How to find coordinates of intersection point in xy-plane in this question?

[Win_odd Dhamnekar]

Hello,

Let S be the sphere with radius 1 centered at (0,0,1),and let $S^*$ be S without the “north pole” point (0,0,2). Let(a,b,c) be an arbitrary point on $S^*$. Then the line passing through (0,0,2) and (a,b,c) intersects the xy-plane at some point (x,y,0), as in following figure . Now, how to find this point (x,y,0) in terms of a,b and c?

(Note: Every point in the xy-plane can be matched with a point on $S^*$, and vice versa, in this manner. This method is called stereographic projection, which essentially identifies all of $\mathbb{R^2}$ with a "punctured" sphere.




My attempt:-
Z coordinate of the intersection point is obviously zero. How to find X and Y coordinates of the intersection point. I assume $S^*$ is any point on/ inside sphere S.

 

[lev888]

My attempt:-
Z coordinate of the intersection point is obviously zero.

Sorry, I wouldn't call this an attempt.
1. Construct the equation of the line based on the 2 points.
2. Given equations of a line and a plane it should be easy to find the coordinates of their intersection.

 

[Win_odd Dhamnekar]

Let $P_1=(0,0,2),P_2=(a,b,c)$ be distinct points in $\mathbb{R^3}$ and let $r_1=(0,0,2),r_2=(a,b,c)$. Then the line L through $P_1$ and $P_2$ has the following vector form equation.

$(0,0,2)+t*(a,b,c-2),$ for $-\infty < t < \infty$

So, parametric form of equation of line L
x=at, y=bt, z=2+t*(c-2) where $t= \frac{2}{2-c}$ since equation of xy-plane is z=0.

So, the point (x,y,0) in terms of a,b,c is $(\frac{2a}{2-c},\frac{2b}{2-c},0)$.

 

[Steven G]

The point (a, b, c) is NOT inside the sphere simply because the problem stated that (a, b, c) is on S*

 

[Win_odd Dhamnekar]

The point (a, b, c) is NOT inside the sphere simply because the problem stated that (a, b, c) is on S*

So, what is the meaning of " Let $S^*$ be S without a "north pole" point (0,0,2)" . Would you explain?

 

[Steven G]

Let $P_1=(0,0,2),P_2=(a,b,c)$ be distinct points in $\mathbb{R^3}$ and let $r_1=(0,0,2),r_2=(a,b,c)$. Then the line L through $P_1$ and $P_2$ has the following vector form equation.

$(0,0,2)+t*(a,b,c-2),$ for $-\infty < t < \infty$

So, parametric form of equation of line L
x=at, y=bt, z=2+t*(c-2) where $t= \frac{2}{2-c}$ since equation of xy-plane is z=0.

So, the point (x,y,0) in terms of a,b,c is $(\frac{2a}{2-c},\frac{2b}{2-c},0)$.

No, (a, b, c) are not distinct points in R^3! (a, b, c) is ONE point in R^3. Is (3,5) two points in R^2, that is in the x-y plane. If yes, then please plot them.

(a, b, c) is NOT a SINGLE point in R^3. OK, it is! But it is not just any point in R^3. It lies on the sphere! What is the equation of the sphere? Because (a, b, c) better be on that sphere! You even said (incorrectly) that (a, b, c) is on the sphere or in the sphere, but then you ran away from that. At best you found that given any point (a, b, c) in R^3, that the line from this point through (0, 0, 2) crosses the the x-y plane at the point you stated.

 

[Steven G]

So, what is the meaning of " Let $S^*$ be S without a "north pole" point (0,0,2)" . Would you explain?

S* is ALL the points on the sphere except the point (0, 0, 2).

 

[lev888]

So, what is the meaning of " Let $S^*$ be S without a "north pole" point (0,0,2)" . Would you explain?

We have to exclude it because the following would not make sense: "Then the line passing through (0,0,2) and (a,b,c) intersects the xy-plane at some point (x,y,0)"