Using vectors, find the value of `lambda` such that the points `(lambda,-10),(1,-1,3)a n d(3,5,3)` are collinear.

Explanation

Let the points are A(k, -10, 3), B(1,-1, 3) and C(3, 5, 3).
So `vec(AB)=vec(OB)-vec(OA)`
`=(hati-hatj+3hatk)-(khati-10hatj+3hatk)`
`=(1-k)hati+(-1+10)hatj+(3-3)hatk`
`=(1-k)hati+9hatj+0hatk`
`:.|vec(AB)|=sqrt((1-k)^(2)+(9)^(2)+0)=sqrt((1-hatk)^(2)+81)`
Similarly, `vec(BC)=vec(OC)-vec(OB)`
`=(3hati+5hatj+3hatk)-(hati-hatj+3hatk)`
`=2hati+6hatj+0hatk`
`:.|vec(BC)|=sqrt(2^(2)+6^(2)+0)=2sqrt(10)`
and `vec(AC)=vec (OC)-vec(OA)`
`=(3hati+5hatj+3hatk)-(khati-10hatj+3hatk)` `=(3-k)hati+15hatj+0hatk`
`:.|vec(AC)|=sqrt((3-k)^(2)+225)`
If A, B and C are collinear, then sum of modulus of any two vectors will be equal to the modulus of thridn vectors
For `|vec(AB)|+|vec(BC)|=|vec(AC)|`
`impliessqrt((1-k)^(2)+81)+2sqrt(10)=sqrt((3-k)^(2)+225)`
`sqrt((3-k)^(2)+225)-sqrt((1-k)^(2)+81)=2sqrt(10)`
`impliessqrt(9+k^(2)-6k+225)-sqrt(1+k^(2)-2k+81)=2sqrt(10)`
`impliessqrt(k^(2)-6k+234)-2sqrt(10)=sqrt(k^(2)-2k+82)`
`impliesk^(2)-6k+234+40-2sqrt(k^(2)-6k+234).2sqrt(10)=k^(2)-2k+82`
`impliesk^(2)-6k+234+40-k^(2)+2k-82=4sqrt(10)sqrt(k^(2)+234-6k)`
`implies-4k+192=4sqrt(10)sqrt(k^(2)+234-6k)`
`implies-k+48=sqrt(10)sqrt(k^(2)+234-6k)`
On squaring both sides, we get
`48xx48+k^(2)-96k=10(k^(2)+234-6k)`
`impliesk^(2)-96k-10k^(2)+60k=-48xx48+2340`
`implies-9k^(2)-36k=-48xx48+2340`
`implies(k^(2)+4k)=+16xx16xx-260` [dividing by 9 in both sides]
`impliesk^(2)+4k=-4`
`k^(2)+4k+4=0`
`implies(k+2)^(2)=0`
`:.k=-2`