Exercise 8.1 - Chapter 8 Vector Algebra–I 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Vector Algebra - I
Ex 8.1
Question 1 .
Represent graphically the displacement of
(i) $45 \mathrm{~cm} 30^{\circ}$ north of east
(ii) $80 \mathrm{~km}, 60^{\circ}$ south of west
Solution:
(i) $45 \mathrm{~cm} 300$ north of east

$text { (ii) } 80 \mathrm{~km} 60^{\circ} \text { south of west }$

Question 2.
Prove that the relation $\mathrm{R}$ defined on the set $\mathrm{V}$ of all vectors by $\vec{a} \mathrm{R} \vec{b}$ if $\vec{a}=\vec{b}$ is an equivalence relation on $\mathrm{V}$.
Solution:
$\vec{a} \mathrm{R} \vec{b}$ is given as $\vec{a}=\vec{b}$
(i) $\vec{a}=\vec{a} \Rightarrow \vec{a} \mathrm{R} \vec{a}$
(i.e.,) the relation is reflexive.
(ii) $\vec{a}=\vec{b} \Rightarrow \vec{b}=\vec{a}$
(i.e.,) $\vec{a} \mathrm{R} \vec{b}-\vec{b} \mathrm{R} \vec{a}$
So, the relation is symmetric.
(iii) $\vec{a}=\vec{b} ; \vec{b}=\vec{c} \Rightarrow \vec{a}=\vec{c}$
(i.e,,) $\vec{a} \mathrm{R} \vec{b} ; \vec{b} \mathrm{R} \vec{c} \Rightarrow \vec{a} \mathrm{R} \vec{c}$
So the given relation is transitive
So, it is an equivalence relation.
Question 3.
Let $\vec{a}$ and $\vec{a}$ be the position vectors of the points $\mathrm{A}$ and B. Prove that the position vectors of the points
which trisects the line segment $\mathrm{AB}$ are $\frac{\vec{a}+2 \vec{b}}{3}$ and $\frac{\vec{b}+2 \vec{a}}{3}$.
Solution:

Given $\overrightarrow{\mathrm{OA}}=\vec{a}$ and $\overrightarrow{\mathrm{OB}}=\vec{b}$
Join $=\mathrm{AB}$
The points $\mathrm{P}$ and $\mathrm{Q}$ are on $\mathrm{AB}$ such that $\mathrm{AP}=\mathrm{PQ}=\mathrm{QB}$
To find the postion vertors of $P$ and $Q$
Now
$A P=P Q=Q B$
$\therefore \quad \mathrm{AP}=\frac{1}{2} \mathrm{~PB}$
OR
$2 \mathrm{AP}=\mathrm{PB}$
$\Rightarrow \quad \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{1}{2}$
(i.e.,) $\mathrm{AP}: \mathrm{PB}=1: 2$
$\Rightarrow \mathrm{P}$ divides $\mathrm{AB}$ internally in the ratio $1: 2$
$
\overrightarrow{\mathrm{OP}}=\frac{1(\vec{b})+2(\vec{a})}{1+2}
$
$
=\frac{2 \vec{a}+\vec{b}}{3}
$
IIIrly
$\mathrm{AQ}: \mathrm{QB}=2: 1$
\begin{tabular}{lcccc} $(\vec{a})$ & 2 & & 1 & $(\vec{b})$ \\ \hline $\mathrm{A}$ & $\mathrm{Q}$ & $\mathrm{B}$ \end{tabular}
$
\therefore \quad \overrightarrow{\mathrm{OQ}}=\frac{2(\vec{b})+(\vec{a})}{2+1}=\frac{\vec{a}+2 \vec{b}}{3}
$
So, the position vector of the point of trisection are $\frac{\vec{a}+2 \vec{b}}{3}$ and $\frac{2 \vec{a}+\vec{b}}{3}$.
Question 4.
If $\mathrm{D}$ and $\mathrm{E}$ are the midpoints of the sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$, prove that

$\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{DC}}=\frac{3}{2} \overrightarrow{\mathrm{BC}}$

Solution:
$
\begin{aligned}
& \text { In } \triangle \mathrm{ABC}, \overrightarrow{\mathrm{OA}}=\vec{a} ; \overrightarrow{\mathrm{OB}}=\vec{b}, \overrightarrow{\mathrm{OC}}=\vec{c} \\
& \mathrm{D}=\text { mid point of } \mathrm{AB} \\
& \Rightarrow \\
& \overrightarrow{\mathrm{OD}}=\frac{\vec{a}+\vec{b}}{2} \\
& \mathrm{E}=\operatorname{mid} \text { point of } \mathrm{AC} \\
& \Rightarrow \\
& \overrightarrow{\mathrm{OE}}=\frac{\vec{a}+\vec{c}}{2} \\
& \overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{DC}}=\overrightarrow{\mathrm{BD}}+\overrightarrow{\mathrm{DE}}+\overrightarrow{\mathrm{DE}}+\overrightarrow{\mathrm{EC}} \\
& =\frac{\overrightarrow{\mathrm{BA}}}{2}+2 \overrightarrow{\mathrm{DE}}+\frac{\overrightarrow{\mathrm{AC}}}{2} \\
& =\frac{\vec{a}-\vec{b}}{2}+\vec{c}-\vec{b}+\frac{\vec{c}-\vec{a}}{2} \\
& =\frac{3}{2}(\vec{c}-\vec{b})=\frac{3}{2} \overrightarrow{\mathrm{BC}} \\
& =\mathrm{RHS} \\
&
\end{aligned}
$
Note: We have used $\overrightarrow{\mathrm{DE}}=\frac{1}{2} \overrightarrow{\mathrm{BC}}$ from Basic proportionality theorem.
Question 5.
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.
Solution:

In $\triangle \mathrm{ABC}$
$
\begin{aligned}
\overrightarrow{\mathrm{OA}} & =\vec{a} \\
\overrightarrow{\mathrm{OB}} & =\vec{b} \text { and } \\
\overrightarrow{\mathrm{OC}} & =\vec{c}
\end{aligned}
$

$
\begin{aligned}
\mathrm{D} & =\text { mid point of } \mathrm{AB}=\overrightarrow{\mathrm{OD}}=\frac{\vec{a}+\vec{b}}{2} \\
\mathrm{E} & =\text { mid point of } \mathrm{AC}=\overrightarrow{\mathrm{OE}}=\frac{\vec{a}+\vec{c}}{2} \\
\overrightarrow{\mathrm{DE}} & =\overrightarrow{\mathrm{OE}}-\overrightarrow{\mathrm{OD}}=\frac{\vec{a}+\vec{c}}{2}-\frac{\vec{a}+\vec{b}}{2} \\
& =\frac{\vec{a}+\vec{c}-\vec{a}-\vec{b}}{2}=\frac{\vec{c}-\vec{b}}{2} \\
& =\frac{\overrightarrow{\mathrm{BC}}}{2} \\
\overrightarrow{\mathrm{DE}} & =\frac{\overrightarrow{\mathrm{BC}}}{2} \Rightarrow \overrightarrow{\mathrm{DE}} \| \text { to } \overrightarrow{\mathrm{BC}} \text { and half of } \mathrm{BC} .
\end{aligned}
$
Question 6.
Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Solution:

$\mathrm{ABCD}$ is a quadrilateral with $\overrightarrow{\mathrm{OA}}=\vec{a} ; \overrightarrow{\mathrm{OB}}=\vec{b}, \overrightarrow{\mathrm{OC}}=\vec{c}$ and $\overrightarrow{\mathrm{OD}}=\vec{d}$

$
\begin{aligned}
& \mathrm{P}=\text { mid point of } \mathrm{AB} \Rightarrow \overrightarrow{\mathrm{OP}}=\frac{\vec{a}+\vec{b}}{2} \\
& \mathrm{Q}=\text { mid point of } \mathrm{BC} \Rightarrow \overrightarrow{\mathrm{OQ}}=\frac{\vec{b}+\vec{c}}{2} \\
& \mathrm{R}=\text { mid point of } \mathrm{CD} \Rightarrow \overrightarrow{\mathrm{OR}}=\frac{\vec{c}+\vec{d}}{2} \\
& \mathrm{~S}=\text { mid point of } \mathrm{DA} \Rightarrow \overrightarrow{\mathrm{OS}}=\frac{\vec{d}+\vec{a}}{2}
\end{aligned}
$
To prove: $\mathrm{P} \mathrm{Q} \mathrm{R} \mathrm{S}$ is a parallelogram.

$
\begin{aligned}
\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}} & =\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}+\vec{b}}{2}=\frac{\vec{b}+\vec{c}-\vec{a}-\vec{b}}{2}=\frac{\vec{c}-\vec{a}}{2} \\
\overrightarrow{\mathrm{SR}}=\overrightarrow{\mathrm{OR}}-\overrightarrow{\mathrm{OS}} & =\frac{\vec{c}+\vec{d}}{2}-\frac{\vec{a}+\vec{d}}{2}=\frac{\vec{c}+\vec{d}-\vec{a}-\vec{d}}{2}=\frac{\vec{c}-\vec{a}}{2} \\
& \Rightarrow \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}} \\
\overrightarrow{\mathrm{PS}}=\overrightarrow{\mathrm{OS}}-\overrightarrow{\mathrm{OP}} & =\frac{\vec{d}+\vec{a}}{2}-\frac{\vec{a}+\vec{b}}{2}=\frac{\vec{d}-\vec{b}}{2} \\
\overrightarrow{\mathrm{QR}}=\overrightarrow{\mathrm{OR}}-\overrightarrow{\mathrm{OQ}} & =\frac{\vec{c}+\vec{d}}{2}-\frac{\vec{c}+\vec{b}}{2}=\frac{\vec{d}-\vec{b}}{2} \\
& \Rightarrow \overrightarrow{\mathrm{PS}}=\overrightarrow{\mathrm{QR}}
\end{aligned}
$
In a quadrilateral when opposite sides are equal and parallel it is a parallelogram So, PQRS is a parallelogram, from (1) and (2).
Question 7.
If $\vec{a}$ and $\vec{b}$ represent a side and a diagonal of a parallelogram, find the other sides and the other diagonal.

Solution:

$\mathrm{OABC}$ is a parallelogram where
$
\overrightarrow{\mathrm{OA}}=\vec{a} \text { and } \overrightarrow{\mathrm{OB}}=\vec{b}
$

$
\begin{aligned}
& \text { Now } \quad \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}} \\
& \text { (i.e.,) } \vec{b}=\vec{a}+\overrightarrow{\mathrm{AB}} \Rightarrow \vec{b}-\vec{a}=\overrightarrow{\mathrm{AB}} \\
& \therefore \quad \overrightarrow{\mathrm{AB}}=\vec{b}-\vec{a} ; \overrightarrow{\mathrm{OA}}=\vec{a} \Rightarrow \overrightarrow{\mathrm{CB}}=\vec{a} \\
& \overrightarrow{\mathrm{AB}}=\vec{b}-\vec{a} \Rightarrow \overrightarrow{\mathrm{OC}}=\vec{b}-\vec{a} \\
& \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=(\vec{b}-\vec{a})+(-\overrightarrow{\mathrm{CB}}) \\
& =\vec{b}-\vec{a}-\vec{a} \\
& =\vec{b}-2 \vec{a} \\
&
\end{aligned}
$

So the sides of the parallelogram are $\vec{a}, \vec{b}-\vec{a}, \vec{a}, \vec{b}-\vec{a}$ and the diagonals are $\vec{b}$ and $\vec{b}-2 \vec{a}$
Question 8.
If $\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}$, prove that the points $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are collinear.
Solution:

$
\begin{aligned}
& \overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{PQ}} \\
& \overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}=\overrightarrow{\mathrm{QR}} \\
& \overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}} \text { (given) } \\
& \therefore \quad \therefore \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{QR}} \Rightarrow \overrightarrow{\mathrm{PQ}} \| \overrightarrow{\mathrm{QR}} \\
&
\end{aligned}
$
But $\mathrm{Q}$ is a common point.
$\Rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{R}$ are collinear.
Question 9.
If $\mathrm{D}$ is the midpoint of the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$, prove that $\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A C}}=2 \overrightarrow{\mathbf{A D}}$

Solution:

$\mathrm{D}$ is the midpoint of $\triangle A B C$.

$
\begin{aligned}
& \Rightarrow \quad \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{DC}} \\
& \text { Join AD. } \\
& \text { Now } \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DB}} \\
& \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DC}} \\
& \mathrm{LHS}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DB}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DC}} \\
&=2 \overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DB}}+\overrightarrow{\mathrm{DC}} \\
&=2 \overrightarrow{\mathrm{AD}}+(\overrightarrow{0}) \\
& \therefore \quad[\because \overrightarrow{\mathrm{DB}} \text { and } \overrightarrow{\mathrm{DC}} \text { are equal and opposite vectors). } \\
&=2 \overrightarrow{\mathrm{AD}}=\mathrm{RHS}
\end{aligned}
$
Question 10.
If $\mathrm{G}$ is the centroid of a triangle $\mathrm{ABC}$, prove that $\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0$
Solution:
For any triangle $\mathrm{ABC}$,
$\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0$
Now $\mathrm{G}$ is the centroid of $\triangle A B C$, which divides the medians $(A D, B E$ and $C F$ ) in the ratio $2: 1$.

$
\begin{aligned}
\Rightarrow \overrightarrow{\mathrm{AG}}: \overrightarrow{\mathrm{GD}} & =2: 1 \\
\overrightarrow{\mathrm{GA}} & =-\frac{2}{3} \overrightarrow{\mathrm{AD}} \\
\text { Similarly } \overrightarrow{\mathrm{GB}} & =-\frac{2}{3} \overrightarrow{\mathrm{BE}} \text { and } \overrightarrow{\mathrm{GC}}=-\frac{2}{3} \overrightarrow{\mathrm{CF}} \\
\text { Now LHS } & =\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=-\frac{2}{3}[\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}}] \\
& =-\frac{2}{3}\left[\overrightarrow{\mathrm{AB}}+\frac{1}{2} \overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{BC}}+\frac{1}{2} \overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{CA}}+\frac{1}{2} \overrightarrow{\mathrm{AB}}\right] \\
& =-\frac{2}{3}\left\{\frac{3}{2}(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}})\right\} \\
& =-1(\overrightarrow{0})=\overrightarrow{0}=\mathrm{RHS} .
\end{aligned}
$
Method 2: $\mathrm{G}$ is the centroid of a triangle ABC.
Let $\mathrm{O}$ be the origin
$
\therefore \overrightarrow{\mathrm{OG}}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}
$

$
\begin{aligned}
& \overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OG}}+\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OG}}+\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OG}} \\
& =\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}-3 \overrightarrow{\mathrm{OG}} \\
& =\vec{a}+\vec{b}+\vec{c}-3 \overrightarrow{\mathrm{OG}} \\
& =\vec{a}+\vec{b}+\vec{c}-3\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right) \\
& \text { (i.e.) } \overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0 \\
&
\end{aligned}
$
Question 11.
Let $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ be the vertices of a triangle. Let $\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ be the midpoints of the sides $\mathrm{BC}, \mathrm{CA}$, and $\mathrm{AB}$ respectively. Show that $\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}}=\overrightarrow{0}$
Solution:
In $\triangle A B C, D, E, F$ are the midpoints of $\mathrm{BC}, \mathrm{CA}$ and $\mathrm{AB}$ respectively.

To Prove:
Now

$
\begin{aligned}
\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}} & =\overrightarrow{0} \\
\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BD}} & =\overrightarrow{\mathrm{AB}}+\frac{1}{2} \overrightarrow{\mathrm{BC}} \\
\overrightarrow{\mathrm{BE}}=\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CE}} & =\overrightarrow{\mathrm{BC}}+\frac{1}{2} \overrightarrow{\mathrm{CA}} \\
\overrightarrow{\mathrm{CF}}=\overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{AF}} & =\overrightarrow{\mathrm{CA}}+\frac{1}{2} \overrightarrow{\mathrm{AB}} \\
\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}} & =\frac{3}{2}(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}) \\
& =\frac{3}{2}(\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CA}}) \\
& =\frac{3}{2}(\overrightarrow{0})=\overrightarrow{0}
\end{aligned}
$

Question 12 .
If $\mathrm{ABCD}$ is a quadrilateral and $\mathrm{E}$ and $\mathrm{F}$ are the midpoints of $\mathrm{AC}$ and $\mathrm{BD}$ respectively, then prove that $\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}}=4 \overrightarrow{\mathrm{EF}}$
Solution:
$\mathrm{ABCD}$ is a quadrilateral in which $\mathrm{E}$ and $\mathrm{F}$ are the midpoints of $\mathrm{AC}$ and $\mathrm{BD}$ respectively.

$
\text { Now } \begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{EF}}+\overrightarrow{\mathrm{FB}} \\
\overrightarrow{\mathrm{AD}} & =\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{EF}}+\overrightarrow{\mathrm{FD}} \\
\overrightarrow{\mathrm{CB}} & =\overrightarrow{\mathrm{CE}}+\overrightarrow{\mathrm{EF}}+\overrightarrow{\mathrm{FB}} \\
\overrightarrow{\mathrm{CD}} & =\overrightarrow{\mathrm{CE}}+\overrightarrow{\mathrm{EF}}+\overrightarrow{\mathrm{FD}}
\end{aligned}
$
So, $\quad \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CD}}=4 \overrightarrow{\mathrm{EF}}+2(\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{CE}})+2(\overrightarrow{\mathrm{FB}}+\overrightarrow{\mathrm{FD}})$
$
=4 \overrightarrow{\mathrm{EF}}+\overrightarrow{0}+\overrightarrow{0}=4 \overrightarrow{\mathrm{EF}}
$
$[\because \overrightarrow{\mathrm{AE}}$ and $\overrightarrow{\mathrm{CE}}$ are equal and opposite vectors and $\overrightarrow{\mathrm{FB}}$ and $\overrightarrow{\mathrm{FD}}$ are equal and opposit vectors $]$.