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 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 3i-3j-6k, then I subbed coordinates of point A into i, j and k and I got x-y-2z = -7, however the answer was x-y-2z = -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=2-t, z=3+2t I know the vector/ direction vector is i-j+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
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 Originally Posted by Ter 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 3i-3j-6k, then I subbed coordinates of point A into i, j and k and I got x-y-2z = -7, however the answer was x-y-2z = -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=2-t, z=3+2t I know the vector/ direction vector is i-j+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!
1) The desired plane is not through A, but through P as defined.
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
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Re: Vectors

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 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 3i-3j-6k
I presume you know then that the equation of the plane is 3x- 3y- 6k= C where you can put the coordinates of P in for x, y, and z, to determine C.

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 "x-difference", from x= 1 to x= 4, is 3, so the distance from A to P is 1. The "y-difference", from y= 4 to y=1 is -3 so the (signed) distance from A to P is -1. The "z-difference", 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
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 Originally Posted by Ter 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.
P = xA + (1-x)B, AP:PB (=1/2) = (1-x)/x (exercise for you - verify), calculate x, find P and derive plane through P.

 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 (4-x,1-y,-4-z)=2(x-1,y-4,z-2). 4-x=2(x-1) -> x=2, 1-y=2(y-4) -> y=3, -4-z=2(z-2) -> z=0 So, P=(2,3,0). Now plug in to x-y-2z=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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