May 16th, 2017, 01:14 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 106 Thanks: 1  Shortest Distance
The shortest distance from the sphere $x^2+y^2+z^22x4y6z+11=0$ 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 $ax+ by+ czd$ / $\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 
May 17th, 2017, 03:28 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,405 Thanks: 611 
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.

May 17th, 2017, 03:49 AM  #3 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 106 Thanks: 1 
[QUOTE=Lalitha183;570237]The shortest distance from the sphere $x^2+y^2+z^22x4y6z+11=0$ 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$\displaystyle \frac{ ax+ by+ czd } {\sqrt {a^2 + b^2 + c^2} }$ to get the distance and the final answer is $\displaystyle \sqrt 3 $ Please let me know the steps if it is wrong!! 
May 17th, 2017, 05:46 AM  #4 
Member Joined: Jan 2016 From: Athens, OH Posts: 40 Thanks: 24 
Your sphere has equation $$(x1)^2+(y2)^2+(z3)^23=0$$ 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 dr; 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. 
May 17th, 2017, 10:20 PM  #5  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 106 Thanks: 1  Quote:
Then substituting these values in the plane :$ x+y+z3=0 $ $1+t+2+t+3+t3=0$ $3+3t=0$ $t=1$ 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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