Question

Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `bar r=(mbarb+nbara)/(m+n)`. Hence find the position vector of R which divides the line segment joining the points A(1, –2, 1) and B(1, 4, –2) internally in the ratio 2 : 1.

Answer 1

Consider a line segment AB. 

Let R be any point on it such that point R divides AB internally in the ratio m : n.
bar(OA) = bara, bar(OR) = barr and bar(OB) = barb are the position vectors of points A, R, B respectively. Since point R divides AB internally in the ratio m : n,

`(AR)/(RB)=m/n`

`n(AR)=m(RB)`

`bar(AR) and bar(RB)` are in the same direction.

`therefore n(bar(AR))=m(bar(RB))`

`n(barr-bara)=m(barb-barr)`

`nbarr-nbara=mbarb-mbarr`

`nbarr+mbarr=mbarb+nbara`

`barr(m+n)=mbarb+nbara`

`barr=(mbarb+nbara)/(m+n)`    ...(1)

This is the section formula for internal division

Let P.V. of point `A bara=hati-2hatj+hatk`

P. V. of point `B barb=hati+4hatj-2hatk`

Given, `m/n=2/1`

Now, `bar r=(2(hati+4hatj-2hatk)+1(hati-2hatj+hatk))/(2+1)`  ...From (1)

`bar r=(3hati+6hatj-3hatk)/3`

P. V. of R is `bar r=hati+2hatj-hatk`