Exercise 23.2
Question 1
If P, Q and R are three collinear points such that $\overrightarrow{P Q}=\vec{a}$ and $\overrightarrow{Q R}=\vec{b}$. Find the vector $\overrightarrow{P R}$.
Sol :
As P, Q and R are three collinear points.
Hence, $\overrightarrow{\mathrm{PR}}=\overrightarrow{\mathrm{PQ}}+\overrightarrow{\mathrm{QR}}$ as shown in above fig
And given $\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{QR}}=\overrightarrow{\mathrm{b}}$
Therefore, $\overrightarrow{\mathrm{PR}}=\overrightarrow{\mathrm{PQ}}+\overrightarrow{\mathrm{QR}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$
Question 2
Give condition that three vectors $\vec{a}, \vec{b}$, and $\vec{c}$ form the three sides of a triangle. What are the other possibilities?
Sol :
Given that, $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are three sides of a triangle.
Hence from the above figure we get,
$\mathrm{AB}=\overrightarrow{\mathrm{a}}, \mathrm{BC}=\overrightarrow{\mathrm{b}}$ and $\mathrm{AC}=\overrightarrow{\mathrm{c}}$
So, $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$=AB+BC+CA=AC+CA
[Since AB+BC=AC]
=AC-AC=0 [Since CA=-AC]
Triangle law says that, if vectors are represented in magnitude and direction by the two sides of a triangle is same order, then their sum is represented by the third side took in reverse order. Thus,
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{c}}$ or $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}}$
Question 3
If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors having the same initial point. What are the vectors represented by $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ ?
Sol :
Here, it is given that $\vec{a}$ and $\vec{b}$ are two non-collinear vectors having the same initial point.
Let $\vec{a}=\overrightarrow{A B}$ and $\vec{b}=\overrightarrow{A D}$, So we can draw a parallelogram ABCD as above.
By the properties of parallelogram
$\overrightarrow{B C}=\vec{b}$ and $\overrightarrow{D C}=\vec{a}$
In ΔABC,
Using triangle law,
$\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}$
$\vec{a}+\vec{b}=\overrightarrow{A C}$...(i)
In ΔABD,
Using triangle law,
$\vec{b}+\overrightarrow{D B}=\vec{a}$
$\overrightarrow{D B}=\vec{a}-\vec{b}$...(ii)
From equation (i) and (ii), we get that
$\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are diagonals of a parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$
Question 4
If $\vec{a}$ is a vector and m is a scalar such that $m \vec{a}=\overrightarrow{0}$, then what are the alternatives for m and $\vec{a}$ ?
Sol :
Given $\vec{a}$ is a vector and m is a scalar such that $m \vec{a}=\overrightarrow{0}$
Let $\vec{a}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}$ then according to the given question
$m \vec{a}=\overrightarrow{0}$
$\Rightarrow \mathrm{m}\left(\mathrm{a}_{1} \hat{\imath}+\mathrm{b}_{1} \hat{\jmath}+\mathrm{c}_{1} \hat{\mathrm{k}}\right)=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}}$
$\Rightarrow\left(\mathrm{ma}_{1} \hat{\imath}+\mathrm{mb}_{1} \hat{\jmath}+\mathrm{mc}_{1} \hat{\mathrm{k}}\right)=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}}$
Compare the coefficients of $\hat{1}, \hat{\jmath}, \hat{k}$ we get
$\mathrm{ma}_{1}=0 \Rightarrow \mathrm{m}=0$ or $\mathrm{a}_{1}=0$
Similarly, $\mathrm{mb}_{1}=0 \Rightarrow \mathrm{m}=0$ or $\mathrm{b}_{1}=0$
And $m c_{1}=0 \Rightarrow m=0$ or $c_{1}=0$
From the above three conditions ,
m=0 or $a_{1}=b_{1}=c_{1}=0$
$\Rightarrow \mathrm{m}=0$ or $\overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \hat{\imath}+\mathrm{b}_{1} \hat{\jmath}+c_{1} \hat{\mathrm{k}}$
$=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}}=0$
Hence the alternatives for m and $\vec{a}$ are m=0 or $\vec{a}=0$
Question 5
If $\vec{a} \vec{b}$ are two vectors, then write the truth value of the following statements:
(i) $\vec{a}=-\vec{b} \Rightarrow|\vec{a}|=|\vec{b}|$
(ii) $|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\pm \vec{b}$
(iii) $|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\vec{b}$
Sol :
(i)
Let $\vec{a}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}$
$\vec{b}=a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}$
Given that, a=-b
$a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}=-a_{2} \hat{i}-b_{2} \hat{j}-c_{2} \hat{k}$
Comparing the coefficients of i, j, k in LHS and RHS,
$a_{1}=-a_{2}$...(1)
$b_{1}=-b_{2}$...(2)$c_{1}=-c_{2}$...(3)
$|\vec{a}|=\sqrt{{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}$
Using (1),(2) and (3),
$|\vec{a}|=\sqrt{\left(-a_{2}\right)^{2}+\left(-b_{2}\right)^{2}+\left(-c_{2}\right)^{2}}$
$|\vec{a}|=\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}$
$\therefore|\vec{a}|=|\vec{b}|$
(ii)
Given a and b are two vectors such that $|\vec{a}|=|\vec{b}|$
It means magnitude of vector $\vec{a}$ is equal to the magnitude of vector $\vec{b}$, but we cannot conclude anything about the direction of the vector.
So,it is false that
$|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\pm \vec{b}$
(iii)
Given for any vector $\vec{a}$ and $\vec{b}$
$|\vec{a}|=|\vec{b}|$
It means magnitude of the vector $\vec{a}$ and $\vec{b}$ are equal but we cannot say any thing about the direction of the vector $\vec{a}$ and $\vec{b}$. And we know that $\vec{a}=\vec{b}$ means magnitude and same direction. So, it is false.
Question 6
ABCD is a quadrilateral. Find the sum of the vectors $\overrightarrow{B A}, \overrightarrow{B C}, \overrightarrow{C D}$ and $\overrightarrow{D A}$.
Sol :
Here it given that ABCD is a quadrilateral.
In ΔADC, using triangle law, we get
$\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{C A}$...(i)
In ΔABC, using triangle law, we get
$\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{B A}$...(ii)
Put value of $\overrightarrow{C A}$ in equation (ii)
$\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}$
Adding $\overrightarrow{B A}$ on both the sides,
$\overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}+\overrightarrow{B A}$
$\therefore \overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=2 \overrightarrow{B A}$
Question 7
ABCDE is a pentagon, prove that
(i) $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D E}+\overrightarrow{E A}=0$
Sol :
Given: ABCDE is a pentagon as shown below
Adding (i), (ii) and (iii), we get
$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{AC}}$
$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{AD}}$
$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}$
$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DE}}=-\overrightarrow{\mathrm{EA}}~[\text{ as } \overrightarrow{\mathrm{AE}}=-\overrightarrow{\mathrm{EA}}]$
$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DE}}+\overrightarrow{\mathrm{EA}}=0$
Hence Proved
(ii) $\overrightarrow{A B}+\overrightarrow{A E}+\overrightarrow{B C}+\overrightarrow{D C}+\overrightarrow{E D}+\overrightarrow{A C}=3 \overrightarrow{A C}$
Sol :
Given: ABCDE is a pentagon as shown below
Consider ΔABC and apply triangle law of vector, we get
$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}$....(i)
