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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 P(2a→+b→) and Q(a→-3b→) externally in the ratio 1: 2. Also, show that P is the mid

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Question

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are `P(2veca + vecb)` and `Q(veca - 3vecb)` externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Sum
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Solution

It is given that `vec(OP) = 2veca + vecb, vec(OQ) = veca - 3vecb`

Given that point R divides a line segment joining two points P and Q in the ratio 1:2, we get using the section formula.

`vec("OR") = (2(2veca + vecb) - (veca - 3vecb))/((2 - 1))`

`= (4veca + 2vecb - veca + 3vecb)/1 = 3veca + 5`

Therefore, the position vector of point R is `3veca + 5vecb`

Position vector of midpoint of RQ = `((vec(OQ) + vec(OR)))/2`

= `((veca - 3vecb) + (3veca + 5vecb))/2`

= `2veca + vecb`

= `vec(OP)`

Therefore, P is the midpoint of the line segment RQ.

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Chapter 10: Vector Algebra - Exercise 10.5 [Page 458]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 10 Vector Algebra
Exercise 10.5 | Q 9 | Page 458
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