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April 7th, 2014, 11:37 AM   #11
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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 ?
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April 7th, 2014, 11:47 AM   #12
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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.
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April 7th, 2014, 01:05 PM   #13
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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
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April 7th, 2014, 03:25 PM   #14
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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.
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April 7th, 2014, 04:48 PM   #15
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I think you have an error in your value for $a$.
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April 8th, 2014, 06:50 AM   #16
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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 ?
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April 8th, 2014, 08:18 AM   #17
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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.
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Last edited by v8archie; April 8th, 2014 at 08:26 AM.
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April 8th, 2014, 08:49 AM   #18
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Yeah, I have done the question again and got a positive value ! Thanks anyway !
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