# Vectors in 3D

• Nov 17th 2011, 06:56 AM
kandyfloss
Vectors in 3D
I started solving this by taking the first point as x1,y1,z1 and second one as x2,y2,z2
and so the distance between these points to B is *sqrt*14.
Although this method doesn't seem to be working. Any help/hint on how to proceed with this question will be much appreciated.
• Nov 17th 2011, 07:12 AM
Plato
Re: Vectors in 3D
You need to find where the y-axis insects the sphere $(x+1)^2+(y+1)^2+(z-2)^2=14~.$
Recall that the y-axis is $(0,t,0)~.$
• Nov 17th 2011, 07:41 AM
kandyfloss
Re: Vectors in 3D
So i put y=t? and x=0 and z=0?
• Nov 17th 2011, 08:10 AM
Plato
Re: Vectors in 3D
Quote:

Originally Posted by kandyfloss
So i put y=t? and x=0 and z=0?

Indeed, yes. Solve for t.
• Nov 17th 2011, 08:15 AM
kandyfloss
Re: Vectors in 3D
Quote:

Originally Posted by Plato
Indeed, yes. Solve for t.

So i get t(^2) + 2t -13=0
And it seems my values for t are going to be in fraction, and the answer is in whole numbers.
• Nov 17th 2011, 08:30 AM
Plato
Re: Vectors in 3D
Quote:

Originally Posted by kandyfloss
So i get t(^2) + 2t -13=0

I get $t^2+2t-8=0$
• Nov 17th 2011, 08:33 AM
kandyfloss
Re: Vectors in 3D
Quote:

Originally Posted by Plato
I get $t^2+2t-8=0$

oh yes, sorry. I'm kind of out of my mind today. Thanks a lot!