Physics and Mathematics

Question 1:

Question 2:

Yes. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed.

Question 3:

No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ of unequal magnitudes acting in opposite directions. The resultant vector is given by

 $R=\sqrt{A^2+B^2+2AB\cos \theta }$

If two vectors are exactly opposite to each other, then
$$\begin{gathered} \theta =180°, \cos 180°=-1 \\ R=\sqrt{A^2+B^2-2AB} \\ \Rightarrow R=\sqrt{{\left(A-B\right)}^2} \\ \Rightarrow R=\left(A-B\right) \mathrm{or} \left(B-A\right) \end{gathered}$$

From the above equation, we can say that the resultant vector is zero (R = 0) when the magnitudes of the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are equal (A = B) and both are acting in the opposite directions.

Yes, it is possible to add three vectors of equal magnitudes and get zero.
Lets take three vectors of equal magnitudes $\overrightarrow{A, }\overrightarrow{B} \mathrm{and} \overrightarrow{C}$, given these three vectors make an angle of $120°$ with each other. Consider the figure below:


Lets examine the components of the three vectors.
$$\begin{gathered} A_x\mathit{=}A \\ {A}_y\mathit{=}0 \\ {B}_x\mathit{=}\mathit{-}B \cos 60° \\ {B}_y\mathit{=}B \sin \mathit{60}\mathit{°} \\ {C}_x\mathit{=}\mathit{-}C \cos 60° \\ {C}_y\mathit{=}\mathit{-}C \sin 60° \\ \mathrm{Here}\mathit{,}\mathit{ }A\mathit{=}B\mathit{=}C \\ \mathrm{So}, \mathrm{along} \mathrm{the} x-\mathrm{axis} , \mathrm{we} \mathrm{have}: \\ A\mathit{-}\mathit{(}2A \cos 60°\mathit{)}\mathit{=}\mathit{0}\mathit{,}\mathit{ }as\mathit{ }\cos \mathit{ }60°\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{ } \\ \Rightarrow B \sin 60°\mathit{-}C \sin 60°\mathit{=}\mathit{0} \end{gathered}$$

Hence, proved.

Question 4:

A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
For any vector $\overrightarrow{A}$, assume that
$$\begin{gathered} \overrightarrow{A}+\overrightarrow{0}=\overrightarrow{A} \\ \overrightarrow{A}-\overrightarrow{0}=\overrightarrow{A} \\ \overrightarrow{A}\times \overrightarrow{0}=\overrightarrow{0} \end{gathered}$$
Again, for any real number $\lambda$, we have:
$\lambda \overrightarrow{0}=\overrightarrow{0}$
The significance of a zero vector can be better understood through the following examples:
The displacement vector of a stationary body for a time interval is a zero vector.
Similarly, the velocity vector of the stationary body is a zero vector.
When a ball, thrown upward from the ground, falls to the ground, the displacement vector is a zero vector, which defines the displacement of the ball.

 

Question 5:

Yes we can add three unit vectors to get a unit vector.
No, the answer does not change if two unit vectors are along the coordinate axes. Assume three unit vectors $\widehat{i,}-\hat{i} \mathrm{and} \hat{j}$ along the positive x-axis, negative x-axis and positive y-axis, respectively. Consider the figure given below:



The magnitudes of the three unit vectors ($\widehat{i,}-\hat{i} \mathrm{and} \hat{j}$ ) are the same, but their directions are different.
So, the resultant of $\hat{i} \mathrm{and} -\hat{i}$ is a zero vector.
Now, $\hat{j}+\overrightarrow{0}=\hat{j}$          (Using the property of zero vector)
∴ The resultant of three unit vectors ($\widehat{i,}-\hat{i} \mathrm{and} \hat{j}$)  is a unit vector ($\hat{j}$).

Question 6:

Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.

Question 7:

Two forces are added using triangle rule, because force is a vector quantity. This statement is more appropriate, because we know that force is a vector quantity and only vectors are added using triangle rule.

Question 8:

No, we cannot add two vectors representing physical quantities of different dimensions. However, we can multiply two vectors representing physical quantities with different dimensions.
Example: Torque, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Question 9:

Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Example: Consider a two dimensional vector $2\hat{i}+0\hat{j}$. This vector has zero components along a line lying along the Y-axis and a nonzero component along the X-axis. The magnitude of the vector is also nonzero.
Now, magnitude of $2\hat{i}+0\hat{j}$ = $\sqrt{2^2+0^2}=2$

Question 10:

The direction of $-\overrightarrow{A}$ is opposite to $\overrightarrow{A}$. So, if vector $\overrightarrow{A}$ and $-\overrightarrow{A}$ make the angles ε₁ and ε₂ with the X-axis, respectively, then ε₁ is equal to ε₂ as shown in the figure:

 
Here, tan ε₁ = tan ε₂
Because these are alternate angles.
Thus, giving tan ε does not uniquely determine the direction of $\overrightarrow{A}$.

Question 11:

No, the vector sum of the unit vectors $\overrightarrow{i}$ and $\overrightarrow{j}$ is not a unit vector, because the magnitude of the resultant of $\overrightarrow{i}$ and $\overrightarrow{j}$ is not one.

Magnitude of the resultant vector is given by
R = $\sqrt{1^2+1^2+\cos 90°}=\sqrt{2}$

Yes, we can multiply this resultant vector by a scalar number $\frac{1}{\sqrt{2}}$ to get a unit vector.

Question 12:

A vector  $\overrightarrow{B}$ such that $\overrightarrow{A}\ne \overrightarrow{B}$, but A = B are as follows:
$$\begin{gathered} \left(\mathrm{i}\right) \overrightarrow{B}=3\overrightarrow{i}-4\overrightarrow{j} \\ \left(\mathrm{ii}\right) \overrightarrow{B}=3\overrightarrow{j}+4\overrightarrow{k} \\ \left(\mathrm{iii}\right) \overrightarrow{B}=3\overrightarrow{k}+4\overrightarrow{i} \\ \left(\mathrm{iv}\right) \overrightarrow{B}=3\overrightarrow{j}-4\overrightarrow{k} \end{gathered}$$

Question 13:

No, we cannot have  $\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{A}·\overrightarrow{B}$ with A ≠ 0 and B ≠ 0. This is because the left hand side of the given equation gives a vector quantity, while the right hand side gives a scalar quantity. However, if one of the two vectors is zero, then both the sides will be equal to zero and the relation will be valid.

Question 14:

If $\overrightarrow{A}\times \overrightarrow{B}=0$, then both the vectors are either parallel or antiparallel, i.e., the angle between the vectors is either $0° \mathrm{or} 180°$.
($\overrightarrow{A}\overrightarrow{B}\sin \theta \hat{n}=0$ $\because$$\sin 0°=\sin 180°=0$)

Both the conditions can be satisfied:
(a) $\overrightarrow{A}=\overrightarrow{B},$ i.e., the two vectors are equal in magnitude and parallel to each other
(b) $\overrightarrow{A}\ne \overrightarrow{B}$, i.e., the two vectors are unequal in magnitude and parallel or anti parallel to each other

Question 1:

If  $\overrightarrow{A}=5\overrightarrow{i}-4\overrightarrow{j} \mathrm{and} \overrightarrow{B}=-7·5\overrightarrow{i}+6\overrightarrow{j}$, then we have $\overrightarrow{B}=k\overrightarrow{A}$ by putting the value of scalar k as $-1.5$.
However, we cannot say that $\frac{\overrightarrow{B}}{\overrightarrow{A}}$ = k, because a vector cannot be divided by other vectors, as vector division is not possible.

Question 2:

(d) it is slid parallel to itself.

A vector is defined by its magnitude and direction. If we slide it to a parallel position to itself, then none of the given parameters, which define the vector, will change.

Let the magnitude of a displacement vector ($\overrightarrow{A}$) directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.

Question 3:

(c) 1, 2, 1

1,2 and 1 may represent the magnitudes of three vectors adding to zero.For example one of the vector of length 1 should make an angle of ${135}^{\circ }$ with x axis and the other vector of length 1 makes an angle of ${225}^{\circ }$ with x axis. The third vector of length 2 should lie along x axis.

Question 4:

(c) α < β if A > B

The resultant of two vectors is closer to the vector with the greater magnitude.
Thus, α < β if A > B

Question 5:

(d) None of these.

All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components. 

Question 6:

(a) along the west

The vector product $\overrightarrow{A}\times \overrightarrow{B}$ will point towards the west. We can determine this direction using the right hand thumb rule.

Question 1:

(b) 14.1 cm²

Area of a circle, A = $\pi r^2$
On putting the values, we get:

$$\begin{gathered} A=\frac{22}{7}\times 2.12\times 2.12 \\ \Rightarrow A=14.1 {\mathrm{cm}}^2 \end{gathered}$$
 
The rules to determine the number of significant digits says that in the multiplication of two or more numbers, the number of significant digits in the answer should be equal to that of the number with the minimum number of significant digits. Here, 2.12 cm has a minimum of three significant digits. So, the answer must be written in three significant digits.

Question 2:

(a) the value of a scalar
(c) a vector
(d) the magnitude of a vector

The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation. However, components of a vector depend on the orientation of the axes.

Question 3:

(b) It is possible to have $\left| \overrightarrow{C} \right|$ < $\left| \overrightarrow{A} \right|$ and $\left| \overrightarrow{C} \right|$ < $\left| \overrightarrow{B} \right|$

Statements (a), (c) and (d) are incorrect.
Given: $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$
Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of $\overrightarrow{A}$ and $\overrightarrow{B}$ or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.

Question 4:

(b) C must be less than $\left| A-B \right|$

Here, we have three vector A, B and C.
$$\begin{gathered} {\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2={\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2+2\overrightarrow{A}.\overrightarrow{B} ...\left(i\right) \\ {\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2={\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2-2\overrightarrow{A}.\overrightarrow{B} ...\left(ii\right) \end{gathered}$$

Subtracting (i) from (ii), we get:
${\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2-{\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2=4\overrightarrow{A}.\overrightarrow{B}$

Using the resultant property $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$, we get:
$$\begin{gathered} {\left|\overrightarrow{C}\right|}^2-{\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2=4\overrightarrow{A}.\overrightarrow{B} \\ \Rightarrow {\left|\overrightarrow{C}\right|}^2={\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2+4\overrightarrow{A}.\overrightarrow{B} \\ \Rightarrow {\left|\overrightarrow{C}\right|}^2={\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2+4\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos 120° \end{gathered}$$

Since cosine is negative in the second quadrant, $C$must be less than $\left|A-B\right|$.

Question 5:

(a) is equal to the sum of the x-components of the vectors    
(b) may be smaller than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.

The x-component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.

Question 1:

(b) equal to AB
(c) less than AB
(d) equal to zero.

The magnitude of the vector product of two vectors $\left| \overrightarrow{A} \right|$ and $\left| \overrightarrow{B} \right|$ may be less than or equal to AB, or equal to zero, but cannot be greater than AB.

Question 2:

From the above figure, we have:
Angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ = 110° − 20° = 90°
$\left|\overrightarrow{\mathrm{A}}\right|=3 \mathrm{m} \mathrm{and} \left|\overrightarrow{\mathrm{B}}\right|= 4 \mathrm{m}$
Magnitude of the resultant vector is given by
$$\begin{gathered} \mathrm{R}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{AB} \cos \mathrm{\theta }} \\ =\sqrt{3^2+4^2+2\times 3\times 4\times \cos 90°} \\ = 5 \mathrm{m} \end{gathered}$$

Let β be the angle between $\overrightarrow{\mathrm{R}} \mathrm{and} \overrightarrow{\mathrm{A}}$.
$$\begin{gathered} \mathrm{\beta }={\tan }^{-1} \left(\frac{A \sin 90°}{A+B \cos 90°}\right) \\ ={\tan }^{-1} \left(\frac{4 \sin 90°}{3+4 \cos 90°}\right) \\ ={\tan }^{-1} \frac{4}{3} \\ ={\tan }^{-1} \left(1.333\right) \\ =53° \end{gathered}$$

Now, angle made by the resultant vector with the X-axis = 53° + 20° = 73°
∴ The resultant $\overrightarrow{\mathrm{R}}$ is 5 m and it makes an angle of 73° with the x-axis.

Question 3:

Angle between $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$, θ = 60° − 30° = 30°
$$\begin{gathered} \left|\overrightarrow{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{B}}\right|=10 \mathrm{units} \\ \mathrm{The} \mathrm{magnitude} \mathrm{of} \mathrm{the} \mathrm{resultant} \mathrm{vector} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{R}=\sqrt{A^2+A^2+2AA\cos \theta } \\ =\sqrt{{10}^2+{10}^2+2\times 10\times 10\times \cos 30°} \\ =\sqrt{200+200 \cos 30°} \\ =\sqrt{200+200\times \frac{\sqrt{3}}{2}} \\ =19.3 \mathrm{units} \end{gathered}$$



Let β be the angle between $\overrightarrow{\mathrm{R}} \mathrm{and} \overrightarrow{\mathrm{A}}$.
$$\begin{gathered} \therefore \mathrm{\beta }={\tan }^{-1} \left(\frac{A \sin 30°}{A+A \cos 30°}\right) \\ \Rightarrow \mathrm{\beta }={\tan }^{-1} \left(\frac{10 \sin 30°}{10+10 \cos 30°}\right) \\ \Rightarrow \mathrm{\beta }={\tan }^{-1} \left(\frac{1}{2+\sqrt{3}}\right)={\tan }^{-1} \left(\frac{1}{3.372}\right) \\ \Rightarrow \mathrm{\beta }={\tan }^{-1} \left(0.26795\right)=15° \end{gathered}$$

Angle made by the resultant vector with the X-axis = 15° + 30° = 45°

∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.

Question 4:

First, we will find the components of the vector along the x-axis and y-axis. Then we will find the resultant x and y-components.  
x-component of $\overrightarrow{\mathrm{A}}=\mathrm{Acos}45°=100 \cos 45°=\frac{100}{\sqrt{2}} \mathrm{unit}$
x-component of $\overrightarrow{\mathrm{B}} = \overrightarrow{\mathrm{B}}\cos 135°=-\frac{100}{\sqrt{2}}$
x-component of $\overrightarrow{\mathrm{C}}$ = $\overrightarrow{\mathrm{C}}$cos315$°$ = 100 cos 315°
                            $=100 \cos 45°=\frac{100}{\sqrt{2}}$
Resultant x-component $=\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$



Now, y-component of $\overrightarrow{\mathrm{A}}=100 \sin 45°=\frac{100}{\sqrt{2}}$
y-component of $\overrightarrow{\mathrm{B}}=100 \sin 135°=\frac{100}{\sqrt{2}}$
y-component of $\overrightarrow{\mathrm{C}} = 100 \sin 315°=-\frac{100}{\sqrt{2}}$
Resultant y-component$=\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$

Magnitude of the resultant$=\sqrt{{\left(\frac{100}{\sqrt{2}}\right)}^2+{\left(\frac{100}{\sqrt{2}}\right)}^2}$
                                           $=\sqrt{10000}=100$
Angle made by the resultant vector with the x-axis is given by
$$\begin{gathered} \tan \mathrm{\alpha }=\frac{y \mathrm{comp}}{x \mathrm{comp}} \\ =\frac{100\sqrt{2}}{100\sqrt{2}}=1 \end{gathered}$$
⇒ α = tan⁻¹ (1) = 45°

∴ The magnitude of the resultant vector is 100 units and it makes an angle of 45° with the x-axis.

Question 5:

Given: $\overrightarrow{a}=4\overrightarrow{i}+3\overrightarrow{j} \mathrm{and} \overrightarrow{b}=3\overrightarrow{i}+4\overrightarrow{j}$

(a) Magnitude of $\overrightarrow{a}$ is given by $\left|\overrightarrow{a}\right|=\sqrt{4^2+3^2}=\sqrt{16+9}=5$

(b)  Magnitude of $\overrightarrow{b}$ is  given by $\left|\overrightarrow{b}\right|=\sqrt{3^2+4^2}=\sqrt{9+16}=5$

(c) $\overrightarrow{a}+\overrightarrow{b}=(4\hat{i}+3\hat{j})+(3\hat{i}+4\hat{j})=(7\hat{i}+7\hat{j})$

∴ Magnitude of vector $\overrightarrow{a}+\overrightarrow{b}$ is given by $\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}$

(d) $\overrightarrow{a}-\overrightarrow{b}=\left(4\overrightarrow{i}+3\overrightarrow{j}\right)-\left(3\overrightarrow{i}+4\overrightarrow{j}\right)=\overrightarrow{i}-\overrightarrow{j}$

∴ Magnitude of vector $\overrightarrow{a}-\overrightarrow{b}$ is given by $\left|\overrightarrow{a}-\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}=\sqrt{2}$

Question 6:

First, let us find the components of the vectors along the x and y-axes. Then we will find the resultant x and y-components.  

x-component of $\overrightarrow{\mathrm{OA}}= 2 \cos 30°=\sqrt{3}$ m
x-component of $\overrightarrow{\mathrm{BC}}$ =1.5cos120°
                               $=-\frac{\left(1.5\right)}{2}=-7.5$ m
x-component of $\overrightarrow{\mathrm{DE}}$ = 1cos270°
                               = 1 × 0 = 0 m

y-component of $\overrightarrow{\mathrm{OA}}$ = 2 sin 30° = 1
y-component of $\overrightarrow{\mathrm{BC}}$ = 1.5 sin 120°
                              $=\frac{\left(\sqrt{3}\times 1.5\right)}{2}=1.3$
y-component of $\overrightarrow{\mathrm{DE}}$ = 1 sin 270° = −1
x-component of resultant ${\mathrm{R}}_x=\sqrt{3}-0.75+0=0.98 \mathrm{m}$

y-component of resultant R_y = 1 + 1.3 − 1 = 1.3 m

$$\begin{gathered} \therefore \mathrm{Resultant}, \mathrm{R}=\sqrt{{\left(R_x\right)}^2+{\left(R_y\right)}^2} \\ = \sqrt{{\left(0.98\right)}^2+{\left(1.3\right)}^2} \\ =\sqrt{0.96+1.69} \\ =\sqrt{2.65} \\ =1.6 \mathrm{m} \end{gathered}$$

If it makes an angle α with the positive x-axis, then
$$\begin{gathered} \tan \mathrm{\alpha }=\frac{y-component}{x-\mathrm{component}} \\ =\frac{1.3}{0.98}=1.332 \end{gathered}$$
∴ α = tan⁻¹ (1.32)

Question 7:

Let the two vectors be $\overrightarrow{a}$ and $\overrightarrow{b}$.
Now, $\left|\overrightarrow{a}\right|=3 \mathrm{and} \left|\overrightarrow{b}\right|=4$

(a) If the resultant vector is 1 unit, then

$$\begin{gathered} \sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2+2.\overrightarrow{a}.\overrightarrow{b} \cos \mathrm{\theta }}=1 \\ \Rightarrow \sqrt{3^2+4^2+2.3.4 \cos \mathrm{\theta }}=1 \end{gathered}$$
Squaring both sides, we get:

$$\begin{gathered} 25+24 \cos \mathrm{\theta }=1 \\ \Rightarrow 24 \cos \mathrm{\theta }=-24 \\ \Rightarrow \cos \mathrm{\theta }=-1 \\ \Rightarrow \mathrm{\theta }=180° \end{gathered}$$

Hence, the angle between them is 180°.

(b) If the resultant vector is 5 units, then

$$\begin{gathered} \sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2+2.\overrightarrow{a}.\overrightarrow{b} \cos \mathrm{\theta }}=5 \\ \Rightarrow \sqrt{3^2+4^2+2.3.4 \cos \mathrm{\theta }}=5 \end{gathered}$$
Squaring both sides, we get:

    25 + 24 cos θ = 25
⇒ 24 cos θ = 0
⇒ cos θ = 90°

Hence, the angle between them is 90°.

(c) If the resultant vector is 7 units, then

$$\begin{gathered} \sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2+2.\overrightarrow{a}.\overrightarrow{b} \cos \mathrm{\theta }}=1 \\ \Rightarrow \sqrt{3^2+4^2+2.3.4 \cos \mathrm{\theta }}=7 \end{gathered}$$
Squaring both sides, we get:

    25 + 24 cos θ = 49,
⇒ 24 cos θ = 24
⇒ cos θ = 1
⇒ θ = cos⁻¹ 1 = 0°

Hence, the angle between them is 0°.

Question 8:

The displacement of the car is represented by $\overrightarrow{\mathrm{AD}}$.
$$\begin{gathered} \overrightarrow{\mathrm{AD}}=2\hat{i}+0.5\hat{j}+4\hat{i} \\ =6\hat{i}+0.5\hat{j} \end{gathered}$$

Magnitude of $\overrightarrow{\mathrm{AD}}$ is given by

$$\begin{gathered} \mathrm{AD}=\sqrt{{\mathrm{AE}}^2+{\mathrm{DE}}^2} \\ =\sqrt{6^2+{\left(0.5\right)}^2} \\ =\sqrt{36+0.25}=6.02 \mathrm{km} \end{gathered}$$



Now, $\tan \mathrm{\theta }=\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{1}{12}$
$\Rightarrow \mathrm{\theta }={\tan }^{-1} \left(\frac{1}{12}\right)$
Hence, the displacement of the car is 6.02 km along the direction ${\tan }^{-1} \left(\frac{1}{12}\right)$ with positive the x-axis.

Question 9:

Consider that the queen is initially at point A as shown in the figure.
Let AB be x ft.
So, DE = (2 $-$ x) ft
In ∆ABC, we have:
$\tan \mathrm{\theta }=\frac{x}{2}$   ...(i)
Also, in ∆DCE, we have:
$\tan \mathrm{\theta }=\frac{\left(2-x\right)}{4}$  ...(ii)
From (i) and (ii), we get:
$$\begin{gathered} \frac{x}{2}=\frac{\left(2-x\right)}{4} \\ \Rightarrow 2\left(2-x\right)=4x \\ \Rightarrow 4-2x=4x \\ \Rightarrow 6x=4 \\ \Rightarrow x=\frac{2}{3}\mathrm{ft} \end{gathered}$$

(a) In ∆ABC, we have:
$\mathrm{AC}=\sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2}$
    $$\begin{gathered} =\sqrt{{\left(\frac{2}{3}\right)}^2+2^2} \\ =\sqrt{\frac{4}{9}+4}=\sqrt{\frac{40}{9}} \\ =\frac{2}{3}\sqrt{10} \mathrm{ft} \end{gathered}$$

(b) In ∆CDE, we have:
DE$=2-\frac{2}{3}=\frac{6-2}{3}=\frac{4}{3} \mathrm{ft}$
CD = 4 ft
      $$\begin{gathered} \therefore \mathrm{CE}=\sqrt{{\mathrm{CD}}^2+{\mathrm{DE}}^2} \\ =\sqrt{4^2+{\left(\frac{4}{3}\right)}^2} \\ =\frac{4}{3}\sqrt{10} \mathrm{ft} \end{gathered}$$

(c) In ∆AGE, we have:
$\mathrm{AE}=\sqrt{{\mathrm{AG}}^2+{\mathrm{GE}}^2}$
    $$\begin{gathered} =\sqrt{2^2+2^2} \\ =\sqrt{8}+2\sqrt{2} \mathrm{ft} \end{gathered}$$

Question 10:

Displacement vector of the mosquito, $\overrightarrow{r}=7\hat{i}+4\hat{i}+3\hat{k}$




(a) Magnitude of displacement$=\sqrt{7^2+4^2+3^2}$
                                                 $=\sqrt{74} \mathrm{ft}$

(b) The components of the displacement vector are 7 ft, 4 ft and 3 ft along the X, Y and Z-axes, respectively.

Question 11:

Given: $\overrightarrow{a}$ is a vector of magnitude 4.5 units due north.

Case (a):
$3\left|\overrightarrow{a}\right|=3\times 4.5=13.5$
∴ $3\overrightarrow{a}$ is a vector of magnitude 13.5 units due north.

Case (b):
$\left|-4\overrightarrow{a}\right|=4\times 1.5=6 \mathrm{units}$
∴$-4\overrightarrow{a}$ is a vector of magnitude 6 units due south.

Question 12:

Let the two vectors be $\left|\overrightarrow{a}\right|= 2 \mathrm{m} \mathrm{and} \left|\overrightarrow{b}\right|=3 \mathrm{m}$.
Angle between the vectors, θ = 60°

(a) The scalar product of two vectors is given by $\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right| \cos \theta °$.

∴ $\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right| \cos 60°$
          $=2\times 3\times \frac{1}{2}=3 {\mathrm{m}}^2$

(b) The vector product of two vectors is given by $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right| \left|\overrightarrow{a}\right| \sin \theta °$.
∴ $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right| \left|\overrightarrow{a}\right| \sin 60°$
             $$\begin{gathered} =2\times 3\times \frac{\sqrt{3}}{2} \\ =3\sqrt{3} {\mathrm{m}}^2 \end{gathered}$$

Question 13:

According to the polygon law of vector addition, the resultant of these six vectors is zero.
Here, a = b = c = d = e = f (magnitudes), as it is a regular hexagon. A regular polygon has all sides equal to each other.
So, $R_x=A \cos 0 + A \cos \frac{\mathrm{\pi }}{3}+A \cos \frac{2\mathrm{\pi }}{3}+A \cos \frac{3\mathrm{\pi }}{3}+ A \cos \frac{4\mathrm{\pi }}{3}+A \cos \frac{5\mathrm{\pi }}{3}=0$
[As the resultant is zero, the x-component of resultant R_x is zero]
$\Rightarrow \cos 0 + \cos \frac{\mathrm{\pi }}{3}+\cos \frac{2\mathrm{\pi }}{3}+\cos \frac{3\mathrm{\pi }}{3}+\cos \frac{4\mathrm{\pi }}{3}+\cos \frac{5\mathrm{\pi }}{5}=0$



Note: Similarly, it can be proven that $\sin 0+\sin \frac{\mathrm{\pi }}{3}+\sin \frac{2\mathrm{\pi }}{3}+\sin \frac{3\mathrm{\pi }}{3}+\sin \frac{4\mathrm{\pi }}{3}+\sin \frac{5\mathrm{\pi }}{3}=0$.

Question 14:

We have:
$$\begin{gathered} \overrightarrow{a}=2\overrightarrow{\mathit{i}}+3\overrightarrow{\mathit{j}}+4\overrightarrow{k} \\ \overrightarrow{b}=3\overrightarrow{\mathit{i}}+4\overrightarrow{\mathit{j}}+5\overrightarrow{k} \end{gathered}$$

Using scalar product, we can find the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.
i.e., $\overrightarrow{\mathit{a}}\mathit{.}\overrightarrow{\mathit{b}}\mathit{=}\left|\overrightarrow{\mathit{a}}\right|\left|\overrightarrow{\mathit{b}}\right|\mathit{ }\cos \theta$
So, $\theta ={\cos }^{-1}\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)$

$$\begin{gathered} ={\cos }^{-1}\left(\frac{2\times 3+3\times 4+4\times 5}{\sqrt{\left(2^2+3^2+4^2\right)} \sqrt{\left(3^2+4^2+5^2\right)}}\right) \\ ={\cos }^{-1}\left(\frac{38}{\sqrt{29} \sqrt{50}}={\cos }^{-1}\frac{38}{\sqrt{1450}}\right) \end{gathered}$$

∴ The required angle is ${\cos }^{-1}\frac{38}{\sqrt{1450}}.$

Question 15:

To prove: $\overrightarrow{\mathrm{A}}.\left(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right)=0$

Proof: Vector product is given by $\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}= \left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right| \sin \hat{n}$.

$\left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right| \sin \hat{n}$ is a vector which is perpendicular to the plane containing $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$. This implies that it is also perpendicular to $\overrightarrow{\mathrm{A}}$. We know that the dot product of two perpendicular vectors is zero.
∴ $\overrightarrow{\mathrm{A}}.\left(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right)=0$

Hence, proved.