Find the values of *k* for which vec(AB)−−→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)−−→OA=(1,0,2) and vec(OB)= (k, -k, 2k)−−→OB=(k,−k,2k),
where k is a constant.
(i) In the case where k=2,k=2, calculate angle AOB.
(ii) Find the values of k for which vec(AB)−−→AB is a unit vector.

1 Answer
Narad T.
May 7, 2018

Please see the explanation below

Explanation:

vec(OA)=<1,0,2>−−→OA=<1,0,2>

vec(OB)= < k,-k, 2k >−−→OB=<k,−k,2k>

"Question (i)"Question (i)

When k=2k=2

vec(OB)= < 2,-2, 4>−−→OB=<2,−2,4>

The angle cos(hat(AOB))=(vec(OA). vec(OB))/(|vec(OA)|*|vec(OB)|)cos(ˆAOB)=−−→OA.−−→OB∣∣∣−−→OA∣∣∣⋅∣∣∣−−→OB∣∣∣

The dot product is

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

The magnitude of vec(OA)−−→OA is

=|vec(OA)| =|<1,0,2>| = sqrt(1^2+0^2+2^2)=sqrt(5)=∣∣∣−−→OA∣∣∣=|<1,0,2>|=√12+02+22=√5

The magnitude of vec(OB)−−→OB is

=|vec(OB)| =|<2,-2,4>| = sqrt(2^2+(-2)^2+4^2)=sqrt(24)=∣∣∣−−→OB∣∣∣=|<2,−2,4>|=√22+(−2)2+42=√24

The angle is

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

"Question (ii)"Question (ii)

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

The unit vector is

|vec(AB)|=1∣∣∣−−→AB∣∣∣=1

The magnitude of vec(AB)−−→AB is

|vec(AB)| = |< k-1, -k, 2k-2 >|∣∣∣−−→AB∣∣∣=|<k−1,−k,2k−2>|

=sqrt((k-1)^2+(-k)^2+(2(k-1)^2))=√(k−1)2+(−k)2+(2(k−1)2)

=sqrt(k^2-2k+1+k^2+4k^2-8k+4)=√k2−2k+1+k2+4k2−8k+4

=sqrt(6k^2-10k+5)=√6k2−10k+5

Therefore,

|vec(AB)| =1∣∣∣−−→AB∣∣∣=1

6k^2-10k+5=16k2−10k+5=1

6k^2-10k+4=6k2−10k+4=

3k^2-5k+2=03k2−5k+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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