My Math Forum intersection of two lines (vectors)

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 April 7th, 2014, 11:37 AM #11 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 Hey, sorry about this, but i'm sort of stuck on the last part of the same question. Find one of the two points on the line L1 whos distance from P is 10. Any hints ?
 April 7th, 2014, 11:47 AM #12 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Using the distance formula, we could state: $\displaystyle \sqrt{(3M+9-6)^2+(-4M+2-6)^2+(-1+1)^2}=10$ Simplify: $\displaystyle \sqrt{9(M+1)^2+16(M+1)^2}=10$ $\displaystyle |5(M+1)|=10$ $\displaystyle |M+1|=2$ Hence, $M$ is a number whose distance on the real number line from -1 is 2: $\displaystyle M=-3,1$ These values of $M$ will give you both points. Thanks from marks2014
 April 7th, 2014, 01:05 PM #13 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 Just working through another question here, wondering whether you could check my answer ? The plane P contains point A (1,2,3) and is parallel to u=2i +3j and v= i +2j - k. Assume the vector normal to plane is written as (3,a,b). Find suitable a and b values. I get a=-2 b=-1. Cartesian coordinates of 3x -2y -z = -4 Distance from origin to nearest point as -4/sqrt(14) Sorry if i'm annoying you here but i havent seen much of this before today
 April 7th, 2014, 03:25 PM #14 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs I recommend creating new threads for new problems...this will increase the chances that more of our helpers will answer you. when I look at the list of new posts and I see a thread that already has many replies, I assume help is already being given and so I am less likely to look at that thread right away, whereas if I see a thread with zero replies I will look at it more quickly. I really don't have time at the moment to help with this latest problem...not until many hours from now.
 April 7th, 2014, 04:48 PM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra I think you have an error in your value for $a$.
 April 8th, 2014, 06:50 AM #16 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 I found 'a' by the dot product of the normal and u, considering u is parallel to the plane it should be perpendicular to u. (3,a,b).(2,3,0) i get 6+3a=0 3a=-6 a=-2 Can you tell me where you think I have went wrong ?
 April 8th, 2014, 08:18 AM #17 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra My apologies. What you have wrong is that you are talking to someone who apparently can't do simple algebra. You are correct. You do somehow have a negative distance though. Thanks from marks2014 Last edited by v8archie; April 8th, 2014 at 08:26 AM.
 April 8th, 2014, 08:49 AM #18 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 Yeah, I have done the question again and got a positive value ! Thanks anyway !

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