Q 42. - 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. 

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

Step-by-Step Solution

Answer

The answer is skew lines.

Step 1: Given information

The two lines,

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

Step 2: Calculation

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

To determine whether two lines are parallel, 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 $r_1\left(t\right)=(3-t,7+t,4+5t)$ the direction vector is, $d_1=(-1,1,5)$ where $d_1$ is the direction vector

For the line equation $\left(2\right)$ that is for $r_2\left(t\right)=(3-t,4+3t,5-2t)$ the direction vector is $d_2=(-1,3,-2)$ where data-custom-editor="chemistry" $d_2$ 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 $r_1\left(t\right)=(3-t,7+t,4+5t)$ and $r_2\left(u\right)=(3-t,4+3t,5-2t)$ are not parallel lines. Now calculate the point of intersection of two lines.

To calculate the intersection point of two equations, replace the parameter $t$ by $u$ in the equation $\left(2\right)$ that is $r_2\left(t\right)=(3-t,4+3t,5-2t)$

Then,

$r_2\left(u\right)=\left(3-u_{1}4+3u_{5}5-2u\right)$

To calculate the intersection point of two equations, equate the values of equations.

That is,

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

Step 3: Calculation

Take the equation (3) that is $3-t=3-u$ - Add $-3\mathrm{cn}$ both sides of the equation.

$$\begin{gathered} 3-t-3=3-u-3 \\ -l=-u \\ t=u \end{gathered}$$

Substitute $t=u$ in the equation (4) that is $7+t=4+3 u$ -

Then,

$7+u=4+3 u$

Add $-2 r,-4$ on both sides of the equation.

$$\begin{gathered} 7+u-u-4=4+3u-u-4 \\ 7-4=3n-n \end{gathered}$$

Thus,

$u=\frac{3}{2}$

Substitute $u=\frac{3}{2}$ in the equation $4+5 t=5-2 u t$

$$\begin{gathered} 4+5·\frac{3}{2}=5-2·\frac{3}{2} \\ \frac{23}{2}=2 \end{gathered}$$

The answer does not satisfy the equation. As a result, the lines do not cross. 

Therefore, the answer is skew lines.