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
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}