\(\mathrm{A}\) s position vector is \(\underline{\mathrm{a}}+2 \underline{\mathrm{b}} \mathrm{P}\) s position vector is \(\mathrm{a} \cdot \mathrm{P}\) is division point of \(\underline{A B}\) from \(A\) in the ratio \(2: 3\).find \(B\) s position vector is (a) \(2 \underline{a}-\underline{b}\) (b) \(\underline{b}-2 \underline{a}\) (c) \(\underline{a}-3 \underline{b}\) (d) \(\underline{b}\)

When a point P divides a line segment with endpoints A and B in ratio m:n, the position vector of point P can be expressed using the section formula: \(\underline{p}= \frac{m\underline{b} + n\underline{a}}{m+n}\) In this exercise, point P divides line segment AB in the ratio 2:3, so we can write: \(\underline{p}= \frac{2\underline{b} + 3\underline{a}}{2+3}\)

We are given that the position vector of point A is \(\underline{a} + 2\underline{b}\) and the position vector of point P is \(\underline{a}\cdot \underline{p}\). Plug these values into the section formula: \(\underline{a}\cdot \underline{p} = \frac{2\underline{b} + 3(\underline{a} + 2\underline{b})}{2+3}\) Now, solve the equation for \(\underline{b}\): \(\underline{a}\cdot \underline{p} = \frac{2\underline{b} + 3\underline{a} + 6\underline{b}}{5}\) \(\underline{a}\cdot \underline{p} = \frac{8\underline{b} + 3\underline{a}}{5}\) Multiply both sides by 5: \(5\underline{a}\cdot \underline{p} = 8\underline{b} + 3\underline{a}\) Subtract 3 \(\underline{a}\) from both sides: \(2\underline{a}\cdot \underline{p} = 8\underline{b}\) Divide both sides by 8: \(\frac{2}{8}\underline{a}\cdot \underline{p} = \underline{b}\) \(\frac{1}{4}\underline{a}\cdot \underline{p} = \underline{b}\) This solution does not match any of the options in the exercise. There might be a mistake in the given position vector for point P, since it is unclear how the dot product is used in the given expression. Without further clarification, the solution cannot be matched to the given options.