# Find the points of intersection of the line x = 3+2t, y =5+7t, z = -3+t, that is, I(t) = 3+2t, 5+7t, -3+t, with the coordinate planes.

Find the points of intersection of the line $x=3+2t,y=5+7t,z=-3+t$, that is, $I\left(t\right)=3+2t,5+7t,-3+t$, with the coordinate planes.

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yunitsiL

The equations of the coordinate planes is given below:
$yz:x=0,zx:y=0,xy:z=0$
Consider in xy plane z=0.
Substitute 0 for z in the equation of line as given below,
$z=0⇒-3+t=0$
t=3, now $x=3+2t$
At $t=3$,
$x=3+2\ast 3=3+6=9$
At t=3, $y=5+7\ast 3=5+21=26$
Point is (9,26,0)
Consider in yz plane x=0.
Substitute 0 for x in the equation of line as given below,
$x=0⇒3+2t=0$
2t=-3. now
At $t=-\frac{3}{2}$
$y=5+7\cdot \left(-\frac{3}{2}\right)=\frac{10-21}{2}=-\frac{11}{2}$
At $t=-\frac{3}{2}$
$z=-3-\frac{3}{2}=\frac{-6-3}{2}=-\frac{9}{2}$
Point is $\left(0,-\frac{11}{2},-\frac{9}{2}\right)$
Consider in zx plane y=0.
Substitute 0 for y in the equation of line as given below,
$y=0⇒5+7t=0$
7t=-5, now
At $t=-\frac{5}{7}$
$x=3+2\cdot \left(-\frac{5}{7}\right)=\frac{21-10}{7}=\frac{11}{7}$
At $t=-\frac{5}{7}$
$z=-3-\frac{5}{7}=-21-\frac{5}{7}=-\frac{26}{7}$
Point is $\left(\frac{11}{7},0,-\frac{26}{7}\right)$
Thus, the point of intersections on the coordinate planes are as given below:
xy:(9,26,0)
yz:$\left(0,\frac{11}{2},\frac{9}{2}\right)$
zx:$\left(\frac{11}{7},0,-\frac{26}{7}\right)$