Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2a^{\to}+b^{\to}) and (a^{\to}-3b^{\to}) respectively, externally in the ratio 1:2. Also, show that P is the midpoint of the line segment R.

CBSE Class 12 Maths 2010 Solved Paper - Question 17

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↖{→})

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 e

(2a↖{→}+b↖{→})

Position vector of point Q is

(a↖{→}-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}

{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 =

{\Point \vector \of \R + \Position \vector \of \q}/2

{(3a↖{→}+5b↖{→})+(a↖{→}-3b↖{→})}/2

(2a↖{→}+b↖{→})

= Position vector of P (given)
Hence proved.

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