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the distance from the point i + 2j + 6k to the straight line through the point (2,3,4) and parallel to the vector 6 i+ 3j  4k.
 one year ago
 one year ago
the distance from the point i + 2j + 6k to the straight line through the point (2,3,4) and parallel to the vector 6 i+ 3j  4k.
 one year ago
 one year ago

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SpacelimbusBest ResponseYou've already chosen the best response.2
Seems like a work for the dot product to me.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
so i converted all of it into cartesian form. wrote the line as (x2)/6=(y3)/3=(z+4)/4 then i put all of it =r. found out random points. then found out direction ratios of the perpendicular. through condition of perpendicularity found r. and found those random points. and the found the distance. the answer i am getting is wrong.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
dw:1360345934670:dw
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
careful, a line in 3D space shouldn't be written in cartesian form,
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[\Large \vec{v} \cdot \vec{SP}= 0 \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[\Large 0S \in g \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
dw:1360346169283:dw
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
well it can be written in cartesian form with direction ratios.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
well if you want to do it like that, then unfortunately I can't help you. I haven't seen such a thing, what I would agree with, is saying that you can always set up an equation of a plane that includes the line and from there figure out the distance. But as soon as I see something of the form x+y+z=c, I think of a plane, not of a line.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I am confused for many reasons: "the distance from the point i + 2j + 6k..." but i +2j + 6k is not a point. and the distance to which point on the other line? the minimum distance I presume?
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
hmm. i just want to confirm if i am interpreting the question right. its saying the line is parallel to the vector 6 i+ 3j  4k and not that it wants the distance parallel to this vector?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
Mathematical definition is always considering it's normal, orthogonal point. That is a distance. So if you have a point given, as I tried to sketch it in the above picture, then it's distance to the line is the one where you start from the line, draw a right angle through that point and figure out the distance.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
@TuringTest well its the position vector as they call it. its basically a point.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
ok, I think I understand the question better now, thanks
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
@Spacelimbus well i still have the doubt. does it ask for the distance parallel to that vector? or the line is parallel to that vector? 6 i+ 3j  4k.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
the line is parallel to the vector in my interpretation
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
To set up an equation in the 3DSpace, you need a direction vector, an a point, that's how I interpret it, because that satisfies the equation: \[\Large r_x=0P + \lambda \vec{v} \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
I guess @sirm3d and I agree then, 6i+4j4k is a direction vector of the parametric line
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
as @Spacelimbus said, it's the dot product at work here.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I agree as well, for what that's worth :p
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
hmm. and whats wrong with the method i'm using?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
dw:1360346938724:dw
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
stealing @sirm3d superior model
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
dw:1360347140886:dw \[\cos \theta = \frac{\vec{PR} \cdot \vec{SR}}{\vec{PR}\vec{SR}}\\\vec{PS}=\vec{PR}\sin\theta\]
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
i think i can do it now. thank you guys.
 one year ago
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