Question

Find the points where the line

$\mathbf{r}\left(t\right)=⟨4-3t,8+7t,5+t⟩,-\infty <t<\infty ,$

intersects each of the coordinate planes

Short answer

$\text{The required answer is}\left(0,\frac{52}{3},\frac{19}{3}\right)\left(\frac{52}{7},0,\frac{27}{7}\right)(19,-27,0)$

Step-by-step solution

Step 1:Given information

The line is

$\mathbf{r}\left(t\right)=⟨4-3t,8+7t,5+t⟩,-\infty <t<\infty$

Calculation

$\text{Consider the line}\mathcal{L}\text{determined by the equations}r\left(t\right)=(4-3t,8+7t,5+t),\infty \le t\le \infty \text{.}$

The objectlve is to find the points where the line intersects each of the coordinate planes.

Take the line$ℒ$equation.

$\text{Then}r\left(t\right)=(4-3t,8+7t,5+t)$

$\text{So,}x=4-3t,y=8+7t,z=5+t$

$\text{When the line intersect in}yz\text{-plane the value of}x=0\text{.}$

$\text{Now substitute}x=0\text{in}x=4-3t\text{.}$

$0=4-3t$

$4=-3t$

$\Rightarrow t=\frac{4}{3}$

$\text{Substitute}t=\frac{4}{3}\text{in}r\left(t\right)=(4-3t,8+7t,5+t)\text{.}$

$r\left(\frac{4}{3}\right)=\left(0,8+7·\frac{4}{3},5+\frac{4}{3}\right)$

$r\left(\frac{4}{3}\right)=\left(0,\frac{52}{3},\frac{19}{3}\right)$

$\text{Thus the point in}yz\text{-plane is}\left(0,\frac{52}{3},\frac{19}{3}\right)\text{.}$

$\text{When the line intersect in}zx\text{-plane the value of}y=0\text{.}$

$\text{Now substitute}y=0\text{in}y=8+7t\text{.}$

$0=8+7t$

$-8=7t$

$\Rightarrow t=\frac{-8}{7}$

$\text{Substitute}t=\frac{-8}{7}\text{in}r\left(t\right)=(4-3t,8+7t,5+t)\text{.}$

$r\left(\frac{-8}{7}\right)=\left(4-3·\frac{-8}{7},0,5+\frac{-8}{7}\right)$

$r\left(\frac{-8}{7}\right)=\left(4+\frac{24}{7},0,5-\frac{8}{7}\right)$

$r\left(\frac{-8}{7}\right)=\left(\frac{28+24}{7},0,\frac{35-8}{7}\right)$

$\text{Thus the point in}zx\text{-plane is}\left(\frac{52}{7},0,\frac{27}{7}\right)\text{.}$

When the line intersects in xy-plane the value of$z=0.$

Now substitute $z=0$in $z=5+t.$

$0=5+t$

$\Rightarrow t=-5$

$\text{Substitute}t=-5\text{in}r\left(t\right)=(4-3t,8+7t,5+t)\text{.}$

$r(-5)=(4-3·-5,8+7·-5,0)$

$r(-5)=(4+15,8-35,0)$

$r(-5)=(19,-27,0)$

$\text{Thus the point in}xy\text{-plane is}(19,-27,0)\text{.}$

Thus the points where the line intersects the coordinate planes $yz,zx,xy$are,

$\left(0,\frac{52}{3},\frac{19}{3}\right)\left(\frac{52}{7},0,\frac{27}{7}\right)(19,-27,0)$