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 May 27th, 2017, 08:56 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 10:56 PM. May 27th, 2017, 11:00 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,390 Thanks: 2015 Is there a point that lies in the given plane and on the given line? May 28th, 2017, 11:12 AM   #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 10:10 PM. May 29th, 2017, 04:55 AM   #4
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 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 10:10 PM. May 29th, 2017, 10:25 AM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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 09:59 PM. Tags line, parallel, vector Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post LadyLisa Geometry 1 July 2nd, 2015 04:13 AM tx3 Algebra 4 May 23rd, 2013 11:47 PM harsha973 Algebra 11 November 8th, 2012 08:48 AM fran1942 Calculus 5 April 2nd, 2012 08:48 PM

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