Addition and Subtraction of Vectors Questions in English

(D) The resultant $R$ of two vectors $A$ and $B$ lies in the range $|A-B| \leq R \leq |A+B|$.
For option $A$: $|1-2| \leq R \leq |1+2| \implies 1 \leq R \leq 3$. Since $3$ is in the range,the resultant can be $3$.
For option $B$: $|2-5| \leq R \leq |2+5| \implies 3 \leq R \leq 7$. Since $3$ is in the range,the resultant can be $3$.
For option $C$: $|3-6| \leq R \leq |3+6| \implies 3 \leq R \leq 9$. Since $3$ is in the range,the resultant can be $3$.
For option $D$: $|4-8| \leq R \leq |4+8| \implies 4 \leq R \leq 12$. The range is $[4, 12]$. Since $3$ is not in this range,the resultant can never be $3$ units.

(D) Let the two forces be $F_1$ and $F_2$. The angle between them is $\theta = 120^{\circ}$.
Let the resultant $R = 10 \,kg$-wt be perpendicular to force $F_1$.
Using the formula for the angle $\alpha$ that the resultant makes with force $F_1$:
$\tan \alpha = \frac{F_2 \sin \theta}{F_1 + F_2 \cos \theta}$
Since the resultant is perpendicular to $F_1$, $\alpha = 90^{\circ}$, so $\tan 90^{\circ} = \infty$.
This implies the denominator must be zero: $F_1 + F_2 \cos 120^{\circ} = 0$.
$F_1 + F_2 (-0.5) = 0 \implies F_1 = 0.5 F_2 \implies F_2 = 2 F_1$.
The magnitude of the resultant is given by $R^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos 120^{\circ}$.
$10^2 = F_1^2 + (2 F_1)^2 + 2 F_1 (2 F_1) (-0.5)$.
$100 = F_1^2 + 4 F_1^2 - 2 F_1^2 = 3 F_1^2$.
$F_1^2 = \frac{100}{3} \implies F_1 = \frac{10}{\sqrt{3}} \,kg$-wt.
Thus, the force perpendicular to the resultant is $\frac{10}{\sqrt{3}} \,kg$-wt.

(D) The maximum resultant magnitude of two vectors $A$ and $B$ is given by $R_{max} = A + B$.
The minimum resultant magnitude of two vectors $A$ and $B$ is given by $R_{min} = A - B$.
According to the problem,$R_{max} = n \cdot R_{min}$.
Substituting the expressions,we get $A + B = n(A - B)$.
Expanding the equation: $A + B = nA - nB$.
Rearranging the terms to group $A$ and $B$: $A - nA = -nB - B$.
$A(1 - n) = -B(n + 1)$.
Dividing both sides by $B(1 - n)$: $\frac{A}{B} = \frac{-(n + 1)}{1 - n}$.
Multiplying the numerator and denominator by $-1$: $\frac{A}{B} = \frac{n + 1}{n - 1}$.

(B) Let the three vectors be $\vec{A}$,$\vec{B}$,and $\vec{C}$. The magnitude of each vector is $A = B = C = 3 \sqrt{1.5} = 3 \sqrt{\frac{3}{2}}$.
Since the angle between any two vectors is $\frac{\pi}{3}$ $(60^\circ)$,we can place them in a 3D coordinate system.
Let $\vec{A} = A \hat{i}$.
Since the angle between $\vec{A}$ and $\vec{B}$ is $60^\circ$,$\vec{B} = A(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = A(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})$.
For the third vector $\vec{C}$,it must make a $60^\circ$ angle with both $\vec{A}$ and $\vec{B}$. This implies $\vec{C}$ lies in the plane formed by $\vec{A}$ and $\vec{B}$ or out of it. However,in 3D space,three vectors with $60^\circ$ between each pair form the edges of a regular tetrahedron.
The resultant vector $\vec{R} = \vec{A} + \vec{B} + \vec{C}$.
The magnitude squared is $R^2 = A^2 + B^2 + C^2 + 2(AB \cos 60^\circ + BC \cos 60^\circ + CA \cos 60^\circ)$.
$R^2 = 3A^2 + 2(3 \cdot A^2 \cdot \frac{1}{2}) = 3A^2 + 3A^2 = 6A^2$.
Given $A = 3 \sqrt{1.5} = 3 \sqrt{\frac{3}{2}}$,then $A^2 = 9 \cdot \frac{3}{2} = 13.5$.
$R^2 = 6 \times 13.5 = 81$.
$R = \sqrt{81} = 9$ units.

(B) Let the two vectors be $\overrightarrow{A} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\overrightarrow{B} = 2 \hat{i} - 7 \hat{j} - 4 \hat{k}$.
The resultant vector $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} = (2+2) \hat{i} + (3-7) \hat{j} + (4-4) \hat{k} = 4 \hat{i} - 4 \hat{j} + 0 \hat{k}$.
The magnitude of the resultant vector is $R = \sqrt{(4)^2 + (-4)^2 + (0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.
The angle $\theta$ made by the resultant vector with the $x$-axis is given by $\cos \theta = \frac{R_x}{R}$.
Here,$R_x = 4$.
So,$\cos \theta = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = 45^{\circ}$.

(C) Let the two forces be $F_1 = 5x$ and $F_2 = 3x$.
Given the resultant $R = 35 \ N$ and the angle $\theta = 60^{\circ}$.
The formula for the resultant of two vectors is $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$.
Substituting the values: $35 = \sqrt{(5x)^2 + (3x)^2 + 2(5x)(3x) \cos 60^{\circ}}$.
Since $\cos 60^{\circ} = 0.5$,we have $35 = \sqrt{25x^2 + 9x^2 + 30x^2(0.5)}$.
$35 = \sqrt{25x^2 + 9x^2 + 15x^2} = \sqrt{49x^2} = 7x$.
Solving for $x$: $x = 35 / 7 = 5$.
Therefore,$F_1 = 5 \times 5 = 25 \ N$ and $F_2 = 3 \times 5 = 15 \ N$.

(A) Let the two vectors be $\vec{A}$ and $\vec{B}$,where $|\vec{A}|$ is the smaller magnitude.
Given: $|\vec{A}| + |\vec{B}| = 18$ and $|\vec{R}| = 12$.
Let $|\vec{A}| = x$,then $|\vec{B}| = 18 - x$.
Since the resultant $\vec{R}$ is at $90^{\circ}$ to the smaller vector $\vec{A}$,we have the relation: $|\vec{R}|^2 + |\vec{A}|^2 = |\vec{B}|^2$.
Substituting the values: $12^2 + x^2 = (18 - x)^2$.
$144 + x^2 = 324 + x^2 - 36x$.
$36x = 324 - 144$.
$36x = 180$.
$x = 5$.
Thus,$|\vec{A}| = 5$ and $|\vec{B}| = 18 - 5 = 13$.

(B) The angle between the vectors is $\theta = 60^{\circ}$.
The angle between the resultant vector and $\vec{a}$ is $\phi = 45^{\circ}$.
According to the parallelogram law of vector addition,the angle $\phi$ that the resultant makes with vector $\vec{a}$ is given by:
$\tan(\phi) = \frac{|\vec{b}| \sin(\theta)}{|\vec{a}| + |\vec{b}| \cos(\theta)}$
Substituting the given values $|\vec{b}| = 2$,$\theta = 60^{\circ}$,and $\phi = 45^{\circ}$:
$\tan(45^{\circ}) = \frac{2 \sin(60^{\circ})}{|\vec{a}| + 2 \cos(60^{\circ})}$
Since $\tan(45^{\circ}) = 1$,$\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$,and $\cos(60^{\circ}) = \frac{1}{2}$:
$1 = \frac{2 \times (\frac{\sqrt{3}}{2})}{|\vec{a}| + 2 \times (\frac{1}{2})}$
$1 = \frac{\sqrt{3}}{|\vec{a}| + 1}$
$|\vec{a}| + 1 = \sqrt{3}$
$|\vec{a}| = \sqrt{3} - 1$ units.

(D) Let vector $P = P \hat{i}$.
Since vector $Q$ has a magnitude of $10 \ m$,let $Q = 10 \cos \theta \hat{i} + 10 \sin \theta \hat{j}$.
The resultant vector $R = P + Q = (P + 10 \cos \theta) \hat{i} + (10 \sin \theta) \hat{j}$.
Given that the resultant is directed along the $Y$-axis,the $X$-component must be zero:
$P + 10 \cos \theta = 0 \Rightarrow 10 \cos \theta = -P$.
The magnitude of the resultant is $R = 10 \sin \theta$.
We are given $R = 2P$,so $10 \sin \theta = 2P \Rightarrow 5 \sin \theta = P$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$(P/5)^2 + (-P/10)^2 = 1$.
$P^2/25 + P^2/100 = 1$.
$(4P^2 + P^2) / 100 = 1 \Rightarrow 5P^2 = 100$.
$P^2 = 20 \Rightarrow P = \sqrt{20} = 2 \sqrt{5} \ m$.

(D) Let the initial velocity be $\vec{v}_i$ and the final velocity be $\vec{v}_f$. The magnitudes are given as $|\vec{v}_i| = \sqrt{5} \ m/s$ and $|\vec{v}_f| = 2\sqrt{5} \ m/s$.
The magnitude of the change in velocity is given by $|\Delta \vec{v}| = |\vec{v}_f - \vec{v}_i| = 5 \ m/s$.
Using the law of vector subtraction,the magnitude of the change in velocity is given by:
$|\Delta \vec{v}|^2 = |\vec{v}_f|^2 + |\vec{v}_i|^2 - 2|\vec{v}_f||\vec{v}_i| \cos \theta$,where $\theta$ is the angle between $\vec{v}_i$ and $\vec{v}_f$.
Substituting the given values:
$5^2 = (2\sqrt{5})^2 + (\sqrt{5})^2 - 2(2\sqrt{5})(\sqrt{5}) \cos \theta$
$25 = 20 + 5 - 2(2 \times 5) \cos \theta$
$25 = 25 - 20 \cos \theta$
$0 = -20 \cos \theta$
$\cos \theta = 0$
$\theta = 90^{\circ}$.
Thus,the angle between the initial and final velocities is $90^{\circ}$.

(C) Let the two forces in vector form be $\vec{A}$ and $\vec{B}$.
Their vector sum is $(\vec{A} + \vec{B})$ and their vector difference is $(\vec{A} - \vec{B})$.
Since the vector sum and vector difference are perpendicular to each other,their dot product must be zero:
$(\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = 0$
Expanding the dot product:
$\vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{A} - \vec{B} \cdot \vec{B} = 0$
Since the dot product is commutative,$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$,so the terms cancel out:
$|\vec{A}|^2 - |\vec{B}|^2 = 0$
$|\vec{A}|^2 = |\vec{B}|^2$
$|\vec{A}| = |\vec{B}|$
Therefore,the two forces are equal in magnitude.

(B) Based on the vector addition rule (triangle law) in the provided figure:
In $\triangle PQR$,the vectors $B$ and $E$ are in sequence,and $A$ is the resultant vector closing the triangle in the opposite direction. Thus,$A = B + E$.
Rearranging this gives $E = A - B$,which implies $-E = -(A - B) = -A + B$.
In $\triangle PSR$,the vectors $A$ and $D$ are in sequence,and $C$ is the resultant vector. Thus,$C = A + D$.
Rearranging this gives $A = C - D$,which can be written as $A = -D + C$.
Comparing these results with the given options,option $(b)$ is correct.

(A) Given equations are:
$A_1+A_2=5 A_3$ ---$(i)$
$A_1-A_2=3 A_3$ ---(ii)
Adding equations $(i)$ and (ii):
$2 A_1 = 8 A_3 \Rightarrow A_1 = 4 A_3$
Since $A_3 = 2 \hat{i} + 4 \hat{j}$,we have $A_1 = 4(2 \hat{i} + 4 \hat{j}) = 8 \hat{i} + 16 \hat{j}$.
Subtracting equation (ii) from $(i)$:
$2 A_2 = 2 A_3 \Rightarrow A_2 = A_3 = 2 \hat{i} + 4 \hat{j}$.
Now,calculating the ratio of magnitudes:
$\frac{|A_1|}{|A_2|} = \frac{|8 \hat{i} + 16 \hat{j}|}{|2 \hat{i} + 4 \hat{j}|} = \frac{\sqrt{8^2 + 16^2}}{\sqrt{2^2 + 4^2}}$
$= \frac{\sqrt{64 + 256}}{\sqrt{4 + 16}} = \frac{\sqrt{320}}{\sqrt{20}} = \sqrt{\frac{320}{20}} = \sqrt{16} = 4$.

(A) Given vectors are $r_1 = 2 \hat{x}$ and $r_2 = 2 \hat{y}$.
Their sum is $r_1 + r_2 = 2 \hat{x} + 2 \hat{y}$.
The magnitude of a vector $A = a \hat{x} + b \hat{y}$ is given by $|A| = \sqrt{a^2 + b^2}$.
Therefore,$|r_1 + r_2| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8}$.
Simplifying $\sqrt{8}$,we get $\sqrt{4 \times 2} = 2 \sqrt{2}$.

(A) Let the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force.
Given: $F_1 + F_2 = 25$ (Equation $1$)
The resultant $R$ is perpendicular to $F_1$. The magnitude of the resultant is $R = 10 \,N$.
Using the triangle law of vector addition, since $R \perp F_1$, we have a right-angled triangle formed by $F_1$, $R$, and $F_2$ (as the hypotenuse).
By Pythagoras theorem: $F_2^2 = F_1^2 + R^2$
$F_2^2 - F_1^2 = 10^2 = 100$
$(F_2 - F_1)(F_2 + F_1) = 100$
Substitute $F_1 + F_2 = 25$ into the equation:
$(F_2 - F_1)(25) = 100$
$F_2 - F_1 = 4$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$2F_2 = 29 \implies F_2 = 14.5 \,N$
Subtracting Equation $2$ from Equation $1$:
$2F_1 = 21 \implies F_1 = 10.5 \,N$
Thus, the two forces are $14.5 \,N$ and $10.5 \,N$.

(A) Let the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force. Let $F_1 = x$.
Given that the sum of magnitudes is $16 \,N$, so $F_2 = 16 - x$.
The resultant force $R = 8 \,N$ is normal to the smaller force $F_1$.
Using the vector triangle or the property of the resultant, we have the relation: $F_2^2 = F_1^2 + R^2$.
Substituting the values: $(16 - x)^2 = x^2 + 8^2$.
Expanding the equation: $256 + x^2 - 32x = x^2 + 64$.
Subtracting $x^2$ from both sides: $256 - 32x = 64$.
Rearranging to solve for $x$: $32x = 256 - 64 = 192$.
$x = \frac{192}{32} = 6 \,N$.
Therefore, the smaller force $F_1 = 6 \,N$ and the larger force $F_2 = 16 - 6 = 10 \,N$.
The forces are $6 \,N$ and $10 \,N$.

(B) Let the magnitudes of the two vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = a$.
The resultant vector $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B}$ makes an angle $\alpha$ with vector $\overrightarrow{A}$,given by the formula:
$\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$
Since $A = B = a$,we have:
$\tan \alpha = \frac{a \sin \theta}{a + a \cos \theta} = \frac{\sin \theta}{1 + \cos \theta}$
Using trigonometric identities $\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$ and $1 + \cos \theta = 2 \cos^2(\frac{\theta}{2})$:
$\tan \alpha = \frac{2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})}{2 \cos^2(\frac{\theta}{2})} = \tan(\frac{\theta}{2})$
Therefore,$\alpha = \frac{\theta}{2}$.
This shows that the resultant vector bisects the angle between the two vectors of equal magnitude.

(C) Let the origin be $O(0,0)$.
$1$. The person moves $30 \,m$ North: Position becomes $(0, 30)$.
$2$. Then moves $20 \,m$ East: Position becomes $(20, 30)$.
$3$. Finally moves $30 \sqrt{2} \,m$ in South-West direction. South-West direction makes an angle of $45^{\circ}$ with the South and West axes. The components of this displacement are:
$\Delta x = -30 \sqrt{2} \cos(45^{\circ}) = -30 \sqrt{2} \times \frac{1}{\sqrt{2}} = -30 \,m$
$\Delta y = -30 \sqrt{2} \sin(45^{\circ}) = -30 \sqrt{2} \times \frac{1}{\sqrt{2}} = -30 \,m$
New position = $(20 - 30, 30 - 30) = (-10, 0)$.
The displacement from the origin $(0, 0)$ to $(-10, 0)$ is $10 \,m$ along the West direction.