Q 43. - Calculus

Question

determine whether the given pairs of lines are parallel, identical, intersecting, or skew. If the lines are parallel, compute the distance between them. If the lines intersect, find the point of intersection and the angle at which the

lines intersect. 

$$\begin{gathered} r_1\left(t\right)=(5t+2,-4t,t-7) \\ {r}_2\left(t\right)=(-3t+4,-t+12,-2t-1) \end{gathered}$$

Step-by-Step Solution

Answer

The answer is Lines are intersecting at $(-8,8,-9)$

Step 1: Given information

Consider the two lines,

$$\begin{gathered} r_1\left(t\right)=(5t+2,-4t,t-7) \dots \dots \left(1\right) \\ {r}_2\left(t\right)=(-3t+4,-t+12,-2t-1) \dots \dots \left(2\right) \end{gathered}$$

Step 2: Calculation

It's your job to figure out whether the lines are parallel or intersecting.

To determine whether two lines are parallel, we must first determine the direction vectors for both equations. The lines are parallel if one equation's direction vector is a scalar multiple of another equation's direction vector.

For the line equation $\left(1\right)$ that is for $r_1\left(t\right)=(5t+2,-4t,t-7)$ the direction vector is, $d_1=(5,-4,1)$ where $d_1$ is the direction vector

For the line equation $\left(2\right)$ that is for $r_2\left(t\right)=(-3t+4,-t+12,-2t-1)$ the direction vector is $d_2=(-3,-1,-2)$ where $d_2$ is the direction vector

Here the direction vectors $d_1,d_2$ are not scalar multiples of each other. Thus, the lines are not parallel.

Thus, the lines $r_1\left(t\right)=(5t+2,-4t,t-7)$ and $r_2\left(t\right)=(-3t+4,-t+12,-2t-1)$ are not parallel lines.

Calculate the junction location of two lines now.

To calculate the intersection point of two equations, replace the parameter $t$ by $u$ in the equation $\left(2\right)$ that is $r_2\left(t\right)=(-3t+4,-t+12,-2t-1)$

Then,

$r_2\left(u\right)=(-3u+4,-u+12,-2u-1)$

Equivalent the values of the equations to find the intersection point of two equations.

That is,

$$\begin{gathered} r_1\left(t\right)=(5t+2,-4t,t-7)\text{ and }{r}_2\left(u\right)=(-3u+4,-u+12,-2u-1) \\ 5t+2=-3u+4\dots \dots .\left(3\right) \\ -4t=-u+12\dots \dots \left(4\right) \\ t-7=-2u-1 \dots \dots .\left(5\right) \end{gathered}$$

Step 3: Calculation

Take the equation $\left(4\right)$ that is $-4 t=-u+12$

Divide by $-4$ on both sides of the equation.

$$\begin{gathered} \frac{-4t}{-4}=\frac{-u+12}{-4} \\ t=\frac{-u+12}{-4} \end{gathered}$$

Substitute $t=\frac{-u+12}{-4}$ in the equation $\left(3\right)$ that is $5 t+2=-3 u+4$

Then,

$$\begin{gathered} 5·\frac{-u+12}{-4}+2=-3u+4 \\ \frac{-5u+60}{-4}+2=-3u+4 \\ \frac{-5u+60-8}{-4}=-3u+4 \\ -5u+52=12u-16 \end{gathered}$$

Add $5 u, 16$ on both sides of the equation.

$$\begin{gathered} -5u+52+5u+16=12u-16+5u+16 \\ 52+16=12u+5u \\ 68=17u \end{gathered}$$

Thus,

$u=4$

Substitute $u=4$ in the equation $\left(4\right)$ that is $-4 t=-u+12$

$$\begin{gathered} -4t=-4+12 \\ -4t=8 \\ t=-2 \end{gathered}$$

Step 4: Calculation

Substitute $t=-2$ in the equation $r_1\left(t\right)=(5t+2,-4t,t-7)$

$$\begin{gathered} r_1(-2)=(5·-2+2,-4·-2,-2-7) \\ {r}_1(-2)=(-10+2,8,-9) \\ {r}_1(-2)=(-8,8,-9) \end{gathered}$$

Substitute $u=4$ in the equation $r_2\left(u\right)=(-3u+4,-u+12,-2u-1)$

$$\begin{gathered} r_2\left(4\right)=(-3·4+4,-4+12,-2·4-1) \\ {r}_2\left(4\right)=(-12+4,-4+12,-8-1) \\ {r}_2\left(4\right)=(-8,8,-9) \end{gathered}$$

Thus the intersecting point of the lines $r_1\left(t\right)=(5t+2,-4t,t-7)$ and $r_2\left(t\right)=(-3t+4,-t+12,-2t-1)$ is equals to $(-8,8,-9)$

Therefore, the answer is Lines are intersecting at $(-8,8,-9)$