If M and N are the mid-points of the sides BC | vedclass

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

Question

(C) Let $\vec{AB} = \vec{a}$ and $\vec{AD} = \vec{b}$. Since $ABCD$ is a parallelogram,$\vec{BC} = \vec{AD} = \vec{b}$ and $\vec{CD} = \vec{AB} = \vec

Vedclass · Accepted Answer

$\frac{3}{2} AC$

Vedclass · Answer

(C) Let $\vec{AB} = \vec{a}$ and $\vec{AD} = \vec{b}$. Since $ABCD$ is a parallelogram,$\vec{BC} = \vec{AD} = \vec{b}$ and $\vec{CD} = \vec{AB} = \vec{a}$.Given that $M$ is the mid-point of $BC$,we have $\vec{BM} = \frac{1}{2} \vec{BC} = \frac{1}{2} \vec{b}$.Given that $N$ is the mid-point of $CD$,we have $\vec{DN} = \frac{1}{2} \vec{CD} = \frac{1}{2} \vec{a}$.In $	riangle ABM$,by triangle law of vector addition:$\vec{AM} = \vec{AB} + \vec{BM} = \vec{a} + \frac{1}{2} \vec{b}$.In $	riangle ADN$,by triangle law of vector addition:$\vec{AN} = \vec{AD} + \vec{DN} = \vec{b} + \frac{1}{2} \vec{a}$.Adding these two vectors:$\vec{AM} + \vec{AN} = (\vec{a} + \frac{1}{2} \vec{b}) + (\vec{b} + \frac{1}{2} \vec{a})$$= (1 + \frac{1}{2}) \vec{a} + (1 + \frac{1}{2}) \vec{b}$$= \frac{3}{2} \vec{a} + \frac{3}{2} \vec{b} = \frac{3}{2} (\vec{a} + \vec{b})$.Since $\vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b}$,we get:$\vec{AM} + \vec{AN} = \frac{3}{2} \vec{AC}$.

If $M$ and $N$ are the mid-points of the sides $BC$ and $CD$ respectively of a parallelogram $ABCD$,then $AM + AN$ equals

Similar Questions

Find a vector of magnitude $5$ units,and parallel to the resultant of the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}.$

If $\vec{a}, \vec{b}, \vec{c}$ are $3$ vectors such that $|\vec{a}|=5, |\vec{b}|=8, |\vec{c}|=11$ and $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

The position vector of a point at a distance of $3\sqrt{11}$ units from $i - j + 2k$ on a line passing through the points $i - j + 2k$ and $3i + j + k$ is

$A$ vector $\vec{a}$ has components $2p$ and $1$ with respect to a rectangular Cartesian system. The system is rotated through a certain angle about the origin in the anti-clockwise sense. If $\vec{a}$ has components $p+1$ and $1$ with respect to the new system,then:

If the vectors represented by the sides $AB$ and $BC$ of the regular hexagon $ABCDEF$ are $a$ and $b$ respectively,then the vector represented by $\overrightarrow{AE}$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module