February 8th, 2013, 01:04 AM  #1 
Member Joined: Apr 2012 Posts: 90 Thanks: 0  Vectors
Hi Please help me with these questions, I really really don't know how to do them! 1.) Find the equation of the plane perpendicular to the line connecting A(1,4,2) and B(4,1,4) and containing P such that AP:PB = 1:2 I got the vector AB which is 3i3j6k, then I subbed coordinates of point A into i, j and k and I got xy2z = 7, however the answer was xy2z = 1! 2a.) Prove that vectors XY+YO+OZ+ZX = 0 I know that the resultant vector will be XX, will that be equal to 0? I'm confused though, why, how does XX look like? All I know is that a x a = 0 in space. 2b.) If Vectors AO + OB = Vectors BO + OC then A, B, C are collinear. I know to prove that points are collinear through component forms of the vectors.. am I supposed to make up component vectors here? 3.) Find the equation of the plane containing A(3,2,1) and the line x=1+t, y=2t, z=3+2t I know the vector/ direction vector is ij+2k from the parametric equations. I know that I have two position vectors, 3i+2j+k and i+2j+3k. But then how do I continue? Thank you very much I really appreciate this, Thank you! 
February 8th, 2013, 12:17 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,607 Thanks: 616  Re: Vectors Quote:
2a) If both end points of a vector are the same, the vector has 0 length. 2b) yes. 3) Form two vectors from A to 2 different points on the line. The cross product is the direction of the normal to the plane.  
February 9th, 2013, 03:40 AM  #3 
Member Joined: Apr 2012 Posts: 90 Thanks: 0  Re: Vectors
Thank you so much I got the hang of 2 and 3! However, could you show me the steps for 1? Im really lost in that one.

February 9th, 2013, 03:19 PM  #4  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Vectors Quote:
So the question is just to find P. P divides AB so that the distance from P to B is twice the distance from A to P thats what "AP:PB= 1:2 means. The "xdifference", from x= 1 to x= 4, is 3, so the distance from A to P is 1. The "ydifference", from y= 4 to y=1 is 3 so the (signed) distance from A to P is 1. The "zdifference", from z= 2 to z= 4 is 6 so the (signed) difference from A to P is 2. That is P is (1+ 1, 4 1, 2 2)= (2, 5, 0). Mathman has given you a different, but equivalent, way to do this problem. It all boils down to "proportions".  
February 9th, 2013, 03:22 PM  #5  
Global Moderator Joined: May 2007 Posts: 6,607 Thanks: 616  Re: Vectors Quote:
 
April 23rd, 2014, 08:13 PM  #6 
Newbie Joined: Apr 2014 From: Lippo Cikarang, Indonesia Posts: 1 Thanks: 0  think simple
For (1) Let P=(x,y,z). PB=2AP. Then (4x,1y,4z)=2(x1,y4,z2). 4x=2(x1) > x=2, 1y=2(y4) > y=3, 4z=2(z2) > z=0 So, P=(2,3,0). Now plug in to xy2z=d and find d. You will get d=1. 
June 7th, 2017, 08:05 AM  #7 
Newbie Joined: Jun 2017 From: india Posts: 1 Thanks: 0 
can u plzz explain the steps of 2b i do not understand how to solve


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