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The position vector of the point which divides the join of points2 \vec{a}-3 \vec{b}_{\text {and }} \vec{a}+\vec{b}  in the ratio 3 : 1 is


\\A.\frac{3 \vec{a}-2 \vec{b}}{2}\\ B. \frac{7 \overrightarrow{\mathrm{a}}-8 \overrightarrow{\mathrm{b}}}{4} \\C.\frac{3 \overrightarrow{\mathrm{a}}}{4} \\D. \frac{5 \vec{a}}{4}

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D)

We are given points  2 \vec{a}-3 \vec{b}_{\text {and }} \vec{a}+\vec{b}

Let these points be

A(2 \vec{a}-3 \vec{b})_{a n d} B(\vec{a}+\vec{b}).

Also, given in the question that,

A point divides AB in the ratio of 3 : 1.

Let this point be C.

⇒ C divides AB in the ratio = 3 : 1

We need to find the position vector of C.

We know the position vector of a point C dividing the line segment joining the points P and Q, whose position vectors are p and q in the ratio m : n internally is given by,

\text { Position vector }=\frac{\mathrm{m} \overrightarrow{\mathrm{q}}+\mathrm{n} \overrightarrow{\mathrm{p}}}{\mathrm{m}+\mathrm{n}}

According to the question, here

m : n = 3 : 1

⇒ m = 3 and n = 1

\\ Also, $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$\\ And $\overrightarrow{\mathrm{p}}=2 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}

\\ $Substituting these values in the formula above, we get\\ Position vector of $C=\frac{3(\vec{a}+\vec{b})+1(2 \vec{a}-3 \vec{b})}{3+1}$ $\\=\frac{3 \vec{a}+3 \vec{b}+2 \vec{a}-3 \vec{b}}{4}$.

\begin{aligned} &=\frac{3 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{b}}}{4}\\ &=\frac{5 \vec{a}}{4}\\ &\text { Thus, position vector of the point is } \frac{5 \vec{a}}{4} \text { . } \end{aligned}

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