Questions & Answers: "vectors"

Given that the resultant of the three vectors is zero, we have $\vec{A} + \vec{B} + \vec{C} = 0$, which implies $\vec{A} + \vec{B} = -\vec{C}$. Taking the magnitude on both sides: $|\vec{A} + \vec{B}|^2 = |-\vec{C}|^2 \implies A^2 + B^2 + 2AB\cos\theta = C^2$, where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$. Substituting the given values: $3^2 + 4^2 + 2(3)(4)\cos\theta = 5^2 \implies 9 + 16 + 24\cos\theta = 25 \implies 25 + 24\cos\theta = 25 \implies 24\cos\theta = 0 \implies \cos\theta = 0$. Therefore, $\theta = 90^\circ$. The options $180^\circ$, $0^\circ$, and $60^\circ$ are incorrect because they would yield resultant magnitudes different from $5$ units.

A unit vector has a magnitude of exactly $1$. Therefore, the magnitude of the given vector is $\sqrt{(0.5)^2 + (-0.8)^2 + c^2} = 1$. Squaring both sides gives $0.25 + 0.64 + c^2 = 1 \implies 0.89 + c^2 = 1 \implies c^2 = 1 - 0.89 = 0.11 \implies c = \sqrt{0.11}$. The other options incorrect because they do not satisfy the unit magnitude condition: $0.25 + 0.64 + 0.39 \neq 1$, $0.25 + 0.64 + 0.01^2 \neq 1$, and $0.25 + 0.64 + 0.01 \neq 1$.

To find the angle between the two vectors, we can use the dot product formula: $\vec{A} \cdot \vec{B} = AB \cos\theta$. Let's compute the dot product: $\vec{A} \cdot \vec{B} = (3)(3) + (4)(4) + (5)(-5) = 9 + 16 - 25 = 0$. Since the dot product is zero, the angle $\theta$ between them must satisfy $\cos\theta = 0$, which means $\theta = 90^\circ$. The other options are incorrect because a non-zero dot product is required for any angle other than $90^\circ$.

For three vectors to have a minimum resultant of $0\text{ N}$, they must be able to form a closed triangle, which requires the sum of the two smaller magnitudes to be greater than or equal to the third ($2 + 3 = 5 < 6$, which fails this triangle inequality). The maximum possible resultant is when all three act in the same direction: $2 + 3 + 6 = 11\text{ N}$ (making option 3 incorrect). The minimum resultant occurs when the two smaller forces act in the same direction, directly opposing the largest force, resulting in $6\text{ N} - (2\text{ N} + 3\text{ N}) = 1\text{ N}$. Thus, the minimum magnitude is $1\text{ N}$ (making option 1 correct, and options 2 and 4 incorrect).

By vector cross product definition, we know that $\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})$. If the given condition holds, then $\vec{A} \times \vec{B} = -(\vec{A} \times \vec{B}) \implies 2(\vec{A} \times \vec{B}) = 0 \implies \vec{A} \times \vec{B} = 0$. This means the magnitude $AB\sin\theta = 0$. Since $\vec{A}$ and $\vec{B}$ are non-zero vectors, $\sin\theta = 0 \implies \theta = 0$ or $\pi$ radians. The other options are incorrect because they would yield a non-zero cross product where $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A} \neq \vec{B} \times \vec{A}$.

The projection of vector $\vec{A}$ on $\vec{B}$ is given by the formula $\text{proj}_{\vec{B}}\vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$. Calculating the dot product: $\vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) = 1 - 1 = 0$. Since the dot product is zero (the vectors are perpendicular), the projection is $\frac{0}{|\vec{B}|} = 0$. Hence, option 1 is correct, and the other options are mathematically incorrect.

The area of a parallelogram is given by the magnitude of the cross product of its adjacent sides: $\text{Area} = |\vec{P} \times \vec{Q}|$. Computing the cross product: $\vec{P} \times \vec{Q} = (2\hat{i} + 3\hat{j}) \times (\hat{i} + 4\hat{j}) = 8(\hat{i} \times \hat{j}) + 3(\hat{j} \times \hat{i}) = 8\hat{k} - 3\hat{k} = 5\hat{k}$. The magnitude of $5\hat{k}$ is $5$. Hence, the area is $5$ units. Other options represent computational errors in evaluating the cross product magnitude.

Squaring both sides of the given equation $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$, we get: $A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 - 2AB\cos\theta \implies 4AB\cos\theta = 0$. Since $\vec{A}$ and $\vec{B}$ are non-zero, $\cos\theta = 0$, which yields $\theta = 90^\circ$. The other options are incorrect because any angle other than $90^\circ$ would make the magnitudes of the sum and difference unequal.

Three vectors are coplanar if their scalar triple product is zero: $[\vec{A} \; \vec{B} \; \vec{C}] = 0$. This can be evaluated using the determinant of their components: $\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & m & 5 \end{vmatrix} = 0$. Expanding along the first row: $2(10 - (-3m)) - (-1)(5 - (-9)) + 1(m - 6) = 0 \implies 2(10 + 3m) + 1(14) + m - 6 = 0 \implies 20 + 6m + 14 + m - 6 = 0 \implies 7m + 28 = 0 \implies m = -4$. Thus, $m = -4$ is correct, and other values of $m$ will make the scalar triple product non-zero, meaning the vectors would not be coplanar.

The component of $\vec{F}$ along the direction of $\vec{d}$ is given by $F_d = \vec{F} \cdot \hat{d} = \frac{\vec{F} \cdot \vec{d}}{|\vec{d}|}$. First, we find the magnitude of $\vec{d}$: $|\vec{d}| = \sqrt{3^2 + 4^2} = 5$. Next, we compute the dot product: $\vec{F} \cdot \vec{d} = (5)(3) + (3)(4) = 15 + 12 = 27$. Finally, $F_d = \frac{27}{5} = 5.4\text{ N}$. Thus, option 1 is correct, and the other options do not match this calculation.

A vector perpendicular to both $\vec{A}$ and $\vec{B}$ is given by their cross product: $\vec{C} = \vec{A} \times \vec{B} = (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = -\hat{i} \times \hat{j} + \hat{j} \times \hat{i} = -\hat{k} - \hat{k} = -2\hat{k}$. A unit vector in this direction is $\frac{\vec{C}}{|\vec{C}|} = \frac{-2\hat{k}}{2} = -\hat{k}$. Since unit vectors can be in either direction perpendicular to the plane, $\hat{k}$ is also a valid unit vector perpendicular to both. Options 2 and 3 lie in the same $xy$-plane and thus cannot be perpendicular to the vectors which also lie in the $xy$-plane.

Let the east direction be along the positive x-axis ($\hat{i}$), north be along the positive y-axis ($\hat{j}$), and vertically upward be along the positive z-axis ($\hat{k}$). The displacement vector of the body is $\vec{s} = 4\hat{i} + 3\hat{j} + 12\hat{k}$. The magnitude of this displacement vector is $|\vec{s}| = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13\text{ m}$. The option $19\text{ m}$ is the simple distance path length ($4+3+12=19$), which is incorrect for displacement. $5\text{ m}$ only considers the horizontal displacement ($4\hat{i} + 3\hat{j}$), neglecting the height. Thus, $13\text{ m}$ is the correct total displacement magnitude.

For a body to be in equilibrium under three forces, the resultant of any two forces must be equal in magnitude and opposite in direction to the third force: $\vec{F}_1 + \vec{F}_2 = -\vec{F}_3 \implies |\vec{F}_1 + \vec{F}_2| = |\vec{F}_3|$. Since the angle between $\vec{F}_1$ ($5\text{ N}$) and $\vec{F}_2$ ($12\text{ N}$) is $90^\circ$, their resultant is $R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ N}$. Thus, the third force must have a magnitude of $13\text{ N}$ to keep the system in equilibrium. Other options represent algebraic sum ($17\text{ N}$) or difference ($7\text{ N}$), which are only valid if the angle between them is $0^\circ$ or $180^\circ$, respectively.

We are given $|\vec{A} \times \vec{B}| = \sqrt{3}(\vec{A} \cdot \vec{B}) \implies AB\sin\theta = \sqrt{3}AB\cos\theta$. Dividing both sides by $AB\cos\theta$, we get $\tan\theta = \sqrt{3} \implies \theta = 60^\circ$. Now, the magnitude of the vector sum is given by $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} = \sqrt{A^2 + B^2 + 2AB\cos(60^\circ)} = \sqrt{A^2 + B^2 + 2AB(0.5)} = (A^2 + B^2 + AB)^{1/2}$. Therefore, option 1 is correct.

Let the vertical downward direction be along $-\hat{j}$ and the horizontal direction of the girl's movement be along $\hat{i}$. Thus, the velocity of the rain is $\vec{v}_r = -4\hat{j}$ and the velocity of the girl is $\vec{v}_g = 3\hat{i}$. The velocity of the rain relative to the girl is $\vec{v}_{rg} = \vec{v}_r - \vec{v}_g = -3\hat{i} - 4\hat{j}$. The magnitude of this relative velocity is $|\vec{v}_{rg}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5\text{ km/h}$. Option 2 ($7\text{ km/h}$) and option 3 ($1\text{ km/h}$) are algebraic additions/subtractions which do not apply to perpendicular vector quantities.