P is the point of intersection of the diagona | vedclass

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

Question

(D) We know that in a parallelogram,the diagonals bisect each other. Therefore,$P$ is the midpoint of both diagonals $AC$ and $BD$.Using the section f

Vedclass · Accepted Answer

$4\,\overrightarrow{OP}$

Vedclass · Answer

(D) We know that in a parallelogram,the diagonals bisect each other. Therefore,$P$ is the midpoint of both diagonals $AC$ and $BD$.Using the section formula for the midpoint,for diagonal $AC$ with midpoint $P$,we have:$\overrightarrow{OA} + \overrightarrow{OC} = 2\overrightarrow{OP}$ ......$(i)$Similarly,for diagonal $BD$ with midpoint $P$,we have:$\overrightarrow{OB} + \overrightarrow{OD} = 2\overrightarrow{OP}$ ......$(ii)$Adding equations $(i)$ and $(ii)$,we get:$(\overrightarrow{OA} + \overrightarrow{OC}) + (\overrightarrow{OB} + \overrightarrow{OD}) = 2\overrightarrow{OP} + 2\overrightarrow{OP}$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 4\overrightarrow{OP}$

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $O$ is any point,then $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = $

Similar Questions

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $S$ is any point in space and $\vec{SA} + \vec{SB} + \vec{SC} + \vec{SD} = \lambda \vec{SP}$,then $\lambda$ equals

If the length of a vector is $21$ and its direction ratios are $2, -3, 6$,then its direction cosines are:

$ABC$ is an isosceles triangle right-angled at $A$. Forces of magnitude $2\sqrt{2}$,$5$,and $6$ act along $\overrightarrow{BC}$,$\overrightarrow{CA}$,and $\overrightarrow{AB}$ respectively. The magnitude of their resultant force is

If $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + \hat{j} - \hat{k}$,and $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{p} = 2\vec{a} + \vec{b}$ and $\vec{q} = \vec{a} + 2\vec{b}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

Let $H$ be the orthocentre of an acute angled $\triangle ABC$ and $O$ be its circumcenter. Then,$\vec{HA} + \vec{HB} + \vec{HC}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module