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May 16th, 2017, 01:14 AM   #1
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Shortest Distance

The shortest distance from the sphere $x^2+y^2+z^2-2x-4y-6z+11=0$ to the plane $x+y+z=3$ is equal to
A) $\displaystyle \sqrt 3 $
B) 2 $\displaystyle \sqrt 3 $
C) 3 $\displaystyle \sqrt 3 $
D) 4 $\displaystyle \sqrt 3 $

After making the sphere in standard form, I got the point (1,2,3) and then I used the formula $ax+ by+ cz-d$ / $\displaystyle \sqrt (a^2 + b^2 + c^2) $ to get the distance and the final answer is $\displaystyle \sqrt 3 $

Let me know If I'm wrong
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May 17th, 2017, 03:28 AM   #2
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What do you mean by "I got the point (1,2,3)"? Do you mean that is the center of the sphere? If so, then, since the shortest distance from a point to a plane is always along a line perpendicular to the plane, first find the equation of the line with direction the direction of the normal vector, <1, 1, 1>, passing through point (1, 2, 3). Then determine where that line intersects both sphere and plane and find the distance between those points.
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May 17th, 2017, 03:49 AM   #3
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[QUOTE=Lalitha183;570237]The shortest distance from the sphere $x^2+y^2+z^2-2x-4y-6z+11=0$ to the plane $x+y+z=3$ is equal to
A) $\displaystyle \sqrt 3 $
B) 2 $\displaystyle \sqrt 3 $
C) 3 $\displaystyle \sqrt 3 $
D) 4 $\displaystyle \sqrt 3 $

After making the sphere in standard form, I got the values of $ (h,k,l) = (1,2,3) $ and I took $a,b,c,d$ values from the plane then I used the formula$\displaystyle \frac{ ax+ by+ cz-d } {\sqrt {a^2 + b^2 + c^2} }$ to get the distance and the final answer is $\displaystyle \sqrt 3 $

Please let me know the steps if it is wrong!!
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May 17th, 2017, 05:46 AM   #4
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Your sphere has equation
$$(x-1)^2+(y-2)^2+(z-3)^2-3=0$$
That is, the radius of the sphere is $\sqrt{3}$ with center (1,2,3).

Now generally, let $\pi$ be any plane and S any sphere with center C and radius r. Also let d be the distance from C to $\pi$. If d > r, the distance from S to $\pi$ is d-r; otherwise $\pi$ meets S and the distance is 0. This is easier to see in two dimensions.

As you computed, the distance $d=\sqrt3$. So the distance from your plane to the sphere is 0. In fact, the tangent plane to your sphere at the point (0,1,2), is your plane x+y+z=3. So the distance is obviously 0. None of the answers are correct.
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May 17th, 2017, 10:20 PM   #5
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Quote:
Originally Posted by johng40 View Post
Your sphere has equation
$$(x-1)^2+(y-2)^2+(z-3)^2-3=0$$
That is, the radius of the sphere is $\sqrt{3}$ with center (1,2,3).

Now generally, let $\pi$ be any plane and S any sphere with center C and radius r. Also let d be the distance from C to $\pi$. If d > r, the distance from S to $\pi$ is d-r; otherwise $\pi$ meets S and the distance is 0. This is easier to see in two dimensions.

As you computed, the distance $d=\sqrt3$. So the distance from your plane to the sphere is 0. In fact, the tangent plane to your sphere at the point (0,1,2), is your plane x+y+z=3. So the distance is obviously 0. None of the answers are correct.
I will write the line equation with the point and the directional vector : $x= 1+t; y= 2+t; z= 3+t$.
Then substituting these values in the plane :$ x+y+z-3=0 $
$1+t+2+t+3+t-3=0$
$3+3t=0$
$t=-1$
back to the point :
$x= 0$ ; $y=1$; $z= 2$;
now, distance between $(1,2,3)$ and $(0,1,2)$ is $\displaystyle \sqrt {1+1+1} $ which is equals to $\displaystyle \sqrt 3$.

I got the process. Thank you
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