# vectorsWatch

This discussion is closed.
#1
relative to the origin the points A and B have position vectors

1
5
-1

and
6
3
-6

respectively
find the shortest distance from A to the line OB

thanks
0
13 years ago
#2
(Original post by sarah12345)
relative to the origin the points A and B have position vectors

1
5
-1

and
6
3
-6

respectively
find the shortest distance from A to the line OB
thanks
The line OB has equation r=k(6i+3j-6k) or (simplified) r=i(2t)+j(t)+k(-2t) so an arbitrary point on the line has coordinates [2t,t,-2t]
A has coordinates (1,5,-1)
The square of the distance between A and the arbitrary point is:
d^2=(2t-1)^2+(t-5)^2+(-2t+1)^2
d^2=9t^2-18t+27
d^2=9(t^2-2t+3)
d^2=9[(t-1)^2+2]
d^2(min)=9(2)
d(min)=3rt2
Sorry for the confusion and thanks for pointing out the error.
0
13 years ago
#3
i used scalar product n got 3√2
the previous method seems right too
?!
0
13 years ago
#4
i'll post my method..

OA= i + 5j - k
OB= 6i + 3j - 6k

The shortest distance from OB to A is the perpendicular distance..calling the point of contact of perpendicular from A to OB 'C'

CA= i + 5j - k - λ(6i + 3j -6k)

CA.OB=0
6(1-6λ) + 3(5-3λ) - 6(-1+6λ) = 0
27 - 81λ = 0
so λ=1/3

sub in CA to get
CA= -i + 4j + k
|CA|= √ (1 + 16 +1)
|CA|= √18 = 3√2
0
13 years ago
#5
(Original post by Gaz031)
...A has coordinates (1,5,1)...
(Should be (1,5,-1))
0
13 years ago
#6
(Original post by C4>O7)
(Should be (1,5,-1))
0
13 years ago
#7
(Original post by Vijay1)
I should probably have noticed it earlier when you got an answer much more neat and 'elegant' than mine. Still, the P6 review exercise beckons.
0
13 years ago
#8
(Original post by Gaz031)
I should probably have noticed it earlier when you got an answer much more neat and 'elegant' than mine. Still, the P6 review exercise beckons.
i did the question that was posted
so we agree on that now yeah
0
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