Multiplication of Vectors Questions in English

(A) The angle $\theta$ between two vectors $\vec A$ and $\vec B$ is given by the formula: $\cos \theta = \frac{\vec A \cdot \vec B}{|A||B|}$.
First,calculate the dot product $\vec A \cdot \vec B = (3)(3) + (4)(4) + (5)(-5) = 9 + 16 - 25 = 0$.
Next,calculate the magnitudes $|A| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ and $|B| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$.
Substituting these values into the formula: $\cos \theta = \frac{0}{\sqrt{50} \cdot \sqrt{50}} = \frac{0}{50} = 0$.
Since $\cos \theta = 0$,the angle $\theta = 90^\circ$.

(B) The angle $\theta$ between two vectors is found using the dot product formula: $\cos \theta = \frac{\vec F_1 \cdot \vec F_2}{|\vec F_1| |\vec F_2|}$.
First,calculate the dot product: $\vec F_1 \cdot \vec F_2 = (5 \times 10) + (10 \times -5) + (-20 \times -15) = 50 - 50 + 300 = 300$.
Next,calculate the magnitudes: $|\vec F_1| = \sqrt{5^2 + 10^2 + (-20)^2} = \sqrt{25 + 100 + 400} = \sqrt{525} \approx 22.91$.
$|\vec F_2| = \sqrt{10^2 + (-5)^2 + (-15)^2} = \sqrt{100 + 25 + 225} = \sqrt{350} \approx 18.71$.
Now,substitute these into the formula: $\cos \theta = \frac{300}{\sqrt{525} \times \sqrt{350}} = \frac{300}{\sqrt{183750}} \approx \frac{300}{428.66} \approx 0.70$.
Since $\cos 45^\circ \approx 0.707$,the angle $\theta$ is approximately $45^\circ$.

(C) Let the two vectors be $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ and $\overrightarrow{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$.
Since the vectors are perpendicular,their dot product must be zero,i.e.,$\overrightarrow{A} \cdot \overrightarrow{B} = 0$.
Calculating the dot product:
$(2\hat{i} + 3\hat{j} + 8\hat{k}) \cdot (-4\hat{i} + 4\hat{j} + \alpha\hat{k}) = 0$
$(2)(-4) + (3)(4) + (8)(\alpha) = 0$
$-8 + 12 + 8\alpha = 0$
$4 + 8\alpha = 0$
$8\alpha = -4$
$\alpha = -4/8 = -0.5$.

(B) The expression given is a scalar triple product of the form $\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{A})$.
By the properties of the cross product,the vector $\overrightarrow{C} = (\overrightarrow{B} \times \overrightarrow{A})$ is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$.
Since $\overrightarrow{C}$ is perpendicular to $\overrightarrow{A}$,the dot product of $\overrightarrow{A}$ and $\overrightarrow{C}$ must be zero.
Mathematically,$\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{A}) = 0$ because the vector $(\overrightarrow{B} \times \overrightarrow{A})$ lies in a plane perpendicular to $\overrightarrow{A}$.

(C) The cross product of two vectors is anti-commutative,meaning $\overrightarrow{A} \times \overrightarrow{B} = -(\overrightarrow{B} \times \overrightarrow{A})$.
Given the condition $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$,we can substitute the anti-commutative property into the equation:
$-(\overrightarrow{B} \times \overrightarrow{A}) = \overrightarrow{B} \times \overrightarrow{A}$.
This implies $2(\overrightarrow{B} \times \overrightarrow{A}) = 0$,or $\overrightarrow{A} \times \overrightarrow{B} = 0$.
The magnitude of the cross product is given by $|\overrightarrow{A} \times \overrightarrow{B}| = |A||B| \sin \theta$,where $\theta$ is the angle between the vectors.
For the cross product to be zero,$\sin \theta$ must be $0$,which occurs when $\theta = 0$ or $\theta = \pi$.
However,the vector product $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$ is only possible if both sides are zero vectors,which happens when $\theta = 0$ or $\theta = \pi$. Among the given options,$\pi$ is the correct choice.

(B) The cross product $\overrightarrow A \times \overrightarrow B$ is calculated using the determinant method:
$\overrightarrow A \times \overrightarrow B = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{vmatrix}$
$= \hat i(1 \times 4 - 2 \times -2) - \hat j(3 \times 4 - 2 \times 2) + \hat k(3 \times -2 - 1 \times 2)$
$= \hat i(4 + 4) - \hat j(12 - 4) + \hat k(-6 - 2)$
$= 8\hat i - 8\hat j - 8\hat k$
Now,the magnitude is given by:
$|\overrightarrow A \times \overrightarrow B | = \sqrt{(8)^2 + (-8)^2 + (-8)^2}$
$= \sqrt{64 + 64 + 64}$
$= \sqrt{3 \times 64}$
$= 8\sqrt 3 $

(D) The cross product $\overrightarrow C = \overrightarrow A \times \overrightarrow B$ results in a vector $\overrightarrow C$ that is perpendicular to the plane containing both $\overrightarrow A$ and $\overrightarrow B$.
Since $\overrightarrow C$ is perpendicular to the plane,it is perpendicular to any vector lying within that plane.
$1$. $\overrightarrow C \perp \overrightarrow A$ and $\overrightarrow C \perp \overrightarrow B$ are true by the definition of the cross product.
$2$. The vector sum $(\overrightarrow A + \overrightarrow B)$ also lies in the same plane as $\overrightarrow A$ and $\overrightarrow B$,so $\overrightarrow C \perp (\overrightarrow A + \overrightarrow B)$ is also true.
$3$. The statement $\overrightarrow C \perp (\overrightarrow A \times \overrightarrow B)$ implies $\overrightarrow C \perp \overrightarrow C$,which is impossible for a non-zero vector as a vector cannot be perpendicular to itself. Therefore,this statement is wrong.

(D) The scalar product (dot product) of two vectors $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$ is given by $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.
Given $\vec{F}_1 = 2\hat{i} + 0\hat{j} + 5\hat{k}$ and $\vec{F}_2 = 0\hat{i} + 3\hat{j} + 4\hat{k}$.
Calculating the dot product: $\vec{F}_1 \cdot \vec{F}_2 = (2)(0) + (0)(3) + (5)(4)$.
$\vec{F}_1 \cdot \vec{F}_2 = 0 + 0 + 20 = 20$.
The magnitude of the scalar product is $20$.

(C) Two vectors are perpendicular if their dot product is zero.
Let the given vector be $\overrightarrow{F} = 4\hat{i} - 3\hat{j}$.
For a vector $\overrightarrow{A}$ to be perpendicular to $\overrightarrow{F}$,$\overrightarrow{F} \cdot \overrightarrow{A} = 0$.
Checking option $C$: $\overrightarrow{A} = 7\hat{k}$.
$\overrightarrow{F} \cdot \overrightarrow{A} = (4\hat{i} - 3\hat{j}) \cdot (7\hat{k}) = 4(0) - 3(0) + 0(7) = 0$.
Since the dot product is $0$,the vector $7\hat{k}$ is perpendicular to $\overrightarrow{F}$.

(D) The dot product of two vectors $\vec{A}$ and $\vec{B}$ is defined as $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$,where $\theta$ is the angle between the vectors.
If the vectors are at right angles,then $\theta = 90^\circ$.
Substituting this into the formula,we get $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos 90^\circ$.
Since $\cos 90^\circ = 0$,the dot product $\vec{A} \cdot \vec{B} = 0$.

(B) Let the two vectors be $\vec{A} = -2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + 2\hat{j} - 4\hat{k}$.
The angle $\theta$ between two vectors is given by the formula $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.
First,calculate the dot product $\vec{A} \cdot \vec{B}$:
$\vec{A} \cdot \vec{B} = (-2)(1) + (3)(2) + (1)(-4) = -2 + 6 - 4 = 0$.
Since the dot product is $0$,we have $\cos \theta = 0$.
Therefore,$\theta = 90^\circ$.

(C) Let $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$.
The dot product is given by $\vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k}) = (1)(0) + (1)(1) + (0)(1) = 1$.
The magnitudes are $|\vec{A}| = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $|\vec{B}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The angle $\theta$ between the vectors is given by $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.
Substituting the values,$\cos \theta = \frac{1}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.

(A) The dot product of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is defined as $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ \cos \theta$,where $\theta$ is the angle between the two vectors.
Given that $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ$.
Substituting the definition,we get $PQ \cos \theta = PQ$.
Dividing both sides by $PQ$ (assuming $P, Q \neq 0$),we get $\cos \theta = 1$.
Since $\cos 0^\circ = 1$,the angle $\theta$ must be $0^\circ$.

(C) The angle $\theta$ between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by the dot product formula: $\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta$.
First,calculate the dot product: $\overrightarrow{A} \cdot \overrightarrow{B} = (5\hat{i} + 5\hat{j}) \cdot (5\hat{i} - 5\hat{j}) = (5 \times 5) + (5 \times -5) = 25 - 25 = 0$.
Since the dot product is $0$,we have $0 = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta$.
This implies $\cos \theta = 0$,which means $\theta = 90^\circ$.

(A) Two vectors are perpendicular if their dot product is zero,i.e.,$\overrightarrow P \cdot \overrightarrow Q = 0$.
Given $\overrightarrow P = a\hat i + a\hat j + 3\hat k$ and $\overrightarrow Q = a\hat i - 2\hat j - \hat k$.
Calculating the dot product: $(a)(a) + (a)(-2) + (3)(-1) = 0$.
This simplifies to: $a^2 - 2a - 3 = 0$.
Factoring the quadratic equation: $(a - 3)(a + 1) = 0$.
This gives $a = 3$ or $a = -1$.
Since the question asks for the positive value of $a$,we have $a = 3$.

(D) The area of a parallelogram formed by two vectors $\overrightarrow A$ and $\overrightarrow B$ is given by the magnitude of their cross product,i.e.,$Area = |\overrightarrow A \times \overrightarrow B|$.
Given $\overrightarrow A = 2\hat i + 3\hat j$ and $\overrightarrow B = \hat i + 4\hat j$.
$\overrightarrow A \times \overrightarrow B = (2\hat i + 3\hat j) \times (\hat i + 4\hat j)$.
Using the cross product rules $(\hat i \times \hat i = 0, \hat j \times \hat j = 0, \hat i \times \hat j = \hat k, \hat j \times \hat i = -\hat k)$:
$\overrightarrow A \times \overrightarrow B = 2(4)(\hat i \times \hat j) + 3(1)(\hat j \times \hat i) = 8\hat k - 3\hat k = 5\hat k$.
The magnitude is $|\overrightarrow A \times \overrightarrow B| = |5\hat k| = 5$.
Thus,the area is $5 \text{ units}^2$.

(D) The vector $\overrightarrow{F}_1$ is along the positive $X$-axis,so it can be represented as $\overrightarrow{F}_1 = a\hat{i}$,where $a > 0$.
The vector product of two vectors is zero if and only if the vectors are collinear (parallel or anti-parallel).
Since $\overrightarrow{F}_1 \times \overrightarrow{F}_2 = 0$,the vector $\overrightarrow{F}_2$ must be parallel or anti-parallel to the $X$-axis.
Therefore,$\overrightarrow{F}_2$ must be of the form $k\hat{i}$,where $k$ is any non-zero scalar.
Comparing this with the given options,$-4\hat{i}$ is the only vector that is parallel to the $X$-axis.
Thus,the correct option is $D$.

(B) The cross product of two vectors is defined as $\vec{A} \times \vec{B} = |A||B| \sin \theta \hat{n}$,where $\theta$ is the angle between the vectors.
Given $\vec{A} \times \vec{B} = 0$,it implies that $|A||B| \sin \theta = 0$.
Since the vectors are non-zero,$\sin \theta = 0$,which means $\theta = 0^\circ$ or $\theta = 180^\circ$.
Therefore,the vectors are parallel to each other.

(B) The cross product of two vectors is anticommutative,which means $\overrightarrow {A} \times \overrightarrow {B} = -(\overrightarrow {B} \times \overrightarrow {A})$.
This indicates that the vector $(\overrightarrow {A} \times \overrightarrow {B})$ is equal in magnitude but opposite in direction to the vector $(\overrightarrow {B} \times \overrightarrow {A})$.
Since the two vectors are anti-parallel to each other,the angle between them is $180^{\circ}$ or $\pi$ radians.

(B) The vector sum $(\overrightarrow P + \overrightarrow Q)$ lies in the plane formed by the vectors $\overrightarrow P$ and $\overrightarrow Q$.
By the definition of the vector product (cross product),the resultant vector $(\overrightarrow P \times \overrightarrow Q)$ is always perpendicular to the plane containing both $\overrightarrow P$ and $\overrightarrow Q$.
Since $(\overrightarrow P + \overrightarrow Q)$ lies in the plane and $(\overrightarrow P \times \overrightarrow Q)$ is perpendicular to the plane,the angle between them is $90^{\circ}$ or $\frac{\pi}{2}$ radians.

(D) Let the magnitudes of the two vectors be $A = 2$ and $B = 3$. The resultant $R$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta} = 1$.
Squaring both sides,we get $2^2 + 3^2 + 2(2)(3) \cos \theta = 1^2$.
$4 + 9 + 12 \cos \theta = 1$.
$13 + 12 \cos \theta = 1$.
$12 \cos \theta = -12$,which gives $\cos \theta = -1$.
Therefore,$\theta = 180^\circ$.
The cross product of two vectors is given by $|\vec{A} \times \vec{B}| = AB \sin \theta$.
Since $\theta = 180^\circ$,$\sin 180^\circ = 0$.
Thus,$|\vec{A} \times \vec{B}| = 2 \times 3 \times 0 = 0$.

(D) Let $\vec{A} = 6\hat{i} + 6\hat{j} - 3\hat{k}$ and $\vec{B} = 7\hat{i} + 4\hat{j} + 4\hat{k}$.
The dot product $\vec{A} \cdot \vec{B} = (6)(7) + (6)(4) + (-3)(4) = 42 + 24 - 12 = 54$.
The magnitude of $\vec{A}$ is $|\vec{A}| = \sqrt{6^2 + 6^2 + (-3)^2} = \sqrt{36 + 36 + 9} = \sqrt{81} = 9$.
The magnitude of $\vec{B}$ is $|\vec{B}| = \sqrt{7^2 + 4^2 + 4^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$.
Using the formula $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$,we get $\cos \theta = \frac{54}{9 \times 9} = \frac{54}{81} = \frac{2}{3}$.
Since $\cos \theta = \frac{2}{3}$,we can find $\sin \theta$ using $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (2/3)^2} = \sqrt{1 - 4/9} = \sqrt{5/9} = \frac{\sqrt{5}}{3}$.
Therefore,$\theta = \sin^{-1}\left(\frac{\sqrt{5}}{3}\right)$.

(B) Let the upward vertical direction be represented by the unit vector $\hat{k}$. Thus,$\vec{A} = A\hat{k}$.
Let the north direction be represented by the unit vector $\hat{j}$. Thus,$\vec{B} = B\hat{j}$.
The vector product is given by $\vec{A} \times \vec{B} = (A\hat{k}) \times (B\hat{j})$.
Using the cross product rules for unit vectors,$\hat{k} \times \hat{j} = -\hat{i}$.
Therefore,$\vec{A} \times \vec{B} = AB(-\hat{i}) = -AB\hat{i}$.
Since the unit vector $\hat{i}$ represents the east direction,$-\hat{i}$ represents the west direction.
Hence,the direction of $\vec{A} \times \vec{B}$ is along the west.

(D) Let $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{k}$.
The dot product is given by $\vec{A} \cdot \vec{B} = (1)(1) + (1)(0) + (0)(-1) = 1$.
The magnitudes are $|\vec{A}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ and $|\vec{B}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$.
The angle $\theta$ between the vectors is given by $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.
Substituting the values,$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.

(B) Given that $|\overrightarrow A \times \overrightarrow B | = |\overrightarrow A \cdot \overrightarrow B |$.
Using the definitions of cross product and dot product:
$|\overrightarrow A | |\overrightarrow B | \sin \theta = |\overrightarrow A | |\overrightarrow B | \cos \theta$.
Dividing both sides by $|\overrightarrow A | |\overrightarrow B | \cos \theta$ (assuming $\overrightarrow A, \overrightarrow B \neq 0$):
$\frac{\sin \theta}{\cos \theta} = 1$.
$\tan \theta = 1$.
Since $\tan 45^\circ = 1$,the angle $\theta = 45^\circ$.

(A) In a right-handed Cartesian coordinate system,the unit vectors $\hat i, \hat j, \hat k$ follow the cyclic order $\hat i \times \hat j = \hat k$,$\hat j \times \hat k = \hat i$,and $\hat k \times \hat i = \hat j$.
For option $A$: $\hat j \times \hat k = \hat i$ is correct.
For option $B$: $\hat i \cdot \hat i = 1$ (not $0$).
For option $C$: $\hat j \times \hat j = 0$ (not $1$).
For option $D$: $\hat k \cdot \hat j = 0$ (since they are orthogonal,not $1$).
Therefore,option $A$ is the correct relation.

(D) Given that $\vec a \cdot \vec b = 0$,this implies that the vector $\vec a$ is perpendicular to the vector $\vec b$.
Given that $\vec a \cdot \vec c = 0$,this implies that the vector $\vec a$ is perpendicular to the vector $\vec c$.
By the definition of the cross product,the vector $\vec b \times \vec c$ results in a vector that is perpendicular to both $\vec b$ and $\vec c$.
Since $\vec a$ is perpendicular to both $\vec b$ and $\vec c$,and $\vec b \times \vec c$ is also perpendicular to both $\vec b$ and $\vec c$,it follows that $\vec a$ must be parallel to $\vec b \times \vec c$.

(C) The area of a parallelogram with diagonals represented by vectors $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
Given $\vec{d_1} = 2\,\hat{i}$ and $\vec{d_2} = 2\,\hat{j}$.
$\vec{d_1} \times \vec{d_2} = (2\,\hat{i}) \times (2\,\hat{j}) = 4(\hat{i} \times \hat{j}) = 4\,\hat{k}$.
$\text{Area} = \frac{1}{2} |4\,\hat{k}| = \frac{1}{2} \times 4 = 2\,\text{square units}$.

(C) Let $\vec A = 2\hat i + 2\hat j - \hat k$ and $\vec B = 6\hat i - 3\hat j + 2\hat k$.
The vector perpendicular to both $\vec A$ and $\vec B$ is given by their cross product $\vec C = \vec A \times \vec B$.
$\vec C = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 2 & -1 \\ 6 & -3 & 2 \end{vmatrix} = \hat i(4 - 3) - \hat j(4 + 6) + \hat k(-6 - 12) = \hat i - 10\hat j - 18\hat k$.
The unit vector $\hat n$ is given by $\frac{\vec C}{|\vec C|}$.
$|\vec C| = \sqrt{1^2 + (-10)^2 + (-18)^2} = \sqrt{1 + 100 + 324} = \sqrt{425} = \sqrt{25 \times 17} = 5\sqrt{17}$.
Therefore,$\hat n = \frac{\hat i - 10\hat j - 18\hat k}{5\sqrt{17}}$.

(B) Let the two vectors representing the sides of the parallelogram be $\vec{A} = 0\hat{i} + 1\hat{j} + 3\hat{k}$ and $\vec{B} = 1\hat{i} + 2\hat{j} - 1\hat{k}$.
The area of a parallelogram formed by two vectors $\vec{A}$ and $\vec{B}$ is given by the magnitude of their cross product: $\text{Area} = |\vec{A} \times \vec{B}|$.
First,calculate the cross product $\vec{A} \times \vec{B}$ using the determinant method:
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 3 \\ 1 & 2 & -1 \end{vmatrix}$
$= \hat{i}((1)(-1) - (3)(2)) - \hat{j}((0)(-1) - (3)(1)) + \hat{k}((0)(2) - (1)(1))$
$= \hat{i}(-1 - 6) - \hat{j}(0 - 3) + \hat{k}(0 - 1)$
$= -7\hat{i} + 3\hat{j} - 1\hat{k}$.
Now,calculate the magnitude of the resulting vector:
$|\vec{A} \times \vec{B}| = \sqrt{(-7)^2 + (3)^2 + (-1)^2}$
$= \sqrt{49 + 9 + 1} = \sqrt{59}$ sq. unit.

(D) We use the distributive property of the cross product:
$(\vec A + \vec B) \times (\vec A - \vec B) = \vec A \times \vec A - \vec A \times \vec B + \vec B \times \vec A - \vec B \times \vec B$
Since the cross product of any vector with itself is zero ($\vec A \times \vec A = 0$ and $\vec B \times \vec B = 0$):
$= 0 - (\vec A \times \vec B) + (\vec B \times \vec A) - 0$
Using the anticommutative property of the cross product $(\vec A \times \vec B = -(\vec B \times \vec A))$:
$= -(\vec A \times \vec B) + (\vec B \times \vec A) = (\vec B \times \vec A) + (\vec B \times \vec A) = 2(\vec B \times \vec A)$

(D) Two vectors are perpendicular if their dot product is zero,i.e.,$\vec A \cdot \vec B = 0$.
Given $\vec A = 5\hat i + 7\hat j - 3\hat k$ and $\vec B = 2\hat i + 2\hat j - a\hat k$.
Calculating the dot product:
$(5\hat i + 7\hat j - 3\hat k) \cdot (2\hat i + 2\hat j - a\hat k) = 0$
$(5)(2) + (7)(2) + (-3)(-a) = 0$
$10 + 14 + 3a = 0$
$24 + 3a = 0$
$3a = -24$
$a = -8$.

(B) The area of a parallelogram defined by two adjacent vectors $\vec{A}$ and $\vec{B}$ is given by the magnitude of their cross product,i.e.,$\text{Area} = |\vec{A} \times \vec{B}|$.
Let $\vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{B} = 3\hat{i} - 2\hat{j} + \hat{k}$.
The cross product is calculated as:
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & 1 \end{vmatrix}$
$= \hat{i}(2(1) - 3(-2)) - \hat{j}(1(1) - 3(3)) + \hat{k}(1(-2) - 2(3))$
$= \hat{i}(2 + 6) - \hat{j}(1 - 9) + \hat{k}(-2 - 6)$
$= 8\hat{i} + 8\hat{j} - 8\hat{k}$
The magnitude is:
$|\vec{A} \times \vec{B}| = \sqrt{8^2 + 8^2 + (-8)^2} = \sqrt{64 + 64 + 64} = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$ square units.

(A) The dot product of two vectors $\vec{A}$ and $\vec{B}$ is defined as $\vec{A} \cdot \vec{B} = |A||B| \cos \theta$,where $\theta$ is the angle between the vectors.
Since the vectors are mutually perpendicular,the angle $\theta = 90^\circ$.
Substituting this into the formula,we get $\vec{A} \cdot \vec{B} = |A||B| \cos 90^\circ$.
Since $\cos 90^\circ = 0$,the dot product is $|A||B| \times 0 = 0$.

(D) Given: $|\vec A \times \vec B| = \sqrt 3 (\vec A \cdot \vec B)$
Using the definitions of cross product and dot product: $AB \sin \theta = \sqrt 3 AB \cos \theta$
Dividing both sides by $AB \cos \theta$ (assuming $A, B \neq 0$): $\tan \theta = \sqrt 3$
Thus,$\theta = 60^\circ$
The magnitude of the resultant vector $|\vec A + \vec B|$ is given by: $|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Substituting $\theta = 60^\circ$: $|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB \cos 60^\circ}$
Since $\cos 60^\circ = 1/2$: $|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB(1/2)}$
Therefore,$|\vec A + \vec B| = (A^2 + B^2 + AB)^{1/2}$

(C) The cross product of two vectors $\vec A$ and $\vec B$ is defined as $\vec A \times \vec B = AB \sin \theta \, \hat{n}$,where $\theta$ is the angle between the vectors and $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec A$ and $\vec B$.
If two vectors are parallel,the angle $\theta$ between them is $0^\circ$.
Since $\sin(0^\circ) = 0$,the magnitude of the cross product becomes $AB \sin(0^\circ) = 0$.
The cross product of two vectors is a vector quantity. Therefore,the result of the cross product of two parallel vectors is the zero vector,denoted by $\vec{0}$.

(A) Given $\overrightarrow{OA} = \overrightarrow{a} = 3\hat{i} - 6\hat{j} + 2\hat{k}$ and $\overrightarrow{OB} = \overrightarrow{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.
The area of triangle $OAB$ is given by $\text{Area} = \frac{1}{2} |\overrightarrow{a} \times \overrightarrow{b}|$.
First,calculate the cross product $\overrightarrow{a} \times \overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & 2 \\ 2 & 1 & -2 \end{vmatrix}$
$= \hat{i}((-6)(-2) - (2)(1)) - \hat{j}((3)(-2) - (2)(2)) + \hat{k}((3)(1) - (-6)(2))$
$= \hat{i}(12 - 2) - \hat{j}(-6 - 4) + \hat{k}(3 + 12)$
$= 10\hat{i} + 10\hat{j} + 15\hat{k}$.
Now,calculate the magnitude $|\overrightarrow{a} \times \overrightarrow{b}|$:
$|\overrightarrow{a} \times \overrightarrow{b}| = \sqrt{10^2 + 10^2 + 15^2} = \sqrt{100 + 100 + 225} = \sqrt{425} = \sqrt{25 \times 17} = 5\sqrt{17}$.
Therefore,the area of $\Delta OAB = \frac{1}{2} \times 5\sqrt{17} = \frac{5}{2}\sqrt{17}$ sq. unit.

(D) The component of vector $\vec{A}$ along vector $\vec{B}$ is given by the formula: $\vec{A}_{\text{along } B} = \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} \right) \hat{B} = \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \right) \vec{B}$.
First,calculate the dot product $\vec{A} \cdot \vec{B} = (3\hat{i} + \hat{j}) \cdot (0\hat{i} + 1\hat{j} + 2\hat{k}) = (3)(0) + (1)(1) + (0)(2) = 1$.
Next,calculate the square of the magnitude of $\vec{B}$: $|\vec{B}|^2 = (0)^2 + (1)^2 + (2)^2 = 1 + 4 = 5$.
Now,substitute these values into the formula: $\vec{A}_{\text{along } B} = \frac{1}{5} (\hat{j} + 2\hat{k})$.
Thus,the correct option is $D$.

(D) The cross product of two vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{A} \times \vec{B} = |A||B| \sin(\theta) \hat{n}$.
If the cross product is zero,then $\sin(\theta) = 0$,which implies $\theta = 0^\circ$ or $\theta = 180^\circ$.
This means the two vectors must be collinear (parallel or anti-parallel).
Given $\vec{F_1}$ is in the direction of the positive $X$-axis,$\vec{F_1} = c\hat{i}$ where $c > 0$.
For $\vec{F_2}$ to be collinear with $\vec{F_1}$,it must be of the form $k\hat{i}$ for any scalar $k$.
Looking at the options,option $D$ is $-4\hat{i}$,which is parallel (anti-parallel) to the $X$-axis.
Therefore,the cross product $\vec{F_1} \times \vec{F_2} = (c\hat{i}) \times (-4\hat{i}) = -4c(\hat{i} \times \hat{i}) = 0$.

(A) Given that $\vec{A} \times \vec{B} = \vec{0}$,it implies that $\vec{A}$ is parallel to $\vec{B}$ $(\vec{A} \parallel \vec{B})$.
Given that $\vec{B} \times \vec{C} = \vec{0}$,it implies that $\vec{B}$ is parallel to $\vec{C}$ $(\vec{B} \parallel \vec{C})$.
Since both $\vec{A}$ and $\vec{C}$ are parallel to the same vector $\vec{B}$,it follows that $\vec{A}$ must be parallel to $\vec{C}$ $(\vec{A} \parallel \vec{C})$.
Therefore,the angle between $\vec{A}$ and $\vec{C}$ is $0$ (or $180^{\circ}$ if they are anti-parallel,but $0$ is the standard result for parallel vectors).

(A) The component of a vector $\vec{r}$ in the direction of another vector $\vec{a}$ is given by the projection of $\vec{r}$ onto $\vec{a}$.
This is a vector quantity defined as the projection vector.
The formula for the projection of $\vec{r}$ onto $\vec{a}$ is given by:
$\text{Component} = (\vec{r} \cdot \hat{a}) \hat{a}$
Since the unit vector $\hat{a} = \frac{\vec{a}}{a}$,we substitute this into the formula:
$\text{Component} = \left( \vec{r} \cdot \frac{\vec{a}}{a} \right) \frac{\vec{a}}{a}$
$\text{Component} = \frac{(\vec{r} \cdot \vec{a}) \vec{a}}{a^2}$
Thus,the correct option is $A$.

(B) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
Given $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$.
First,calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(12 - (-2)) + \hat{k}(-9 - 1) = -2\hat{i} - 14\hat{j} - 10\hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{d_1} \times \vec{d_2}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3}$.
Finally,the area is $\frac{1}{2} \times 10\sqrt{3} = 5\sqrt{3}$ square units.

(D) Let $\vec{C} = \vec{B} \times \vec{A}$.
By the definition of the cross product,the vector $\vec{C}$ is perpendicular to both $\vec{B}$ and $\vec{A}$.
Therefore,$\vec{C} \cdot \vec{A} = 0$,because the dot product of two perpendicular vectors is always zero.
Thus,$(\vec{B} \times \vec{A}) \cdot \vec{A} = 0$.

(D) Let $\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = -3\hat{i} + 0\hat{j} + 6\hat{k}$.
The angle $\theta$ between two vectors is given by $\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.
Calculate the dot product: $\vec{A} \cdot \vec{B} = (2)(-3) + (3)(0) + (1)(6) = -6 + 0 + 6 = 0$.
Since the dot product is $0$,$\cos\theta = 0$,which implies $\theta = 90^\circ$.

(B) Let the vertical upward direction be represented by the unit vector $\hat{k}$ and the north direction be represented by the unit vector $\hat{j}$.
Thus,$\vec{A} = A\hat{k}$ and $\vec{B} = B\hat{j}$.
The vector product is given by $\vec{A} \times \vec{B} = (A\hat{k}) \times (B\hat{j})$.
Using the cross product rules for unit vectors,$\hat{k} \times \hat{j} = -\hat{i}$.
Therefore,$\vec{A} \times \vec{B} = -AB\hat{i}$.
The unit vector $-\hat{i}$ points towards the west. Hence,the vector product is along west.

(B) The unit vector $\hat{n}$ perpendicular to both vectors $\vec{P}$ and $\vec{Q}$ is defined as $\hat{n} = \frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|}$.
We know that the magnitude of the cross product is $|\vec{P} \times \vec{Q}| = PQ \sin \theta$.
Substituting this into the formula,we get $\hat{n} = \frac{\vec{P} \times \vec{Q}}{PQ \sin \theta}$.
Since $\hat{P} = \frac{\vec{P}}{P}$ and $\hat{Q} = \frac{\vec{Q}}{Q}$,we can rewrite the expression as $\hat{n} = \frac{\vec{P}}{P} \times \frac{\vec{Q}}{Q} \cdot \frac{1}{\sin \theta} = \frac{\hat{P} \times \hat{Q}}{\sin \theta}$.

(C) The unit vector $\hat{n}$ perpendicular to both $\vec{A}$ and $\vec{B}$ is given by $\hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}$.
First,calculate the cross product $\vec{A} \times \vec{B}$:
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 0 & 1 & 2 \end{vmatrix} = \hat{i}(1 \times 2 - 0 \times 1) - \hat{j}(3 \times 2 - 0 \times 0) + \hat{k}(3 \times 1 - 1 \times 0) = 2\hat{i} - 6\hat{j} + 3\hat{k}$.
Next,find the magnitude of the cross product:
$|\vec{A} \times \vec{B}| = \sqrt{2^2 + (-6)^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.
Finally,the unit vector is $\hat{n} = \frac{2\hat{i} - 6\hat{j} + 3\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{3}{7}\hat{k}$.

(A) The expression is $(\vec{A} + \vec{B}) \cdot (\vec{A} \times \vec{B})$.
Using the distributive property of the dot product over addition,we get:
$(\vec{A} + \vec{B}) \cdot (\vec{A} \times \vec{B}) = \vec{A} \cdot (\vec{A} \times \vec{B}) + \vec{B} \cdot (\vec{A} \times \vec{B})$.
The term $\vec{A} \cdot (\vec{A} \times \vec{B})$ represents the scalar triple product of vectors $\vec{A}, \vec{A},$ and $\vec{B}$. Since two vectors are identical,the volume of the parallelepiped formed is zero,so $\vec{A} \cdot (\vec{A} \times \vec{B}) = 0$.
Similarly,$\vec{B} \cdot (\vec{A} \times \vec{B})$ represents the scalar triple product of vectors $\vec{B}, \vec{A},$ and $\vec{B}$. Since two vectors are identical,$\vec{B} \cdot (\vec{A} \times \vec{B}) = 0$.
Therefore,the total value is $0 + 0 = 0$.

(D) The cross product of two vectors is anti-commutative,meaning $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
Given the condition $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$,we can substitute the anti-commutative property:
$-(\vec{B} \times \vec{A}) = \vec{B} \times \vec{A}$.
This implies $2(\vec{B} \times \vec{A}) = 0$,so $\vec{B} \times \vec{A} = 0$.
The magnitude of the cross product is given by $|\vec{B} \times \vec{A}| = |B||A| \sin(\theta) = 0$.
Since the vectors are non-zero,$\sin(\theta) = 0$,which means $\theta = 0$ or $\theta = \pi$.
However,for the cross product to satisfy the equality $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$,both sides must be zero vectors. This occurs when the vectors are parallel $(\theta = 0)$ or anti-parallel $(\theta = \pi)$. Among the given options,$\pi$ is the correct choice.