# Vector Algebra (Part-4) | S N Dey | Class 12

In the previous article, we have discussed few Long Answer Type Questions (1-10) and solutions of S N De Vector Algebra Chapter. In this article , we will solve the problems of Long Answer Type questions (11-20) as given in the Chhaya Publication Book of Vector Chapter. So, let's start.

$~11.~$ By vector method show that the figure formed by joining the mid points of a quadrilateral is parallelogram.

Solution.

Let $~ABCD~$ be a parallelogram and the position vectors of $~A,~B,~C,~D~$ are $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ respectively.

Further suppose that $~P,~Q,~R,~S~$ are the mid-points of $~AB,~BC,~CD,~DA~$ respectively.

$~\therefore ~$ the position vectors of $~P,~Q,~R,~S~$ are given by :

$~ \frac{\vec{a}+\vec{b}}{2},~\frac{\vec{b}+\vec{c}}{2},~\frac{\vec{c}+\vec{d}}{2},~\frac{\vec{d}+\vec{a}}{2}.$

$~\text{Now,}~ \vec{PS}\\=\frac{\vec{d}+\vec{a}}{2}-\frac{\vec{a}+\vec{b}}{2}\\=\frac{\vec{d}+\vec{a}-\vec{a}-\vec{b}}{2}\\=\frac{\vec{d}-\vec{b}}{2}\rightarrow(1)$

$~\text{Again,}~\vec{QR}\\=\frac{\vec{c}+\vec{d}}{2}-\frac{\vec{b}+\vec{c}}{2}\\=\frac{\vec{c}+\vec{d}-\vec{b}-\vec{c}}{2}\\=\frac{\vec{d}-\vec{b}}{2} \rightarrow(2)$

Hence, from $~(1),~(2)~$ , it follows that

$~ \vec{PS}=\vec{QR} \\~ \therefore \vec{PS} \parallel \vec{QR} \ ~\text{and}~|\vec{PS}|=|\vec{QR}|\rightarrow(3).$

Hence, from $~(3)~$ it follows that the figure formed by joining the mid points of a quadrilateral is parallelogram.

$~12.~$ Using vector method show that the diagonals of a parallelogram bisect each other.

Solution.

Let $~ABCD~$ be a parallelogram and $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ be the position vectors of the points $~A,~B,~C,~D~$ respectively.

$~\therefore~ \vec{AB}=\vec{DC} \\ \text{or,}~~ \vec{b}-\vec{a}=\vec{c}-\vec{d} \\ \text{or,}~~ \vec{b}+\vec{d}=\vec{c}+\vec{a} \\ \text{or,}~~ \frac{\vec{b}+\vec{d}}{2}=\frac{\vec{c}+\vec{a}}{2} \\ \therefore~ \text{the mid point of }~~ \vec{BD} =\text{ the mid point of }~~ \vec{AC} \rightarrow(1)$

Hence, from $~(1),~$ we can conclude that the diagonals of a parallelogram bisect each other.

$~13.~~ABCD~$ is a parallelogram and $~P~$ is the midpoint of the side $~\overline{BC}.~$ Prove that $~\overline{AC}~$ and $~\overline{DP}~$ meet in a common point of trisection .

Solution.

Let $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ be the position vectors of the points $~A,~B,~C,~D~$ respectively.

$~\because~~ABCD~$ is a parallelogram so,

$~~ \vec{AD}=\vec{BC} \\ \therefore~ \vec{d}-\vec{a}=\vec{c}-\vec{b} \\ \text{or,}~~ \vec{b}+\vec{d}=\vec{a}+\vec{c} \\ \text{or,}~~ \vec{c}+\vec{b}+\vec{d}=\vec{a}+\vec{c}+\vec{c} \\ \text{or,}~~ \vec{c}+\vec{b}+\vec{d}=\vec{a}+2\vec{c}\rightarrow(1)$

Let $~R_1~$ divides $~\overline{DP}~$ in the ratio $~2 :1 ~$ and $~R_2~$ divides $~\overline{AC}~$ in the ratio $~2 :1 ~$

$~\therefore~$ the position vector of $(R_1)$

$~=\frac{2 \cdot\frac{\vec{c}+\vec{b}}{2}+\vec{d}}{2+1}=\frac{\vec{c}+\vec{b}+\vec{d}}{3}$

and the position vector of $(R_2)$

$~=\frac{2 \vec{c}+\vec{a}}{2+1}=\frac{\vec{c}+\vec{b}+\vec{d}}{3}~[\text{By (1)}]$

Hence, $~R_1$ and $~R_2~$ denote the same point and so $~\overline{AC}~$ and $~\overline{DP}~$ meet in a common point of trisection.

$~14.~~ABCD~$ is a parallelogram ; $~P~$ and $~Q~$ are the mid-points of the sides $~\overline{AB}~$ and $~\overline{DC}~$ respectively. Show that $~\overline{DP}~$ and $~\overline{BQ}~$ trisect and are trisected by $~\overline{AC}.$

Solution.

Let $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ be the position vectors of the points $~A,~B,~C,~D~$ respectively.

$~\because~~ABCD~$ is a parallelogram,

$~~\frac{\vec{a}+\vec{c}}{2}=\frac{\vec{b}+\vec{d}}{2} \Rightarrow \vec{a}+\vec{c}=\vec{b}+\vec{d}\rightarrow(1)$

Let us consider that $~R_1~$ divides $~\overline{DP}~$ in the ratio $~ 2:1.$

$~\therefore~$ the position vector of $~R_1~$ is given by :

$=\frac{2 \cdot \frac{\vec{a}+\vec{b}}{2}+\vec{d}}{2+1}\\~~~~=\frac{\vec{a}+\vec{b}+\vec{d}}{3}\\~~~~=\frac{\vec{a}+\vec{a}+\vec{c}}{3}~~[\text{By (1)}]\\~~~~=\frac{2 \cdot \vec{a}+1 \cdot \vec{c}}{2+1}\rightarrow(2)$

By $~(2),~$ we can say that $~R_1~$ divides $~\overline{AC}~$ in the ratio $~ 1:2.$

Again, suppose that $~R_2~$ divides $~\overline{BQ}~$ in the ratio $~ 2:1.$

$~\therefore~$ the position vector of $~R_2~$ is given by :

$=\frac{2 \cdot \frac{\vec{d}+\vec{c}}{2}+\vec{b}}{2+1}\\~~~~=\frac{\vec{d}+\vec{c}+\vec{b}}{3}\\~~~~=\frac{\vec{c}+(\vec{b}+\vec{d})}{3}\\~~~~=\frac{\vec{c}+\vec{a}+\vec{c}}{3}~~[\text{By (1)}]\\~~~~=\frac{2 \cdot \vec{c}+1 \cdot \vec{a}}{2+1}\rightarrow(3)$

By $~(3),~$ we can say that $~R_2~$ divides $~\overline{AC}~$ in the ratio $~ 2:1.$

Hence, $~\overline{DP}~$ and $~\overline{BQ}~$ trisect and are trisected by $~\overline{AC}.$

$~15.~~ABCD~$ is a parallelogram and $~P~$ is the mid-point of $~\overline{DC}~$ . If $~Q~$ is a point on $~\overline{AP},~$ such that $~~\overline{AQ}=\frac 23\overline{AP},~$ show that $~Q~$ lies on the diagonal $~\overline{BD}~$ and $~\overline{BQ}=\frac23 \overline{BD}.$

Solution.

Since $~ABCD~$ is a parallelogram,

$~\vec{AB}=\vec{DC} \\ \therefore ~\vec{b}-\vec{a}=\vec{c}-\vec{d}\\ \text{or,}~~\vec{b}+\vec{d}=\vec{a}+\vec{c}\rightarrow(1)$

Again, since $~P~$ is the mid-point of $~\overline{DC}~$, the position vector of $~P~$ is given by : $~\frac{\vec{c}+\vec{d}}{2}$

If $~Q~$ is a point on $~\overline{AP},~$such that $~~\overline{AQ}=\frac 23\overline{AP},~$ then $~Q~$ divides $~\overline{AP}~$ in the ratio $~2:1.$

So, the position vector of $~Q~$ is given by :

$~~=\frac{2 \cdot \frac{\vec{c}+\vec{d}}{2}+\vec{a}}{2+1}\\~=\frac{\vec{c}+\vec{d}+\vec{a}}{3}\\~=\frac{(\vec{a}+\vec{c})+\vec{d}}{3} \\~=\frac{\vec{b}+\vec{d}+\vec{d}}{3} ~~[\text{By (1)}]\\~= \frac{2 \cdot \vec{d}+1 \cdot\vec{b}}{2+1}\rightarrow(2)$

Hence, by $~(2)~$ we can conclude that $~Q~$ divides $~\overline{BD}~$ in the ratio $~2:1~$ and so finally we can say that $~Q~$ lies on the diagonal $~\overline{BD}~$ and $~\overline{BQ}=\frac23 \overline{BD}.$

$~16.~~D,~E,~F~$ are the midpoints of the sides $~\overline{BC},~\overline{CA}~$ and $~\overline{AB}~$ respectively of the triangle $~ABC.~$ If $~P~$ is any point in the plane of the triangle, show that $~\vec{PA}+\vec{PB}+\vec{PC}=\vec{PD}+\vec{PE}+\vec{PF}.$

Solution.

Let the position vectors of the points $~A,~B,~C~$ with respect to the point $~P~$ be $~~\vec{a},~\vec{b},~\vec{c}~~$ respectively.

So, the position vectors of $~D,~E,~F~$ are $~\frac{\vec{b}+\vec{c}}{2},~\frac{\vec{a}+\vec{c}}{2},~\frac{\vec{a}+\vec{b}}{2}~$ respectively.

$~\therefore~ \vec{PD}+\vec{PE}+\vec{PF}\\~~~~~=\frac{\vec{b}+\vec{c}}{2}+\frac{\vec{a}+\vec{c}}{2}+\frac{\vec{a}+\vec{b}}{2}\\~~~~~=\frac{\vec{b}+\vec{c}+\vec{a}+\vec{c}+\vec{a}+\vec{b}}{2}\\~~~~~=\frac{2(\vec{a}+\vec{b}+\vec{c})}{2}\\~~~~~=\vec{a}+\vec{b}+\vec{c}\\~~~~~=\vec{PA}+\vec{PB}+\vec{PC} ~\text{(proved)}$

$~17.~~C~$ is the midpoint of the line segment $~\overline{AB}~$ and $~O~$ is any point outside $~AB~;~$ show that $~\vec{OA}+\vec{OB}=2~\vec{OC}.$

Solution.

From $~\Delta OAC~$, we get

$~\vec{OA}+\vec{AC}=\vec{OC}\rightarrow(1)$

Again, from $~\Delta BOC~$, we get

$~\vec{OC}+\vec{CB}=\vec{OB}\rightarrow(2)$

$~\vec{OA}+\vec{OB}\\=(\vec{OC}-\vec{AC})+(\vec{OC}+\vec{CB}) ~~[\text{By (1),(2)}]\\= \vec{OC}-\vec{AC}+\vec{OC}+\vec{CB} \\ =2\vec{OC}+\vec{CB}-\vec{AC} \\ =2\vec{OC}+(\vec{CB}-\vec{AC})\\=2\vec{OC}+(\vec{AC}-\vec{AC})~[\because~\vec{CB}=\vec{AC}]\\=2\vec{OC}~\text{(proved)}$

$~18.~~G~$ is a point inside the plane of the triangle $~ABC~;~$ if $~\vec{GA}+\vec{GB}+\vec{GC}=\vec{0}~,~$ then show that $~G~$ is the centroid of the triangle $~ABC.$

Solution.

Let the position vector of $~G~$ with respect to origin be $~\vec{g}~$ and also suppose that the position vectors of $~A,B,C~$ are $~\vec{a},\vec{b},\vec{c}~$ respectively.

$~\vec{GA}+\vec{GB}+\vec{GC}=\vec{0} \\ \text{or,}~~ \vec{a}-\vec{g}+\vec{b}-\vec{g}+\vec{c}-\vec{g}=\vec{0} \\ \text{or,}~~ -3\vec{g}+(\vec{a}+\vec{b}+\vec{c})=\vec{0} \\ \text{or,}~~-3\vec{g}=-(\vec{a}+\vec{b}+\vec{c}) \\ \therefore~ \vec{g}=\frac 13(\vec{a}+\vec{b}+\vec{c})\rightarrow(1)$

Hence, from $~(1)~$ we get the position vector of $~G~$ and so $~G~$ is the centroid of the triangle $~ABC.$

$~19.~$ The diagonals of the parallelogram $~ABCD~$ intersect at $~E.~$ If $~\vec{a},\vec{b},\vec{c}~$ and $~\vec{d}~$ be the position vectors of its vertices with respect to an arbitrary origin $~O,~$ then show that $~\vec{a}+\vec{b}+\vec{c}+\vec{d}=4\vec{OE}.$

Solution.

Since diagonals of the parallelogram $~\vec{AC},~\vec{BD}~$ bisect each other at $~E~$, so the position vector of the midpoint $~E~$ of the diagonal $~\vec{AC}~$ is given by :

$~\vec{OE}=\frac{\vec{a}+\vec{c}}{2} \Rightarrow \vec{a}+\vec{c}=2\vec{OE}\rightarrow(1)$

Again, the position vector of the midpoint $~E~$ of the diagonal $~\vec{BD}~$ is given by :

$~\vec{OE}=\frac{\vec{b}+\vec{d}}{2} \Rightarrow \vec{b}+\vec{d}=2\vec{OE}\rightarrow(2)$

Hence, adding $~(1),~(2)~$ we get,

$~(\vec{a}+\vec{c})+(\vec{b}+\vec{d})=2\vec{OE}+2\vec{OE} \\ \therefore~ \vec{a}+\vec{b}+\vec{c}+\vec{d}=4\vec{OE} ~~\text{(showed)}$

$~20.~$ By vector method prove that the straight line joining the midpoints of two non-parallel sides of a trapezium is parallel to the parallel sides and half of their sum.

Solution.

Let $~ABCD~$ be a trapezium and position vectors of $~A,~B,~C,~D~$ with respect to origin ($~O~$) are $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ respectively and $~\vec{AD}\parallel \vec{BC}~~\text{(say)}.$

$\text{Since}~\vec{AD} \parallel \vec{BC}~,~\lambda \vec{AD}=\vec{BC}\\ \Rightarrow \lambda(\vec{d}-\vec{a})=(\vec{c}-\vec{b})\rightarrow(1)$

The position vector of $~E~$ (midpoint of $~\vec{AB}~$) is given by $~\frac12(\vec{a}+\vec{b}).$

The position vector of $~F~$ (midpoint of $~\vec{DC}~$) is given by $~\frac12(\vec{d}+\vec{c}).$

$~\vec{EF}\\=\frac 12(\vec{d}+\vec{c})-\frac 12(\vec{a}+\vec{b})\\=\frac 12[(\vec{d}-\vec{a})+(\vec{c}-\vec{b})]\\=\frac 12[(\vec{d}-\vec{a})+\lambda(\vec{d}-\vec{a})]\\=(\vec{d}-\vec{a}) \times \frac 12(\lambda +1)~~[\text{By (1)}]\\=\frac 12(\lambda+1) \cdot \vec{AD} \rightarrow(2)$

From $~(1),~$ we can say $~\vec{EF} \parallel \vec{AD}~$ but $~\vec{AD}\parallel \vec{BC}~$ and so $~\vec{AD}\parallel \vec{BC} \parallel \vec{EF}$

$~\vec{EF}\\=\frac 12(\vec{d}+\vec{c})-\frac 12(\vec{a}+\vec{b})\\=\frac 12[(\vec{d}-\vec{a})+(\vec{c}-\vec{b})]\\=\frac 12(\vec{AD}+\vec{BC})\rightarrow(3)$

So, by $~(3)~$ we get, $~|\vec{EF}|=\frac 12|\vec{AD}+\vec{BC}|$

Hence the result follows.

WBCHSE | HS 2022 MATH QUESTION PAPER | PART-A

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