Mix Examples-Vectors Questions in English

(A) When $n$ equal forces of magnitude $F$ act at a point in the same plane such that the angle between consecutive forces is equal,the angle between them is given by $\theta = \frac{360^\circ}{n}$.
Here,$n = 5$ and $F = 10 \, N$.
The angle between consecutive forces is $\theta = \frac{360^\circ}{5} = 72^\circ$.
According to the polygon law of vector addition,if a set of vectors acting at a point can be represented by the sides of a closed polygon taken in order,their resultant is zero.
Since these five equal forces are symmetric and distributed equally in a plane,they form a closed regular pentagon when placed head-to-tail.
Therefore,the resultant force is $0 \, N$.

(B) Let the resultant vector be $R$. Given that $R \perp A$,the angle between $R$ and $A$ is $90^\circ$.
From the vector addition formula,the angle $\alpha$ that the resultant $R$ makes with vector $A$ is given by $\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$.
Since $\alpha = 90^\circ$,$\tan 90^\circ$ is undefined,which implies the denominator must be zero: $A + B \cos \theta = 0$,so $\cos \theta = -\frac{A}{B}$.
The magnitude of the resultant is $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Given $R = \frac{B}{2}$,we have $\frac{B^2}{4} = A^2 + B^2 + 2AB \cos \theta$.
Substituting $\cos \theta = -\frac{A}{B}$ into the equation: $\frac{B^2}{4} = A^2 + B^2 + 2AB(-\frac{A}{B}) = A^2 + B^2 - 2A^2 = B^2 - A^2$.
Rearranging gives $A^2 = B^2 - \frac{B^2}{4} = \frac{3B^2}{4}$,so $A = \frac{\sqrt{3}}{2}B$.
Finally,$\cos \theta = -\frac{A}{B} = -\frac{\sqrt{3}}{2}$.
Therefore,$\theta = 150^\circ$.

(C) Let the angle between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ be $\alpha$. The magnitude of the resultant $\overrightarrow{R}$ is given by:
$R^2 = P^2 + Q^2 + 2PQ \cos \alpha \quad \dots(1)$
When $\overrightarrow{Q}$ is doubled to $2\overrightarrow{Q}$,let the new resultant be $\overrightarrow{R_1}$. The magnitude squared is:
$R_1^2 = P^2 + (2Q)^2 + 2P(2Q) \cos \alpha = P^2 + 4Q^2 + 4PQ \cos \alpha \quad \dots(2)$
Given that the new resultant $\overrightarrow{R_1}$ is perpendicular to $\overrightarrow{P}$,the vector sum $\overrightarrow{R_1} = \overrightarrow{P} + 2\overrightarrow{Q}$ implies that $\overrightarrow{P} \cdot \overrightarrow{R_1} = 0$. Since $\overrightarrow{R_1} = \overrightarrow{P} + 2\overrightarrow{Q}$,we have $\overrightarrow{P} \cdot (\overrightarrow{P} + 2\overrightarrow{Q}) = 0$,which gives $P^2 + 2PQ \cos \alpha = 0$,or $2PQ \cos \alpha = -P^2 \quad \dots(3)$
Substitute equation $(3)$ into equation $(1)$:
$R^2 = P^2 + Q^2 + (-P^2) = Q^2$
Therefore,$R = Q$.

(C) Given $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$,which implies $\overrightarrow{B} = \overrightarrow{C} - \overrightarrow{A}$.
Since $\overrightarrow{C} \perp \overrightarrow{A}$,the vectors $\overrightarrow{C}$ and $-\overrightarrow{A}$ form a right-angled triangle with $\overrightarrow{B}$ as the hypotenuse.
Thus,$B^2 = C^2 + A^2$.
Given $|\overrightarrow{A}| = |\overrightarrow{C}|$,we substitute $C = A$ into the equation:
$B^2 = A^2 + A^2 = 2A^2$,so $B = \sqrt{2}A$.
Using the law of cosines for the vector addition $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$:
$C^2 = A^2 + B^2 + 2AB \cos \theta$,where $\theta$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.
Substituting $C^2 = A^2$ and $B^2 = 2A^2$:
$A^2 = A^2 + 2A^2 + 2A(\sqrt{2}A) \cos \theta$.
$A^2 = 3A^2 + 2\sqrt{2}A^2 \cos \theta$.
$-2A^2 = 2\sqrt{2}A^2 \cos \theta$.
$\cos \theta = -\frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{3\pi}{4} \text{ radians}$.

(D) Let the magnitudes of the two vectors be $A$ and $B$.
The maximum magnitude of the resultant is given by $R_{\max} = A + B = 17$.
The minimum magnitude of the resultant is given by $R_{\min} = |A - B| = 7$.
Adding these two equations: $(A + B) + (A - B) = 17 + 7 \implies 2A = 24 \implies A = 12$.
Substituting $A = 12$ into the first equation: $12 + B = 17 \implies B = 5$.
When the two vectors are at right angles to each other $(\theta = 90^\circ)$,the magnitude of their resultant $R$ is given by $R = \sqrt{A^2 + B^2}$.
Substituting the values: $R = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ units.

(B) Given $P = Q = R$.
Case $1$: $\vec{P} + \vec{Q} = \vec{R} \implies \vec{Q} = \vec{R} - \vec{P}$.
Squaring both sides: $Q^2 = R^2 + P^2 - 2RP \cos \theta_1$.
Since $P = Q = R$,we have $P^2 = P^2 + P^2 - 2P^2 \cos \theta_1 \implies P^2 = 2P^2(1 - \cos \theta_1) \implies 1 = 2 - 2 \cos \theta_1 \implies 2 \cos \theta_1 = 1 \implies \cos \theta_1 = 1/2 \implies \theta_1 = 60^\circ$.
Case $2$: $\vec{P} + \vec{Q} + \vec{R} = \vec{0} \implies \vec{P} + \vec{R} = -\vec{Q}$.
Squaring both sides: $P^2 + R^2 + 2PR \cos \theta_2 = Q^2$.
Since $P = Q = R$,we have $P^2 + P^2 + 2P^2 \cos \theta_2 = P^2 \implies 2P^2 + 2P^2 \cos \theta_2 = P^2 \implies 2 \cos \theta_2 = -1 \implies \cos \theta_2 = -1/2 \implies \theta_2 = 120^\circ$.
Comparing the two,$\theta_2 = 2\theta_1$,which implies $\theta_1 = \theta_2 / 2$.

(D) Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. Let $|A| = |B| = x$. Then $|C| = \sqrt{2}x$.
From $\vec{A} + \vec{B} = -\vec{C}$,squaring both sides: $A^2 + B^2 + 2AB \cos \theta_1 = C^2$.
Substituting values: $x^2 + x^2 + 2x^2 \cos \theta_1 = (\sqrt{2}x)^2$.
$2x^2 + 2x^2 \cos \theta_1 = 2x^2 \Rightarrow 2x^2 \cos \theta_1 = 0 \Rightarrow \cos \theta_1 = 0 \Rightarrow \theta_1 = 90^\circ$.
Now,consider $\vec{B} + \vec{C} = -\vec{A}$. Squaring both sides: $B^2 + C^2 + 2BC \cos \theta_2 = A^2$.
Substituting values: $x^2 + 2x^2 + 2(x)(\sqrt{2}x) \cos \theta_2 = x^2$.
$2x^2 + 2\sqrt{2}x^2 \cos \theta_2 = 0 \Rightarrow 2\sqrt{2}x^2 \cos \theta_2 = -2x^2$.
$\cos \theta_2 = -\frac{1}{\sqrt{2}} \Rightarrow \theta_2 = 135^\circ$.
Since $|A| = |B|$,by symmetry,the angle between $\vec{A}$ and $\vec{C}$ is also $\theta_3 = 135^\circ$.
Thus,the angles between the vectors are $90^\circ, 135^\circ, 135^\circ$.

(C) Given $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = 6\hat{i} + 8\hat{j}$.
Magnitudes are $A = \sqrt{3^2 + 4^2} = 5$ and $B = \sqrt{6^2 + 8^2} = 10$.
Check option $(A)$: $\vec{A} \times \vec{B} = (3\hat{i} + 4\hat{j}) \times (6\hat{i} + 8\hat{j}) = (3 \times 8 - 4 \times 6)\hat{k} = (24 - 24)\hat{k} = 0$. This is correct.
Check option $(B)$: $\frac{A}{B} = \frac{5}{10} = \frac{1}{2}$. This is correct.
Check option $(C)$: $\vec{A} \cdot \vec{B} = (3)(6) + (4)(8) = 18 + 32 = 50$. The option states $40$,which is incorrect.
Therefore,the incorrect statement is option $(C)$.

(B) First,calculate the magnitudes of the vectors:
$A = |\overrightarrow A| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$B = |\overrightarrow B| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
Now,evaluate each option:
$A$: $\overrightarrow A \times \overrightarrow B = (3\hat i + 4\hat j) \times (6\hat i + 8\hat j) = (3 \times 8 - 4 \times 6)\hat k = (24 - 24)\hat k = 0$. This is true.
$B$: $\frac{A}{B} = \frac{5}{10} = \frac{1}{2}$. The statement $\frac{A}{B} = \frac{1}{4}$ is false.
$C$: $\overrightarrow A \cdot \overrightarrow B = (3)(6) + (4)(8) = 18 + 32 = 50$. This is true.
$D$: $A = 5$. This is true.
Therefore,the statement that is not true is $\frac{A}{B} = \frac{1}{4}$.

(C) First,calculate the magnitudes of vectors $A$ and $B$:
$|A| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$|B| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
Now,check statement $(b)$: $\frac{|A|}{|B|} = \frac{5}{10} = \frac{1}{2}$. This is correct.
Next,observe that $B = 2(3 \hat{i} + 4 \hat{j}) = 2A$. Since $B$ is a scalar multiple of $A$,the vectors are parallel.
For parallel vectors,the cross product is zero: $A \times B = 0$. This is also correct.
Since both $(a)$ and $(b)$ are correct,the correct option is $(c)$.

(D) When $n$ coplanar forces of equal magnitude $F$ are arranged such that each force makes an equal angle $\Delta\theta$ with the preceding one,they form a closed polygon if the total angle covered is a multiple of $2\pi$.
Here,the number of forces $n = 100$.
The angle between consecutive forces is $\Delta\theta = \pi/50$.
The total angle covered by the forces is $\theta_{total} = n \times \Delta\theta = 100 \times (\pi/50) = 2\pi$.
Since the forces are coplanar and the total angle is $2\pi$,the forces form a closed polygon in vector addition.
Therefore,the vector sum (resultant) of these forces is $0 \ N$.

(A) $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \implies AB \cos \theta = AB \sin \theta \implies \tan \theta = 1 \implies \theta = 45^{\circ}$ or $135^{\circ}$ (if considering magnitude equality). Thus,$A \rightarrow p$.
$(B)$ $\vec{A} \cdot \vec{B} = B^2 \implies AB \cos \theta = B^2 \implies A \cos \theta = B$. If $A=B$,then $\cos \theta = 1 \implies \theta = 0^{\circ}$. Thus,$B \rightarrow q, r$.
$(C)$ $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \implies A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta \implies 4AB \cos \theta = 0 \implies \theta = 90^{\circ}$. Thus,$C \rightarrow s$.
$(D)$ $|\vec{A} \times \vec{B}| = AB \implies AB \sin \theta = AB \implies \sin \theta = 1 \implies \theta = 90^{\circ}$. Thus,$D \rightarrow s$.

(A) Given: $|A| = |B| = x$ and $\theta = 60^{\circ}$.
$(A) |A+B| = \sqrt{x^2 + x^2 + 2x^2 \cos(60^{\circ})} = \sqrt{2x^2 + 2x^2(0.5)} = \sqrt{3x^2} = \sqrt{3}x$. Thus,$(A \rightarrow r)$.
$(B) |A-B| = \sqrt{x^2 + x^2 - 2x^2 \cos(60^{\circ})} = \sqrt{2x^2 - 2x^2(0.5)} = \sqrt{x^2} = x$. Thus,$(B \rightarrow q)$.
$(C) A \cdot B = |A||B| \cos(60^{\circ}) = x \cdot x \cdot 0.5 = \frac{x^2}{2}$. Thus,$(C \rightarrow s)$.
$(D) |A \times B| = |A||B| \sin(60^{\circ}) = x \cdot x \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}x^2$. Thus,$(D \rightarrow p)$.
Therefore,the correct matching is $(A \rightarrow r, B \rightarrow q, C \rightarrow s, D \rightarrow p)$.

$(1)$ For two vectors of equal magnitude,the resultant lies along the angle bisector. If magnitudes are unequal,the resultant lies closer to the larger vector. However,in the context of standard matching problems,the resultant of two vectors is generally associated with the direction between them. Given the options,$(1)$ matches $(a)$ if we assume equal magnitudes or general direction.
$(2)$ The cross product $\overrightarrow A \times \overrightarrow B$ is defined as a vector perpendicular to the plane containing both $\overrightarrow A$ and $\overrightarrow B$ by the right-hand rule. Thus,$(2)$ matches $(c)$.

(C) Given: $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$.
$1$. Scalar Product (Dot Product):
$\overrightarrow{ F } \cdot \overrightarrow{ r } = (2)(3) + (1)(2) + (-1)(-2) = 6 + 2 + 2 = 10$.
$2$. Vector Product (Cross Product):
$\overrightarrow{ F } \times \overrightarrow{ r } = \begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -1 \\ 3 & 2 & -2 \end{vmatrix}$
$= \hat{ i }((1)(-2) - (-1)(2)) - \hat{ j }((2)(-2) - (-1)(3)) + \hat{ k }((2)(2) - (1)(3))$
$= \hat{ i }(-2 + 2) - \hat{ j }(-4 + 3) + \hat{ k }(4 - 3)$
$= 0 \hat{ i } + 1 \hat{ j } + 1 \hat{ k } = \hat{ j } + \hat{ k }$.
$3$. Magnitude of Vector Product:
$|\overrightarrow{ F } \times \overrightarrow{ r }| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
Thus,the magnitudes are $10$ and $\sqrt{2}$.

(B) Given that the particles are at an equal distance from the origin,their magnitudes are equal: $|\overrightarrow{A}| = |\overrightarrow{B}|$.
$|\overrightarrow{A}|^2 = |\overrightarrow{B}|^2 \implies 2^2 + (3n)^2 + 2^2 = 2^2 + (-2)^2 + (4p)^2$.
$4 + 9n^2 + 4 = 4 + 4 + 16p^2 \implies 8 + 9n^2 = 8 + 16p^2 \implies 9n^2 = 16p^2$.
Taking the square root,$3n = \pm 4p$,so $p = \pm \frac{3n}{4}$.
Since the vectors are at a right angle,their dot product is zero: $\overrightarrow{A} \cdot \overrightarrow{B} = 0$.
$(2)(2) + (3n)(-2) + (2)(4p) = 0 \implies 4 - 6n + 8p = 0$.
Case $1$: Substitute $p = \frac{3n}{4}$ into the dot product equation: $4 - 6n + 8(\frac{3n}{4}) = 0 \implies 4 - 6n + 6n = 0 \implies 4 = 0$ (Impossible).
Case $2$: Substitute $p = -\frac{3n}{4}$ into the dot product equation: $4 - 6n + 8(-\frac{3n}{4}) = 0 \implies 4 - 6n - 6n = 0 \implies 12n = 4 \implies n = \frac{1}{3}$.
Therefore,$n^{-1} = \frac{1}{n} = 3$.

(D) The initial displacement vector $\vec{S}_1$ is along the direction of $3 \hat{i} + 4 \hat{j}$. The unit vector is $\hat{u} = \frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^2 + 4^2}} = \frac{3 \hat{i} + 4 \hat{j}}{5}$.
Thus,$\vec{S}_1 = 30 \times \left(\frac{3 \hat{i} + 4 \hat{j}}{5}\right) = 18 \hat{i} + 24 \hat{j} \ m$.
Let the second displacement be $\vec{S}_2 = d \hat{n}$,where $\hat{n}$ is a unit vector perpendicular to $\vec{S}_1$. Since $\vec{S}_1 = 18 \hat{i} + 24 \hat{j}$,a perpendicular vector is proportional to $4 \hat{i} - 3 \hat{j}$.
So,$\vec{S}_2 = k(4 \hat{i} - 3 \hat{j})$. The magnitude is $d = |k| \sqrt{4^2 + (-3)^2} = 5|k|$.
The total displacement $\vec{S}_{net} = \vec{S}_1 + \vec{S}_2 = (18 + 4k) \hat{i} + (24 - 3k) \hat{j}$.
Since the net displacement is along the $x$-axis,the $y$-component must be zero:
$24 - 3k = 0 \implies k = 8$.
Then $\vec{S}_2 = 8(4 \hat{i} - 3 \hat{j}) = 32 \hat{i} - 24 \hat{j}$.
The magnitude $d = |\vec{S}_2| = \sqrt{32^2 + (-24)^2} = \sqrt{1024 + 576} = \sqrt{1600} = 40 \ m$.

(D) Given vectors are $\vec{A}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k}$,$\vec{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k}$,and $\vec{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}$.
First,check the vector addition relation:
$\vec{B}+\vec{C} = (\hat{\imath}-3 \hat{\jmath}+5 \hat{k}) + (2 \hat{\imath}+\hat{\jmath}-4 \hat{k}) = 3 \hat{\imath}-2 \hat{\jmath}+\hat{k} = \vec{A}$.
Thus,$\vec{A}=\vec{B}+\vec{C}$ is satisfied.
Now,calculate the squares of the magnitudes:
$A^2 = |\vec{A}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14$.
$B^2 = |\vec{B}|^2 = 1^2 + (-3)^2 + 5^2 = 1 + 9 + 25 = 35$.
$C^2 = |\vec{C}|^2 = 2^2 + 1^2 + (-4)^2 = 4 + 1 + 16 = 21$.
Observing the values,$B^2 = 35$ and $A^2 + C^2 = 14 + 21 = 35$.
Therefore,$B^2 = A^2 + C^2$ is satisfied.
Hence,the correct option is $\vec{A}=\vec{B}+\vec{C}$ and $B^2=A^2+C^2$.

(C) Let $\vec{B} = B\hat{i}$ and $\vec{A} = A_x\hat{i} + A_y\hat{j}$. Given $|\vec{A}| = 6$,so $A_x^2 + A_y^2 = 36$.
Since the resultant $\vec{R} = \vec{A} + \vec{B} = (A_x + B)\hat{i} + A_y\hat{j}$ is along the $y$-axis,the $x$-component must be zero: $A_x + B = 0$,so $A_x = -B$.
Substituting $A_x = -B$ into the magnitude equation: $(-B)^2 + A_y^2 = 36$,which gives $A_y^2 = 36 - B^2$.
The resultant vector is $\vec{R} = A_y\hat{j}$,so its magnitude is $|\vec{R}| = |A_y| = \sqrt{36 - B^2}$.
Given $|\vec{R}| = 3B$,we have $\sqrt{36 - B^2} = 3B$.
Squaring both sides: $36 - B^2 = 9B^2$,which simplifies to $10B^2 = 36$.
Thus,$B^2 = 3.6$,so $B = \sqrt{3.6}$ units.

(C) First,calculate $\vec{P} = \vec{A} + \vec{B} + \vec{C} = (1-1+2)\hat{i} + (1+1-2)\hat{j} + (3+4-8)\hat{k} = 2\hat{i} + 0\hat{j} - 1\hat{k}$.
Next,calculate $\vec{Q} = \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 3 \\ -1 & 1 & 4 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(4+3) + \hat{k}(1+1) = 1\hat{i} - 7\hat{j} + 2\hat{k}$.
Now,find the dot product $\vec{P} \cdot \vec{Q} = (2)(1) + (0)(-7) + (-1)(2) = 2 + 0 - 2 = 0$.
Since the dot product is $0$,the vectors $\vec{P}$ and $\vec{Q}$ are perpendicular to each other.
Therefore,the angle between them is $90^{\circ}$.

(C) To determine if the vectors form a triangle,we check if their sum is zero or if they can be arranged head-to-tail.
First,calculate the sum of the vectors: $\vec{A} + \vec{B} + \vec{C} = (3+1+2) \hat{i} + (-2-3-1) \hat{j} + (1+5+4) \hat{k} = 6 \hat{i} - 6 \hat{j} + 10 \hat{k} \neq 0$.
Since the sum is not zero,they do not form a closed loop.
Alternatively,check if any vector is the sum of the other two.
$\vec{A} + \vec{B} = 4 \hat{i} - 5 \hat{j} + 6 \hat{k} \neq \vec{C}$.
$\vec{A} + \vec{C} = 5 \hat{i} - 3 \hat{j} + 5 \hat{k} \neq \vec{B}$.
$\vec{B} + \vec{C} = 3 \hat{i} - 4 \hat{j} + 9 \hat{k} \neq \vec{A}$.
Since no vector is the resultant of the other two,these vectors cannot form a triangle.