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CBSE Class 12 Maths 2010 Solved Paper

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Question : 17 of 29
Marks: +1, -0
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2a→+b→)(\overrightarrow{2a} + \overrightarrow{b}) and (a→−3b→)(\overrightarrow{a} - \overrightarrow{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 e (2a→+b→)(\overrightarrow{2a} + \overrightarrow{b})
Position vector of point Q is (a→−3b→)(\overrightarrow{a} - \overrightarrow{3b})
Point R divides the line segment PQ externally in a ratio of 1 : 2.
Position vector of R = 1(a→−3b→)−2(2a→+b→)1−2\frac{1(\overrightarrow{a} - \overrightarrow{3b}) - 2(\overrightarrow{2a} + \overrightarrow{b})}{1-2}
= a→−3b→−4a→−2b→1−2\frac{\overrightarrow{a} - \overrightarrow{3b} - \overrightarrow{4a} - \overrightarrow{2b}}{1-2} = 3a→+5b→\overrightarrow{3a} + \overrightarrow{5b}
Now, we need to show that P is the mid-point of RQ.
So, Position vector of P =
Point vector of R+Position vector of q2\frac{\text{Point vector of R} + \text{Position vector of q}}{2}
= (3a→+5b→)+(a→−3b→)2\frac{(\overrightarrow{3a} + \overrightarrow{5b}) + (\overrightarrow{a} - \overrightarrow{3b})}{2} = (2a→+b→)(\overrightarrow{2a} + \overrightarrow{b}) = Position vector of P (given)
Hence proved.
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