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

The Correct Option is B

Step 1: Recall the section formula for internal division. For a point R dividing PQ in ratio m:n, its position vector is:
\[ \vec{OR} = \frac{n\vec{OP} + m\vec{OQ}}{m+n} \]
Given ratio 1:2 (m=1, n=2), and position vectors:
\[ \vec{OP} = 2\vec{a} - \vec{b} \]
\[ \vec{OQ} = \vec{a} + 3\vec{b} \]

Step 2: Apply the section formula:
\[ \vec{OR} = \frac{2(2\vec{a} - \vec{b}) + 1(\vec{a} + 3\vec{b})}{1+2} \]
\[ = \frac{4\vec{a} - 2\vec{b} + \vec{a} + 3\vec{b}}{3} \]

Step 3: Simplify the expression:
\[ = \frac{5\vec{a} + \vec{b}}{3} \]
\[ = \frac{1}{3}(5\vec{a} + \vec{b}) \]

Conclusion: The position vector of point R is \(\boxed{B}\) \(\left( \frac{1}{3}(5\vec{a} + \vec{b}) \right)\).

Answer 2

Given 

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 \(\dfrac{(5a + b)}{3}\).