OABCD is a pentagon in which the sides OA and | vedclass

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

Question

(D) Given,$\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$.Since $OA \parallel CB$ and $\frac{OA}{CB}=2$,we have $\vec{CB} = \frac{1}{2}\vec{OA} = \frac{1}{2}\vec

Vedclass · Accepted Answer

$6\vec{d}$

Vedclass · Answer

(D) Given,$\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$.Since $OA \parallel CB$ and $\frac{OA}{CB}=2$,we have $\vec{CB} = \frac{1}{2}\vec{OA} = \frac{1}{2}\vec{a}$.Since $OD \parallel AB$ and $\frac{OD}{AB}=\frac{1}{3}$,we have $\vec{AB} = 3\vec{OD} = 3\vec{d}$.Now,we express the required sum in terms of vectors $\vec{a}$ and $\vec{d}$:$\vec{AD} = \vec{OD} - \vec{OA} = \vec{d} - \vec{a}$.$\vec{OC} = \vec{OD} + \vec{DC}$.From the pentagon geometry,$\vec{OC} = \vec{OA} + \vec{AB} + \vec{BC} = \vec{a} + 3\vec{d} - \frac{1}{2}\vec{a} = \frac{1}{2}\vec{a} + 3\vec{d}$.Also,$\vec{DC} = \vec{OC} - \vec{OD} = (\frac{1}{2}\vec{a} + 3\vec{d}) - \vec{d} = \frac{1}{2}\vec{a} + 2\vec{d}$.Summing these: $\vec{AD} + \vec{OC} + \vec{DC} = (\vec{d} - \vec{a}) + (\frac{1}{2}\vec{a} + 3\vec{d}) + (\frac{1}{2}\vec{a} + 2\vec{d})$.$= (-\vec{a} + \frac{1}{2}\vec{a} + \frac{1}{2}\vec{a}) + (\vec{d} + 3\vec{d} + 2\vec{d}) = 0\vec{a} + 6\vec{d} = 6\vec{d}$.

$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. Also,it is given that $\frac{OA}{CB}=2$,$\frac{OD}{AB}=\frac{1}{3}$. If $\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$,then $\vec{AD}+\vec{OC}+\vec{DC}=$

Similar Questions

Let $\vec{a} = 2\hat{i}-\hat{j}-\hat{k}$,$\vec{b} = 5\hat{i}+\hat{j}-2\hat{k}$,and $\vec{c} = -13\hat{i}-11\hat{j}+4\hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $\vec{AB} = \lambda \vec{BC}$ and $\vec{AC} = \mu \vec{CB}$,then find the value of $\lambda + \mu$.

If $\bar{a}, \bar{b}, \bar{c}$ are non-zero vectors such that no two of them are parallel,and $\bar{a} + \bar{b}$ is parallel to $\bar{c}$,and $\bar{b} + \bar{c}$ is parallel to $\bar{a}$,then $\bar{a} + \bar{b} + \bar{c} = $

Let $|\vec{a}| = |\vec{b}| = |\vec{a} - \vec{b}| = 1$,then the angle between $\vec{a}$ and $\vec{b}$ is:

Let $\overrightarrow{OA}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\overrightarrow{OB}=-2 \hat{i}-3 \hat{j}+6 \hat{k}$ be the position vectors of two points $A$ and $B$. If $C$ is a point on the bisector of $\angle AOB$ and $OC=\sqrt{42}$,then $\overrightarrow{OC}=$

If the vectors $\overline{AB}=3 \hat{i}+4 \hat{k}$ and $\overline{AC}=5 \hat{i}-2 \hat{j}+4 \hat{k}$ are the sides of the triangle $ABC$,then the length of the median through $A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module