If D, E and F are the mid-points of the sides | vedclass

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

Question

(B) Let the position vectors of $A, B, C, D, E, F$ be $\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{e}, \overline{f}$ respectivel

Vedclass · Accepted Answer

$\frac{1}{2} \overline{AC}$

Vedclass · Answer

(B) Let the position vectors of $A, B, C, D, E, F$ be $\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{e}, \overline{f}$ respectively.$	herefore \overline{d} = \frac{\overline{b} + \overline{c}}{2}, \overline{e} = \frac{\overline{c} + \overline{a}}{2}, \overline{f} = \frac{\overline{a} + \overline{b}}{2}$.Now,$\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} = (\overline{d} - \overline{a}) + \frac{2}{3}(\overline{e} - \overline{b}) + \frac{1}{3}(\overline{f} - \overline{c})$.Substituting the values:$= \frac{\overline{b} + \overline{c}}{2} - \overline{a} + \frac{2}{3}\left(\frac{\overline{c} + \overline{a}}{2} - \overline{b}ight) + \frac{1}{3}\left(\frac{\overline{a} + \overline{b}}{2} - \overline{c}ight)$.$= \frac{\overline{b} + \overline{c} - 2\overline{a}}{2} + \frac{\overline{c} + \overline{a} - 2\overline{b}}{3} + \frac{\overline{a} + \overline{b} - 2\overline{c}}{6}$.$= \frac{3(\overline{b} + \overline{c} - 2\overline{a}) + 2(\overline{c} + \overline{a} - 2\overline{b}) + (\overline{a} + \overline{b} - 2\overline{c})}{6}$.$= \frac{3\overline{b} + 3\overline{c} - 6\overline{a} + 2\overline{c} + 2\overline{a} - 4\overline{b} + \overline{a} + \overline{b} - 2\overline{c}}{6}$.$= \frac{-3\overline{a} + 3\overline{c}}{6} = \frac{3(\overline{c} - \overline{a})}{6} = \frac{1}{2}(\overline{c} - \overline{a}) = \frac{1}{2} \overline{AC}$.

If $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ of triangle $ABC$ respectively,then $\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} =$

Similar Questions

If $a, b, c$ are three non-coplanar vectors such that $a + b + c = \alpha d$ and $b + c + d = \beta a$,then $a + b + c + d$ is equal to

Write down a unit vector in the $XY$-plane,making an angle of $30^{\circ}$ with the positive direction of the $x$-axis.

$A$ and $B$ are two points. The position vector of $A$ is $6b - 2a$. $A$ point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,then the position vector of $B$ is given by:

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}|=3$,$|\bar{b}|=2$ and $\bar{a} \cdot \bar{b}=5$,then $|\bar{a}-\bar{b}|=$

Find the values of $x, y$ and $z$ so that the vectors $\vec{a} = x \hat{i} + 2 \hat{j} + z \hat{k}$ and $\vec{b} = 2 \hat{i} + y \hat{j} + \hat{k}$ are equal.

Vedclass Test Series

Exam Paper Generator

Online Exam Module