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i have sphere $x^2+y^2+z^2=r^2$ and a vector with points $P=(x_1,y_1,z_1)$ and $Q=(x_2,y_2,z_2)$, i need equation of tangent plane where above two points lyes and touches the sphere in one point only?

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  • $\begingroup$ What have you done so far? $\endgroup$ – almagest Jun 22 '16 at 14:49
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So we want the plane that passes both $P$ and $Q$ and touches the sphere. Let that point, where the sphere and the plane meet be $X(\alpha, \beta, \gamma)$.

Since $X$ is on the sphere, it satisfies $\alpha^2+\beta^2+\gamma^2=r^2$.

Also, the vector $\vec{OX}$ will be the normal vector to the tangent plane. The plane passes $X$ so, the equation of the plane is $$\alpha(x-\alpha)+\beta(y-\beta)+\gamma(z-\gamma)=0$$ Thus,$$\alpha x+\beta y+\gamma z=r^2$$

This plane passes through $P$ and $Q$, so plugging the values in gives us, $$\alpha x_1+\beta y_1+\gamma z_1=r^2$$ $$\alpha x_2+\beta y_2+\gamma z=r_2^2$$

Overall, solve these three equations,

$$\alpha x_1+\beta y_1+\gamma z_1=r^2$$ $$\alpha x_2+\beta y_2+\gamma z_2=r^2$$ $$\alpha^2+\beta^2+\gamma^2=r^2$$

to get the values of $\alpha, \beta, \gamma$.

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  • $\begingroup$ o thank u very much, i really needed this answer, finally i got the solution, thank to u. $\endgroup$ – Dinesh Lama Jun 24 '16 at 18:20
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$$\alpha x_1 +\beta y_1+\gamma z_1=r^2 \ (1) $$ $$\alpha x_2 +\beta y_2+\gamma z_2=r^2 \ (2) $$ $$\alpha^2+\beta^2+\gamma^2=r^2 \ (3) $$

to get the values of $\alpha, \beta, \gamma$.

From (1) we have:

$$\alpha = \frac {r^2 - \beta y_1 - \gamma z_1}{x_1} (4) $$

input $\alpha$ in (2):

$$ \frac{x_2}{x_1}(r^2 - \beta y_1 - \gamma z_1) +\beta y_2+\gamma z_2=r^2$$

thus $$ \beta (y_2 - \frac{y_1 x_2}{x_1}) - \gamma ( \frac {z_1 x_2}{x_1} - z_2) = r^2 (1 - \frac{x_2}{x_1}) $$

or $$ \beta = A \gamma + B \ (5) $$

with $$ A = \frac{\frac{z_1 x_2}{x_1} - z_2} {y_2 - \frac{y_1 x_2}{x_1}} and \ B = \frac{r^2 (1 - \frac{x_2}{x_1})}{y_2 - \frac{y_1 x_2}{x_1}} $$

Reintroducing (5) in (4):

$$ \alpha = \frac{r^2 - (A \gamma + B) y_1 - \gamma z_1}{x_1} $$

$$ \alpha = \frac{\gamma (-A y_1 - z_1) + r^2 - B y_1}{x_1} $$

$$ \alpha = C \gamma + D (6) $$ $$with C = \frac{-A y_1 - z_1}{x_1} and D = \frac{r^2 - B y_1}{x_1} $$

I put (5) and (6) in (3):

$$ (C \gamma + D)^2 + (A \gamma + B)^2 +\gamma^2=r^2 $$

$$ \gamma^2 (C^2 + A2 + 1) + \gamma (2CD +2AB) + D^2 + B^2 - r^2 = 0 $$

by solving this second order equation we have two values for $\gamma$ and we can compute two values for $\beta$ with (5) and two values for $\alpha$ with (6). Unfortunately it does not work, i mean the computed values do not verify $\alpha^2 + \beta^2 + \gamma^2 = r^2$. It means there is at least error in my resolution but I cannot find it. If someone will be kind enough to check

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