Find the values of *k* for which #vec(AB)# is a unit vector.?

The position vector of the points A and B, realative to an origin O, are given by
#vec(OA) = (1, 0, 2)# and #vec(OB)= (k, -k, 2k)#,
where k is a constant.
(i) In the case where# k=2,# calculate angle AOB.
(ii) Find the values of k for which #vec(AB)# is a unit vector.

1 Answer
Narad T.
May 7, 2018

Please see the explanation below

Explanation:

#vec(OA)=<1,0,2>#

#vec(OB)= < k,-k, 2k >#

#"Question (i)"#

When #k=2#

#vec(OB)= < 2,-2, 4>#

The angle #cos(hat(AOB))=(vec(OA). vec(OB))/(|vec(OA)|*|vec(OB)|)#

The dot product is

#(vec(OA). vec(OB))=<1,0,2> . < 2,-2, 4> = 2-0+8=10#

The magnitude of #vec(OA)# is

#=|vec(OA)| =|<1,0,2>| = sqrt(1^2+0^2+2^2)=sqrt(5)#

The magnitude of #vec(OB)# is

#=|vec(OB)| =|<2,-2,4>| = sqrt(2^2+(-2)^2+4^2)=sqrt(24)#

The angle is

#=arccos(10/(sqrt(5)sqrt(24)))=arcos(0.913)=24.09^@#

#"Question (ii)"#

#vec(AB)= < k,-k, 2k > - <1,0,2> = < k-1, -k, 2k-2 > #

The unit vector is

#|vec(AB)|=1#

The magnitude of #vec(AB)# is

#|vec(AB)| = |< k-1, -k, 2k-2 >|#

#=sqrt((k-1)^2+(-k)^2+(2(k-1)^2))#

#=sqrt(k^2-2k+1+k^2+4k^2-8k+4)#

#=sqrt(6k^2-10k+5)#

Therefore,

#|vec(AB)| =1#

#6k^2-10k+5=1#

#6k^2-10k+4=#

#3k^2-5k+2=0#

The discriminant is

#Delta=(-5)^2-4(3)(2)=25-24=1>0#

Therefore,

#k=(5+-1)/(6)#

The solutions are

#S={2/3, 1}#

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