- #1
Matejxx1
- 72
- 1
Homework Statement
Let S be a sphere with the equation ##(x-2)^2+y^2+z^2=2 ## and let p a line which satisfies the condition ## p \in (\Pi \cap \Sigma) ## where ##\Pi## and ##\Sigma## are planes with equations:
##\Pi :x+z=2##
##\Sigma: 5x-2z=3##
a) Show that S and p have exactly one common point
b) find the equation of all planes that contain the point ##(0,0,0)##, are parallel to p , do not contain p and touch s at one point
Homework Equations
plane equation: ## ax+by+cz+d=0##
line equation : ##\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\theta##
distance from a point to a plane : ## d=\left | \frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}\right |## where ##x_0,y_0,z_0## are the coordinates of a point and a,b,c are the directional vectors of a plane
The Attempt at a Solution
Hi I would really appreciate if some could check my answers of this problem (especially b I think I got that one wrong). Also we just started learning Latex in our uni so I apologize for any mistakes
so I though a) was pretty easy. What I did was:
I calculated the directional vector of the line which is equal to ##\vec{p}=\vec{n}_\Pi \times \vec{n}_\Sigma=\begin{bmatrix}
i&j&k\\
1&0&1\\
5&0&-2
\end{bmatrix}=(0,7,0)=(0,1,0)
##
so now I had the directional vector I just needed a point which I got by solving ##x+z=2## and ##5x-2z=3## and got that x=1 and z=1 so :##p:\frac{x}{0}=1=\frac{z}{0}## and ##y=\theta##
now I just pluged in x=1, y=##\theta## and z=1 into the sphere equation and got only one common point ##(1,0,1)## can anybody check if this is correct?b was kinda harder and I think I got it wrong because the answers I got were not pretty
b)
I know that the plane has the point ##(0,0,0)## if we plug this into the plane equation we get ##a(0)+b(0)+c(0)+d=0## which must mean that d =0
then we also know that since the plane is parallel to the line the scalar product of the lines directional vector and the planes directional vector must be 0 therefore ##(\vec{a},\vec{b},\vec{c})\cdot(0,1,0)=0\Rightarrow b=0##
so the plane equation know looks like ##ax+cz=0##
then we also know that the since the plane touches the sphere the shortest distance from the plane to the center is the radius so ## \sqrt{2}=\left | \frac{2a}{\sqrt{a^2+c^2}}\right |## and since ##\sqrt{a^2+c^2}## is the directional vector its length is 1 that means that ##a=\pm\frac{\sqrt{2}}{2}## and since ##\sqrt{a^2+c^2}=1## that means that ##c=\pm\frac{\sqrt{2}}{2}##
so in the end I got 4 different equations:
##\begin{align}
\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}z=0\nonumber\\
-\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}z=0\nonumber\\
\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}z=0\nonumber\\
-\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}z=0\nonumber\\
\end{align}##
Which is totally weird to me since when I tried to visualize the problem I could only find 2 possible solutions at maximum . So having 4 just seems wrong to me
thanks for any help
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