Let ABC be a triangle and bar{a}, bar{b}, bar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $D$ divides $BC$ in the ratio $2:3$, $E$ divides $AC$ in the ratio $2:1$, and $P$ divides $DE$ in the ratio $3:5$ internally.Using the

Vedclass · Accepted Answer

$\frac{1}{8}(2 \bar{a}+3 \bar{b}+3 \bar{c})$

Vedclass · Answer

(A) Given that $D$ divides $BC$ in the ratio $2:3$, $E$ divides $AC$ in the ratio $2:1$, and $P$ divides $DE$ in the ratio $3:5$ internally.Using the section formula, the position vectors are:$\vec{d} = \frac{2\vec{c} + 3\vec{b}}{2+3} = \frac{2\vec{c} + 3\vec{b}}{5} \implies 5\vec{d} = 3\vec{b} + 2\vec{c} \quad (i)$$\vec{e} = \frac{2\vec{a} + 1\vec{c}}{2+1} = \frac{2\vec{a} + \vec{c}}{3} \implies 3\vec{e} = 2\vec{a} + \vec{c} \quad (ii)$Now, $P$ divides $DE$ in the ratio $3:5$, so its position vector $\vec{p}$ is:$\vec{p} = \frac{3\vec{e} + 5\vec{d}}{3+5} = \frac{3\vec{e} + 5\vec{d}}{8}$Substitute the expressions for $5\vec{d}$ and $3\vec{e}$ from equations $(i)$ and $(ii)$:$\vec{p} = \frac{(2\vec{a} + \vec{c}) + (3\vec{b} + 2\vec{c})}{8}$$\vec{p} = \frac{2\vec{a} + 3\vec{b} + 3\vec{c}}{8} = \frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$Thus, the correct option is $(A)$.

Similar Questions

If $\vec{a} + 5\vec{b} = \vec{c}$ and $\vec{a} - 7\vec{b} = 2\vec{c}$,then which of the following is true?

The sum of two forces is $18 \ N$ and the resultant,whose direction is at right angles to the smaller force,is $12 \ N$. The magnitudes of the two forces are:

Classify the following measure as a scalar or a vector:
$30 \, km/hr$

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module