RS Aggarwal Solution Class 12 Chapter 23 Algebra of Vectors Exercise 23.2

 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,

$\overrightarrow{A D}+\overrightarrow{D B}=\overrightarrow{A B}$

$\vec{b}+\overrightarrow{D B}=\vec{a}$

$\overrightarrow{D B}=\vec{a}-\vec{b}$...(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

Consider ΔABC and apply triangle law of vector, we get

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}$....(i)

Similarly, consider ΔACD and apply triangle law of vector, we get

$\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AD}}$...(ii)

And, consider ΔADE and apply triangle law of vector, we get

$\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}$...(iii)

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)

RS Aggarwal Solution Class 12 Chapter 23 Algebra of Vectors Exercise 23.1

Exercise 23.1

Question 1

(i) Represent graphically a displacement of 40 km, 30° east of north.

Sol :

(i) a displacement of 40 km, 30° east of north

Step 1: Draw north, south, east and west as shown below:

Step 2: Plot a line  30° east of north as shown below:

Step 3: Define scale and mark 40km on line $\overrightarrow{\mathrm{OP}}$

Let the scale be 10km = 1cm

∴$\overrightarrow{\mathrm{OP}}$represents the displacement of 40 km, 30° East of North

(ii) Represent graphically a displacement of 50 km, south-east

Sol :

(ii) a displacement of 50 km south - east

Step 1: Draw north, south, east and west as shown below:

Step 2: As the displacement should be south - east, the angle between the displacement and east (or south) will be 45°. Now, plot a line $\overrightarrow{\mathrm{OP}}$ 45° east of south as shown below:

Step 3: Define scale and mark point R such that OR = 50km on line $\overrightarrow{\mathrm{OP}}$. Let the scale be 10km = 1cm

∴$\overrightarrow{\mathrm{OR}}$ represents the displacement of 50 km south – east

(iii) Represent graphically a displacement of 70 km, 40° north of west.

Sol :

(iii) A displacement of 70 km, 40° north of west.

Step 1: Draw north, south, east and west as shown below:

Step 2: Plot a line $\overrightarrow{\mathrm{OP}}$ 40° north of west as shown below:

Step 3: Define scale and mark point R such that OR = 70km on line $\overrightarrow{\mathrm{OP}}$.Let the scale be 10km = 1cm

∴$\overrightarrow{\mathrm{OP}}$  represents the displacement of 70 km, 40o north of west

Question 2

Classify the following measures as scalars and vectors.

(i) 15 kg

Sol :

15 kg - is a scalar quantity as this involves only mass. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

(ii) 20 kg weight

Sol :

20 kg weight - is a vector quantity as it involves both magnitude and direction. Weight is a force which is a vector and has a magnitude and direction.

(iii) 45°

Sol :

45° is a scalar quantity as it involves the only magnitude. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

(iv) 10 metres south-east

Sol :

10 meters south - east is a vector quantity as it involves both magnitude and direction.

(v) 50 m/s²

Sol :

50 m/sec² is a scalar quantity as it involves a magnitude of acceleration. A scalar quantity is a one - dimensional measurement of a quantity.

Question 3

Classify the following as scalar and vector quantities.

(i) Time period

Sol :

Time period - is a scalar quantity as it involves only magnitude. A scalar quantity is a one - dimensional measurement of a quantity. Eg: 10 seconds has only magnitude, i.e., 10 and no direction.

(ii) Distance

Sol :

Distance - is a scalar quantity as it involves only magnitude. A scalar quantity is a one dimensional measurement of a quantity. Eg: 5meters has only magnitude 5 and no direction.

(iii) Displacement

Sol :

Displacement - is vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction.

(iv) Force

Sol :

Force - is a vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction. Eg., 5N downward has magnitude of 5 and direction is downward.

(v) Work

Sol :

Work done - is a scalar quantity as it involves only magnitude and no particular direction. A scalar quantity is a one dimensional measurement of a quantity.

(vi) Velocity

Sol :

Velocity - is a vector quantity as it involves both magnitude as well as direction. Vector quantity has both magnitude and direction. Eg., 5m/s east has magnitude of 5m/s and also direction towards east.

(vii) Acceleration

Sol :

Acceleration is a vector quantity because it involves both magnitude as well as direction.

Question 4

Which vectors are:

(i) Collinear

(ii) Equal

(iii) Coinitial

(iv) Collinear but not equal.

Sol :

(i) Collinear vectors are

$\vec{x}, \vec{z}$ and $\vec{b}$

$\vec{y}, \vec{c}$

$\vec{a}, \vec{d}$

(ii) Equal vectors are

$\vec{y}$ and $\vec{c}$

$\vec{x}$ and $\vec{b}$

$\vec{a}$ and $\vec{d}$

(iii) Coinitial vector are $\vec{a}, \vec{y}$ and $\vec{z}$

(iv) Collinear but not equal

$\vec{b}$ and $\vec{z}$

$\vec{x}$ and $\vec{z}$

Question 5

Answer the following as true or false:

(i) a and b are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii). Zero vector is unique.

(iv) Two vectors having same magnitude are collinear.

(v) Two collinear vectors having the same magnitude are equal.

Sol :

(i) $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{a}}$  are collinear. (True)

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors.

$\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{a}}$ are collinear.

(ii) Two collinear vectors are always equal in magnitude. (False)

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors. Two collinear vectors may point in either same or opposite direction. And they are not necessarily equal in magnitude they can be of different magnitude also.

(iii) Zero vector is unique.(True)

There is only one zero - vector in a vector space. Hence zero vector is unique.

(iv) Two vectors having same magnitude are collinear. (False)

It is not necessary for two vectors having the same magnitude to be parallel to the same line. Hence two vectors having same magnitude need not be collinear.

(v) Two collinear vectors having the same magnitude are equal.(False)

Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points.