Question

The position vectors of two points P and Q are given by $OP=2a-b$ and $OQ=a+3b$ ,respectively.If a point R divides the line joining P and Q internally in the ratio $1:2$ ,then the position vector of the point R is ?

• A. $\dfrac{1}{3}(a-5b)$
• B. $\dfrac{1}{3}(5a+b)$
• C. $\dfrac{1}{3}(a-5b)$
• D. $\dfrac{1}{3}(a-5b)$
• E. $\dfrac{1}{3}(a-b)$

Answer 1

Answer: (B)

$\dfrac{1}{3}(5a+b)$

Answer 2

Given that

Position vector of, $P = OP = 2a - b$

Position vector of, $Q = OQ = a + 3b$

Now, let's find the position vector of R:

Position vector of $R = (2 ×$Position vector of P $+ 1 ×$ Position vector of Q)$/ (1 + 2)$

Position vector of $R = \dfrac{(2 × (2a - b) + 1 × (a + 3b))} {3}$

Position vector of $R = \dfrac{(4a - 2b + a + 3b)} {3}$

Position vector of $R = \dfrac{(5a + b) }{3}$

So, the position vector of point $R$ is $\dfrac{(5a + b)}{3}$. (_Ans)