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: $\displaystyle \sqrt{(3M+96)^2+(4M+26)^2+(1+1)^2}=10$ 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. 
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. 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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