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 May 27th, 2017, 09:56 PM #1 Newbie   Joined: May 2017 From: Tanzania Posts: 8 Thanks: 0 Vector parallel to the line Hi Guys! I was doing a vector math problem and I got stuck on how to find the vector parallel to the line. The question is : Determine if the plane given by -x+2z=10 and the line given by r=(5,2-t,10+4t) are orthogonal, parallel or neither. Now I have already found the normal vector as the first step, but also I have to find the vector that is parallel to the line so as that I can cross it with the normal vector to prove if it they are parallel to each other. Thank you. Last edited by skipjack; May 27th, 2017 at 11:56 PM.
 May 28th, 2017, 12:00 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,433 Thanks: 1462 Is there a point that lies in the given plane and on the given line?
May 28th, 2017, 12:12 PM   #3
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Quote:
 Originally Posted by Techgenius Hi Guys! I was doing a vector math problem and I got stuck on how to find the vector parallel to the line. The question is : Determine if the plane given by -x+2z=10 and the line given by r=(5,2-t,10+4t) are orthogonal, parallel or neither. Now I have already found the normal vector as the first step, but also I have to find the vector that is parallel to the line so as that I can cross it with the normal vector to prove if it they are parallel to each other. Thank you.
I presume that you have determined that a normal vector to the plane is <-1, 0, 2>. Is that what you got? Since your line is given by r = (5, 2 - t, 10 + 4t), taking t = 0, one point on that line is (5, 2, 10). Taking t = 1, another point is (5, 1, 14). The vector from (5, 2, 10) to (5, 1, 14) is <5 - 5, 1 - 2, 14 - 10> = <0, -1, 4> and that is a vector parallel to the line. Notice that <0, -1, 4> are the coefficients of t in "r = 5, 2 - t, 10 + 4t".

In general, a vector in the direction of the line $\displaystyle x= x_0+ At$, $\displaystyle y= y_0+ Bt$, $\displaystyle z= z_0+ Ct$ is <A, B, C>.

Last edited by skipjack; May 29th, 2017 at 11:10 PM.

May 29th, 2017, 05:55 AM   #4
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Quote:
 Originally Posted by Country Boy I presume that you have determined that a normal vector to the plane is <-1, 0, 2>. Is that what you got? Since your line is given by r = (5, 2 - t, 10 + 4t), taking t = 0, one point on that line is (5, 2, 10). Taking t = 1, another point is (5, 1, 14). The vector from (5, 2, 10) to (5, 1, 14) is <5 - 5, 1 - 2, 14 - 10> = <0, -1, 4> and that is a vector parallel to the line. Notice that <0, -1, 4> are the coefficients of t in "r = 5, 2 - t, 10 + 4t". In general, a vector in the direction of the line $\displaystyle x= x_0+ At$, $\displaystyle y= y_0+ Bt$, $\displaystyle z= z_0+ Ct$ is .
Thank you very much Country Boy, you saved me in this. I now get the full picture. Thank you again.
#Slumerican

Last edited by skipjack; May 29th, 2017 at 11:10 PM.

 May 29th, 2017, 11:25 AM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 A tad faster might be: A normal to plane is (-1,0,2) by inspection. Line is r=(5,2,10)+t(0,-1,4) (-1,0,2)x(0,-1,4)=0 [Note by moderator: (-1, 0, 2) × (0, -1, 4) = (2, 4, 1)] Last edited by skipjack; May 29th, 2017 at 10:59 PM.

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