2 TB - Calculus

Question

Finding the equation of the plane containing two intersecting lines: Show that the lines given by \begin{equation}\mathbf{r}1(t)=(3t-4,-4t+1,t) \end{equation} \begin{equation}\mathbf{r}2(t)=(-t+2,2t-9,-2t+7)\end{equation}

intersect, and find the equation of the plane containing the lines. 

Step-by-Step Solution

Answer

The lines will intersect and The equation of the plane containing the lines is

\begin{equation}x + 8y - 5z + 4 = 0 \end{equation}

Step 1. Given Information

The given lines are

\begin{equation}\mathbf{r}1(t)=(3t-4,-4t+1,t) \end{equation} \begin{equation}\mathbf{r}2(t)=(-t+2,2t-9,-2t+7)\end{equation} 

Step 2. Proving the lines intersect

To prove that the lines intersect, we can find the point of intersection by setting the parametric equations equal to each other and solving for t: 

\begin{equation}r1(t)=r2(t)\end{equation}

\begin{equation}(3t-4,-4t+1,t)=(-t+2,2t-9,-2t+7)\end{equation}

Solving the above equation we get t=1.

Substituting t = 1 into either of the parametric equations, we get the point of intersection: 

\begin{equation}r1(1)=(3(1)-4,-4(1)+1,1)=(-1,-3,1)\end{equation}

\begin{equation}r2(1)=(-1(1)+2,2(1)-9,-2(1)+7)=(-1,-3,1)\end{equation}

These points are equal, so the lines intersect at the point (-1, -3, 1). 

Therefore, we have proved that the lines given by r1(t) = (3t - 4, -4t + 1, t) and r2(t) = (-t + 2, 2t - 9, -2t + 7) intersect at the point (-1, -3, 1). 

Step 3. Finding the equation of plane

To find the equation of the plane containing two intersecting lines, first we should find the direction vectors of the two lines. 

The direction vector of a line can be found by taking the derivative of the parametric equations of the line with respect to the parameter t. 

For the first line, the parametric equations are r1(t) = (3t - 4, -4t + 1, t). Taking the derivative with respect to t, we get: 

\begin{equation}r1{}'(t)=(3,-4,1)\end{equation}

So, the direction vector of the first line is (3, -4, 1). 

For the second line, the parametric equations are r2(t) = (-t + 2, 2t - 9, -2t + 7). Taking the derivative with respect to t, we get: 

\begin{equation}r2{}'(t)=(1,2,-2)\end{equation}

So, the direction vector of the second line is (-1, 2, -2). 

Since these two lines intersect, they must be parallel. This means that their direction vectors are proportional. We can check this by taking the ratio of the two direction vectors: 

\begin{equation}(3, -4, 1) / (-1, 2, -2) = -3/1 = -3 \end{equation}

This means that the direction vectors are proportional, so the lines are parallel.

To find the equation of the plane containing the lines, we need a point on each line and the direction vectors of the lines.

We can choose t = 0 for each line to get a point on each line:

\begin{equation}r1(0)=(3(0)-4,-4(0)+1,0)=(-4,1,0)\end{equation}

\begin{equation}r2(0)=(-0+2,2(0)-9,-2(0)+7)=(2,-9,7)\end{equation}

We can use the point-normal form of a plane to find the equation of the plane containing the lines: 

\begin{equation}ax + by + cz + d = 0 \end{equation}

Where (a, b, c) is the normal vector to the plane and d is a constant. 

To find the normal vector, we can take the cross product of the direction vectors of the lines:

\begin{equation}(a, b, c) = (3, -4, 1)\times(-1, 2, -2)\end{equation}

This gives us:

\begin{equation}(a,b,c)=(1,8,-5)\end{equation}

To find d, we can substitute one of the points and the normal vector into the point-normal form: 

\begin{equation}a(-4) + b(1) + c(0) + d = 0 \end{equation}

Substituting (a, b, c) = (1, 8, -5), we get: 

\begin{equation}(-4) + (8) + (0) + d = 0 \end{equation}

Solving for d, we get d=4

So, the equation of the plane containing the lines is: 

\begin{equation}x+8y-5z+4=0\end{equation}

This is the equation of the plane containing the two intersecting lines given by r1(t) = (3t - 4, -4t + 1, t) and r2(t) = (-t + 2, 2t - 9, -2t + 7).