## yrelhan4 Group Title 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

1. Spacelimbus Group Title

Seems like a work for the dot product to me.

2. yrelhan4 Group Title

so i converted all of it into cartesian form. wrote the line as (x-2)/6=(y-3)/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.

3. Spacelimbus Group Title

|dw:1360345934670:dw|

4. Spacelimbus Group Title

careful, a line in 3D space shouldn't be written in cartesian form,

5. Spacelimbus Group Title

$\Large \vec{v} \cdot \vec{SP}= 0$

6. Spacelimbus Group Title

$\Large 0S \in g$

7. Spacelimbus Group Title

|dw:1360346169283:dw|

8. yrelhan4 Group Title

well it can be written in cartesian form with direction ratios.

9. Spacelimbus Group Title

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.

10. TuringTest Group Title

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?

11. yrelhan4 Group Title

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?

12. Spacelimbus Group Title

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.

13. yrelhan4 Group Title

@TuringTest well its the position vector as they call it. its basically a point.

14. TuringTest Group Title

ok, I think I understand the question better now, thanks

15. yrelhan4 Group Title

@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.

16. TuringTest Group Title

the line is parallel to the vector in my interpretation

17. sirm3d Group Title

|dw:1360346673680:dw|

18. Spacelimbus Group Title

To set up an equation in the 3D-Space, you need a direction vector, an a point, that's how I interpret it, because that satisfies the equation: $\Large r_x=0P + \lambda \vec{v}$

19. Spacelimbus Group Title

I guess @sirm3d and I agree then, 6i+4j-4k is a direction vector of the parametric line

20. sirm3d Group Title

as @Spacelimbus said, it's the dot product at work here.

21. TuringTest Group Title

I agree as well, for what that's worth :p

22. yrelhan4 Group Title

hmm. and whats wrong with the method i'm using?

23. Spacelimbus Group Title

|dw:1360346938724:dw|

24. Spacelimbus Group Title

stealing @sirm3d superior model

25. sirm3d Group Title

|dw:1360347140886:dw| $\cos \theta = \frac{\vec{PR} \cdot \vec{SR}}{|\vec{PR}||\vec{SR}|}\\|\vec{PS}|=|\vec{PR}|\sin\theta$

26. yrelhan4 Group Title

i think i can do it now. thank you guys.