Question

Find the points where the line $x=-3t+5,y=4,z=2t+11$intersects the $xy$-plane and $yz$-plane. Explain why the line does not intersect the $xz-$plane.

Short answer

$\text{The point in}xy\text{-plane is}\left(0,4,\frac{43}{3}\right)\text{.}$

Step-by-step solution

Step 1:Given information

$\text{The equations}x=-3t+5,y=4,z=2t+11\text{.}$

Step 2:Calculation

Consider the equations $x=-3t+5,y=4,z=2t+11$

Take the line $$\mathcal{L}$$ equation.

$x=-3t+5,y=4,z=2t+11$

Then, $r\left(t\right)=(-3t+5,4,2t+11)$

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

Now substitute $z=0inz=2t+11.$

$0=2t+11$

$0-11=2t+11-11$

$0-11=2t+11-11$

$-11=2t$

Divide by 2 on both sides of the equation.

$\frac{-11}{2}=\frac{2t}{2}$

$\frac{-11}{2}=t$

$t=\frac{-11}{2}$simplifed

Substitute $$t=\frac{-11}{2}$$ in $$r\left(t\right)=(-3t+5,4,2t+11)$$.

$r\left(-\frac{11}{2}\right)=\left(-3·\frac{-11}{2}+5,4,0\right)$

$r\left(-\frac{11}{2}\right)=\left(\frac{33}{2}+5,4,0\right)$

$r\left(-\frac{11}{2}\right)=\left(\frac{43}{2},4,0\right)$

Thus the point in xy-plane is $\left(\frac{43}{2},4,0\right)$.

When the line intersects in yz-plane the value of $x=0.$

Now substitute$x=0$in$x=-3t+5.$

$0=-3t+5$

$-5=-3t$

$\frac{5}{3}=t$

$\Rightarrow t=\frac{5}{3}$

Substituting $t=\frac{5}{3}\text{in}r\left(t\right)=(-3t+5,4,2t+11)\text{.}$

$r\left(\frac{5}{3}\right)=\left(0,4,2·\frac{5}{3}+11\right)$

$r\left(\frac{5}{3}\right)=\left(0,4,\frac{43}{3}\right)$

$\text{Thus, the point in}xy\text{-plane is}\left(0,4,\frac{43}{3}\right)\text{.}$

The line does not intersect the xz plane as the line is parallel to x-axis. Thus the points where the line intersects the coordinate planes xy, yz are,

$\left(\frac{43}{2},4,0\right)\left(0,4,\frac{43}{3}\right)$