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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.
Solution
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+mbara`
`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`
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