Question

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

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?

Answer 1

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

Explanation:

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|`.

Now, given that, `|vec(AB)|, |vec(BC)| and |vec(CD)|` are in GP.

`:. |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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