Determine a direction vector when a line perpendicular and a point is given

[euro94]

Find the direction vector of the line perpendicular to the line r=[-1,0,1] + t [2,3,-2] and passing through the point (-7,-9,7)

 

[tiny-tim]

welcome to pf!

hi euro94! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[euro94]

I tried finding the dot product of the line's direction vector with a vector from (-7,9,7). I tried (-1+2t, 3t, 1-2t) - (-7,-9,7) and I dotted it with (2,-3,2) and made it equal to zero, and then i got stuck :(

 

[tiny-tim]

I tried finding the dot product of the line's direction vector with a vector from (-7,9,7). I tried (-1+2t, 3t, 1-2t) - (-7,-9,7) and I dotted it with (2,-3,2) and made it equal to zero, and then i got stuck :(

(you mean (2,3,-2)?)

that should work …

what did you get?​

 

[euro94]

I got (6+2t,3t+9,6-2t) and i dotted it with (2,3,-2) and then i made it equal to zero and i got 12+4t+9t+27-12+4t and i solved for t and i got -27/17, but I'm not sure what to do next..

 

[euro94]

i mean i got a value of -27/14 for t, i had a miscalculation, but I am not sure what step to take next

 

[tiny-tim]

hi euro94!

(just got up :zzz:)

i mean i got a value of -27/14 for t, i had a miscalculation, but I am not sure what step to take next

that value of the parameter (t) gives you the foot of the perpendicular …

so put it into [-1,0,1] + t [2,3,-2] to give you the actual coordinates,

and then join that to [-7,-9,7]

 

[euro94]

oppsss, i got t=-3 and i plugged it into the equation and i got that my direction vector is [-7,-9,7] and so my overall equation is r=[-7,-9,7] + t[-7,-9,7]

 

[euro94]

is that normal?

 

[tiny-tim]

you mean it goes through the origin?

can you please write it all out in one go, if you want it checked?

 

[euro94]

okay :)
(-1+2t, 3t, 1-2t) - (-7,-9,7) = (6+2t, 3t+9, -6-2t)
dot product:
(6+2t, 3t+9, -6-2t) dot (2,3,-2) =0
12+4t+9t+27+12+4t =0
51+17t=0
-51/17=t
-3=t
r=[-1,0,1] + 3 [2,3,-2]
= [-7,-9,7]

 

[euro94]

is that right? :)

 

[tiny-tim]

okay :)
(-1+2t, 3t, 1-2t) - (-7,-9,7) = (6+2t, 3t+9, -6-2t)
dot product:
(6+2t, 3t+9, -6-2t) dot (2,3,-2) =0
12+4t+9t+27+12+4t =0
51+17t=0
-51/17=t
-3=t
r=[-1,0,1] + 3 [2,3,-2]
= [-7,-9,7]

hmm … that's weird!

yes, your caluculations are fine

looking back at the original question now, obviously [-7,-9,7] does actually lie on the original line …

so (since we're in three dimensions), asking for "the line perpendicular" makes no sense, since there's an infinite number of them!

 

[euro94]

thank youu :)