# 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.

Question
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.

2020-10-28
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,
$$\displaystyle{z}={0}\Rightarrow-{3}+{t}={0}$$
t=3, now x=3+2t
At t=3,
x=3+2*3=3+6=9
At t=3, y=5+7*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,
$$\displaystyle{x}={0}\Rightarrow{3}+{2}{t}={0}$$
2t=-3. now
At $$\displaystyle{t}=-\frac{{3}}{{2}}$$
$$\displaystyle{y}={5}+{7}\cdot{\left(-\frac{{3}}{{2}}\right)}=\frac{{{10}-{21}}}{{2}}=-\frac{{11}}{{2}}$$
At $$\displaystyle{t}=-\frac{{3}}{{2}}$$
$$\displaystyle{z}=-{3}-\frac{{3}}{{2}}=\frac{{-{6}-{3}}}{{2}}=-\frac{{9}}{{2}}$$
Point is $$\displaystyle{\left({0},-\frac{{11}}{{2}},-\frac{{9}}{{2}}\right)}$$
Consider in zx plane y=0.
Substitute 0 for y in the equation of line as given below,
$$\displaystyle{y}={0}\Rightarrow{5}+{7}{t}={0}$$
7t=-5, now
At $$\displaystyle{t}=-\frac{{5}}{{7}}$$
$$\displaystyle{x}={3}+{2}\cdot{\left(-\frac{{5}}{{7}}\right)}=\frac{{{21}-{10}}}{{7}}=\frac{{11}}{{7}}$$
At t=-5/7ZSK
$$\displaystyle{z}=-{3}-\frac{{5}}{{7}}=-{21}-\frac{{5}}{{7}}=-\frac{{26}}{{7}}$$
Point is $$\displaystyle{\left(\frac{{11}}{{7}},{0},-\frac{{26}}{{7}}\right)}$$
Thus, the point of intersections on the coordinate planes are as given below:
xy:(9,26,0)
yz:$$\displaystyle{\left({0},\frac{{11}}{{2}},\frac{{9}}{{2}}\right)}$$
zx:$$\displaystyle{\left(\frac{{11}}{{7}},{0},-\frac{{26}}{{7}}\right)}$$

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