The points D, E, F divide BC, CA and AB of th | vedclass

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

Question

(B) Let $\overrightarrow{AB} = \vec{c}$ and $\overrightarrow{AC} = \vec{b}$.Then $\overrightarrow{AD} = \frac{4\overrightarrow{AB} + 1\overrightarrow{

Vedclass · Accepted Answer

$2:5$

Vedclass · Answer

(B) Let $\overrightarrow{AB} = \vec{c}$ and $\overrightarrow{AC} = \vec{b}$.Then $\overrightarrow{AD} = \frac{4\overrightarrow{AB} + 1\overrightarrow{AC}}{1+4} = \frac{4\vec{c} + \vec{b}}{5}$.Since $E$ divides $CA$ in ratio $3:2$,$\overrightarrow{AE} = \frac{2\overrightarrow{AC}}{3+2} = \frac{2\vec{b}}{5}$.Thus,$\overrightarrow{BE} = \overrightarrow{AE} - \overrightarrow{AB} = \frac{2\vec{b}}{5} - \vec{c} = \frac{2\vec{b} - 5\vec{c}}{5}$.Since $F$ divides $AB$ in ratio $3:7$,$\overrightarrow{AF} = \frac{3\overrightarrow{AB}}{3+7} = \frac{3\vec{c}}{10}$.Thus,$\overrightarrow{CF} = \overrightarrow{AF} - \overrightarrow{AC} = \frac{3\vec{c}}{10} - \vec{b} = \frac{3\vec{c} - 10\vec{b}}{10}$.Now,$\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = \frac{4\vec{c} + \vec{b}}{5} + \frac{2\vec{b} - 5\vec{c}}{5} + \frac{3\vec{c} - 10\vec{b}}{10} = \frac{8\vec{c} + 2\vec{b} + 4\vec{b} - 10\vec{c} + 3\vec{c} - 10\vec{b}}{10} = \frac{\vec{c} - 4\vec{b}}{10}$.Point $K$ divides $AB$ in ratio $1:3$,so $\overrightarrow{AK} = \frac{1}{4}\overrightarrow{AB} = \frac{\vec{c}}{4}$.Then $\overrightarrow{CK} = \overrightarrow{AK} - \overrightarrow{AC} = \frac{\vec{c}}{4} - \vec{b} = \frac{\vec{c} - 4\vec{b}}{4}$.Ratio = $\frac{(\vec{c} - 4\vec{b})/10}{(\vec{c} - 4\vec{b})/4} = \frac{4}{10} = \frac{2}{5}$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

The points $D, E, F$ divide $BC, CA$ and $AB$ of the triangle $ABC$ in the ratio $1:4, 3:2$ and $3:7$ respectively and the point $K$ divides $AB$ in the ratio $1:3$,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK}$ is equal to

Similar Questions

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non-coplanar vectors. If $m$ and $n$ are scalars such that $\vec{a}+\vec{b}=m \vec{d}-\vec{c}$ and $\vec{b}+\vec{c}=n \vec{a}-\vec{d}$,then $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=$

The vector that is parallel to the vector $2 \hat{i} - 2 \hat{j} - 4 \hat{k}$ and coplanar with the vectors $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ is

Let $\overrightarrow{OA}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\overrightarrow{OB}=\hat{i}-4 \hat{j}-3 \hat{k}$,and $\overrightarrow{OC}=-3 \hat{i}+\hat{j}+2 \hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $G$ is the centroid of triangle $ABC$,then $BC^2+CA^2+AB^2+9(OG)^2=$

Given that $a$ and $b$ are two unit non-collinear vectors,if $u = a - (a \cdot b)b$ and $v = a \times b$,then find $|v| =$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module