mathbf{r}\left(t\right)=⟨3-t,1-t,-t⟩$

Part (c) Step 1: Given information

The line $\mathbf{r}\left(t\right)=⟨3t-5,-2t+7,-4⟩$

Part (c) Step 2: Calculation

The objective is to find the plane orthogonal to the line $\mathbf{r}\left(t\right)=⟨3t-5,-2t+7,-4⟩$

The direction vector of the line $\mathbf{r}\left(t\right)=⟨3t-5,-2t+7,-4⟩$is $\mathbf{d}=⟨3,-2,0⟩$

Consider the plane $2x+3y+5z=6$

The normal vector of the plane $2x+3y+5z=6$ is $\mathbf{N}=⟨2,3,5⟩$

The dot product of $\mathbf{d}=⟨3,-2,0⟩$ and $\mathbf{N}=⟨2,3,5⟩$ is:

$$\begin{gathered} \mathbf{d}·\mathbf{N}=⟨3,-2,0⟩·⟨2,3,5⟩ \\ =6-6+0 \\ =0 \end{gathered}$$

The dot product of $\mathbf{d}=⟨3,-2,0⟩$ and $\mathbf{N}=⟨2,3,5⟩$ is zero. Therefore, the line and the plane are orthogonal.

Therefore, the equation of the plane that is orthogonal to the line $\mathbf{r}\left(t\right)=⟨3t-5,-2t+7,-4⟩$ is $2x+3y+5z=6$