Q .11. - Calculus

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 

Step-by-Step Solution

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 1:Given information

The line is 

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

Step 2: 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.

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

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

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

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

$0=4-3 t$

$4=-3 t$

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

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

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

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

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

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

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

$0=8+7 t$

$-8=7 t$

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

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

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

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

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

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

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,

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