Given that D is a point lying on the line through B and C find 2 possible vectors of point D?

Question

Relative to an origin O, the position vectors of the points A, B and C are given by
#vec (OA) =(8, -6, 5 )#,
#vec (OB) =(-10, 3, -13 )#,
#vec (OC) =(2, -3, -1 )#.
A fourth point, D is such that the magnitudes #|vec(AB)|, |vec(BC)| and |vec(CD)|# are the first, second and third terms respectively of a geometric progression.

Given that D is a point lying on the line through B and C find 2 possible vectors of point D?

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Answer 1 · 2018-05-18T06:24:48

#D(6,-5,3), or, D(-2,-1,-5)#.

We have, #B=B(-10,3,-13) and C=C(2,-3,-1)#.

Hence, the eqn. of line #BC# is given by,

#(x-2)/(-10-2)=(y-(-3))/(3-(-3))=(z-(-1))/(-13-(-1)), i.e., #

#(x-2)/-12=(y+3)/6=(z+1)/-12, or, #

#(x-2)/2=(y+3)/-1=(z+1)/2=t," say, where, "t in RR#.

Because the reqd. point #D# lies on the line #BC#, we must have,

#EE t in RR ; D=D(x,y,z)=D(2t+2,-t-3,2t-1)#.

Now, using this #D# and given #A,B,C#, lt us work out :

#|vec(AB)|=|(-10,3,-13)-(8,-6,5)|=|(-18;9;-18)|#.

#:. |vec(AB)|=9sqrt{(-2)^2+1^2+(-2)^2}=27#,

#|vec(BC)|=18, and, |vec(CD)|=|(2t;-t;2t)|=3|t|#.

#:. |vec(BC)|^2=|vec(AB)|*|vec(CD)|#.

#:. 18^2=(27)(3|t|)#

#:. |t|=(18xx18)/(27xx3)=4#.

#:. t=+-2#.

# t=2 rArr D=D(2(2)+2,-2-3,2(2)-1))=D(6,-5,3)#.

# t=-2" gives, "D=D(-2,-1,-5)#.

Thus, #D(6,-5,3), or, D(-2,-1,-5)#.

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