1. Given vectors

$\vec A = \hat i + \hat j = (1,1), \qquad \vec B = 2\hat i - \hat j = (2,-1)$

We need a coplanar vector $\vec C = (x,y)$ such that

$\vec A\cdot \vec C = \vec B\cdot \vec C = \vec A\cdot \vec B$

2. First compute $\vec A\cdot \vec B$

$\vec A\cdot \vec B = (1)(2) + (1)(-1) = 2-1=1$

So the conditions become:

$\vec A\cdot \vec C = 1, \qquad \vec B\cdot \vec C = 1$

3. Let

$\vec C = x\hat i + y\hat j = (x,y)$

Then

$\vec A\cdot \vec C = x+y = 1$

and

$\vec B\cdot \vec C = 2x-y = 1$

So we solve the system:

$x+y=1$ $2x-y=1$

Adding both equations:

$3x=2 \Rightarrow x=\frac{2}{3}$

Then

$y=1-x=1-\frac{2}{3}=\frac{1}{3}$

Hence,

$\vec C = \frac{2}{3}\hat i + \frac{1}{3}\hat j$

4. Find magnitude of $\vec C$

$|\vec C| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2}$

$= \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}}$

$|\vec C| = \sqrt{\frac{5}{9}}$

5. Option check

• A: $\sqrt{\frac{10}{9}}$ ❌
• B: $\sqrt{\frac{5}{9}}$ ✅
• C: $\sqrt{\frac{20}{9}}$ ❌
• D: $\sqrt{\frac{9}{12}}$ ❌

Therefore, the correct answer is Option B.