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.

${\mathbf{r}}_1\left(t\right)=⟨4+5t,6,7-2t⟩$

${\mathbf{r}}_2\left(t\right)=⟨6-4t,-3+3t,-1+4t⟩$

Short answer

The lines are not parallel andthe intersecting point of the lines is$(-6,6,11)$

Step-by-step solution

Step 1:Given information

The given pairs of lines are

${\mathbf{r}}_1\left(t\right)=⟨4+5t,6,7-2t⟩$

${\mathbf{r}}_2\left(t\right)=⟨6-4t,-3+3t,-1+4t⟩$

Step 2:Calculation

Consider the two lines,

$r_1\left(t\right)=(4+5t,6,7-2t)$....eq(1)

$r_2\left(t\right)=(6-4t,-3+3t,-1+4t)$....eq(2)

For the line equation

(1) that is for $r_1\left(t\right)=(4+5t,6,7-2t)$the direction vector is,

$d_1=(5,0,-2)\text{where}{d}_1\text{is the direction vector}$

For the line equation

$\left(2\right)$that is for $r_2\left(t\right)=(6-4t,-3+3t,-1+4t)$the direction vector is

$d_2=(-4,3,4)\text{. where}{d}_2\text{is the direction vector}$

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

Thus, the lines are not parallel.

Thus, the lines $r_1\left(t\right)=(4+5t,6,7-2t)$and $r_2\left(t\right)=(6-4t,-3+3t,-1+4t)$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 $r_2\left(t\right)=(6-4t,-3+3t,-1+4t)$.

Then,

$r_2\left(u\right)=(6-4u,-3+3u,-1+4u)$

Equate the values of equations,

$r_1\left(t\right)=(4+5t,6,7-2t)\text{and}{r}_2\left(u\right)=(6-4u,-3+3u,-1+4u)$

$4+5t=6-4u$...eq(3)

$6=-3+3u$...eq(4)

$7-2t=-1+4u$...eq(5)

$Taketheequation\left(4\right)thatis6=-3+3u.$

$6+3=3u$

$\Rightarrow u=3$

Substitute $u=3$in the equation

(3) that is$4+5t=6-4u.$

$4+5t=6-4·3$

$4+5t=-6$

$\Rightarrow t=-2$

$\text{Substitute}t=-2\text{in}{r}_1\left(t\right)=(4+5t,6,7-2t)\text{.}$

$r_1(-2)=(4+5·-2,6,7-2·-2)$

$r_1(-2)=(-6,6,11)$

$\text{Now substitute}u=3\text{in the equation}{r}_2\left(u\right)=(6-4u,-3+3u,-1+4u)\text{.}$

$r_2\left(3\right)=(6-4·3,-3+3·3,-1+4·3)$

$r_2\left(3\right)=(-6,6,11)$

$\text{The intersecting point of the lines is}(-6,6,11)\text{.}$