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yrelhan4
 3 years 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.
yrelhan4
 3 years 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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Seems like a work for the dot product to me.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0so 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360345934670:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0careful, a line in 3D space shouldn't be written in cartesian form,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \vec{v} \cdot \vec{SP}= 0 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360346169283:dw

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0well it can be written in cartesian form with direction ratios.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0hmm. 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Mathematical 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.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0@TuringTest well its the position vector as they call it. its basically a point.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0ok, I think I understand the question better now, thanks

yrelhan4
 3 years ago
Best 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.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0the line is parallel to the vector in my interpretation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360346673680:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To 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: \[\Large r_x=0P + \lambda \vec{v} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I guess @sirm3d and I agree then, 6i+4j4k is a direction vector of the parametric line

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0as @Spacelimbus said, it's the dot product at work here.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I agree as well, for what that's worth :p

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0hmm. and whats wrong with the method i'm using?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360346938724:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0stealing @sirm3d superior model

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360347140886:dw \[\cos \theta = \frac{\vec{PR} \cdot \vec{SR}}{\vec{PR}\vec{SR}}\\\vec{PS}=\vec{PR}\sin\theta\]

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0i think i can do it now. thank you guys.
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