Given that point A has the position vector 4i+7j and point B has the position vector 10i+qj, where q is a constant, find a) the vector AB in terms of q b) Given further that the distance AB = 2 sqrt13, find the two possible values of q ?

1 Answer
Steve M
Mar 24, 2018

# bb(vec(AB)) = 6bb(ul(hat i)) + (q-7)bb(ul(hat j)) #

# q=3# or #q=11#

Explanation:

We have:

# bb(vec(OA)) = 4bb(ul(hat i)) + 7bb(ul(hat j)) #
# bb(vec(OB)) = 10bb(ul(hat i)) + qbb(ul(hat j)) \ \ \ \ \ q in RR#

So then:

# bb(vec(AB)) = bb(vec(OB)) - bb(vec(OA)) #
# \ \ \ \ \ \ = (10bb(ul(hat i)) + qbb(ul(hat j))) - (4bb(ul(hat i)) + 7bb(ul(hat j))) #
# \ \ \ \ \ \ = (10-4)bb(ul(hat i)) + (q-7)bb(ul(hat j)) #
# \ \ \ \ \ \ = 6bb(ul(hat i)) + (q-7)bb(ul(hat j)) #

Given that:

# || bb(vec(AB)) || = 2sqrt(13) #

We have:

# sqrt( (6)^2 + (q-7)^2 ) = 2sqrt(13) #

# :. 36 + q^2-14q+49 = (2sqrt(13))^2 #

# :. 36 + q^2-14q+49 = 4 * 13 #

# :. 85 + q^2-14q = 52 #

# :. q^2-14q +33 = 0 #

# :. (q-11)(q-3) = 0 #

Yielding two possible solutions:

# q=3# or #q=11#

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