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Vector Algebra

Question
CBSEENMA12035915
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Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are   2a→ + b→   and   a→ - 3b→  respectively, externally in the ratio 1:2. Also, show that P is the midpoint of the line segment R.

Solution

Position vector of P is  2a→ +b→ Position vector of Q is  a→ -3b→ 

Point R divides the lines segment PQ externally in a ratio of 1 : 2 .

Position vector of R = 1 ( a→ -3b→ ) -2 (  2a→ +b→ ) 1 - 2                               =  a→ -3b→ - 4a→ - 2b→1 - 2                               = 3a→ +5b→

Now, we need to show that P is the mid-point of RQ.

So, position vector of P =  position vector of R +  position vector of Q2                                     =  3a→ + 5b→  +  a→ - 3b→ 2                                     =   2a→ +b→ = Position vector  ( given )

Hence proved.

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