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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