Multiplication of Vectors Questions in English

(A) 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 $\vec{d_1} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{d_2} = 3\hat{i} + 2\hat{j} - \hat{k}$.
First,calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -4 & 3 \\ 3 & 2 & -1 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(-5 - 9) + \hat{k}(10 - (-12)) = -2\hat{i} + 14\hat{j} + 22\hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{d_1} \times \vec{d_2}| = \sqrt{(-2)^2 + 14^2 + 22^2} = \sqrt{4 + 196 + 484} = \sqrt{684}$.
Simplify the square root: $\sqrt{684} = \sqrt{36 \times 19} = 6\sqrt{19}$.
Finally,the area is $\frac{1}{2} \times 6\sqrt{19} = 3\sqrt{19} = \sqrt{9 \times 19} = \sqrt{171} \text{ unit}^2$.

(B) The projection of a vector $\vec A$ onto a vector $\vec B$ is defined as the scalar component of $\vec A$ in the direction of $\vec B$.
Mathematically,this is given by the dot product of $\vec A$ with the unit vector of $\vec B$.
Projection $= |\vec A| \cos \theta$.
Since $\vec A \cdot \vec B = |\vec A| |\vec B| \cos \theta$,we have $|\vec A| \cos \theta = \frac{\vec A \cdot \vec B}{|\vec B|}$.
Since the unit vector $\hat B = \frac{\vec B}{|\vec B|}$,the projection is $\vec A \cdot \hat B$.

(A) Given that $\hat{A}$ is a unit vector,its magnitude is constant,i.e.,$|\hat{A}| = 1$.
Squaring both sides,we get $|\hat{A}|^2 = \hat{A} \cdot \hat{A} = 1$.
Differentiating both sides with respect to time $t$,we get $\frac{d}{dt}(\hat{A} \cdot \hat{A}) = \frac{d}{dt}(1)$.
Using the product rule for the dot product,we have $\frac{d\hat{A}}{dt} \cdot \hat{A} + \hat{A} \cdot \frac{d\hat{A}}{dt} = 0$.
Since the dot product is commutative,this simplifies to $2(\hat{A} \cdot \frac{d\hat{A}}{dt}) = 0$.
Therefore,$\hat{A} \cdot \frac{d\hat{A}}{dt} = 0$.

(A) Given that $|F_1| = |F_2| = F$ and $|F_1 \cdot F_2| = |F_1 \times F_2|$.
Using the definitions of dot and cross products: $F^2 \cos \theta = F^2 \sin \theta$.
This implies $\tan \theta = 1$,so $\theta = 45^{\circ}$.
The magnitude of the resultant vector is given by $|F_1 + F_2| = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$.
Substituting the values: $|F_1 + F_2| = \sqrt{F^2 + F^2 + 2F^2 \cos 45^{\circ}}$.
$|F_1 + F_2| = \sqrt{2F^2 + 2F^2 \left(\frac{1}{\sqrt{2}}\right)} = \sqrt{2F^2 + F^2 \sqrt{2}}$.
$|F_1 + F_2| = F \sqrt{2 + \sqrt{2}}$.

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

(C) Using the vector triple product identity $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$:
$\hat{i} \times (\hat{i} \times \vec{a}) = \hat{i}(\hat{i} \cdot \vec{a}) - \vec{a}(\hat{i} \cdot \hat{i}) = \hat{i}a_x - \vec{a}$
$\hat{j} \times (\hat{j} \times \vec{a}) = \hat{j}(\hat{j} \cdot \vec{a}) - \vec{a}(\hat{j} \cdot \hat{j}) = \hat{j}a_y - \vec{a}$
$\hat{k} \times (\hat{k} \times \vec{a}) = \hat{k}(\hat{k} \cdot \vec{a}) - \vec{a}(\hat{k} \cdot \hat{k}) = \hat{k}a_z - \vec{a}$
Summing these expressions:
$(\hat{i}a_x + \hat{j}a_y + \hat{k}a_z) - 3\vec{a}$
Since $\vec{a} = \hat{i}a_x + \hat{j}a_y + \hat{k}a_z$,the expression becomes:
$\vec{a} - 3\vec{a} = -2\vec{a}$.

(B) The angle $\theta$ between two vectors $\vec A$ and $\vec B$ is given by the formula $\cos \theta = \frac{\vec A \cdot \vec B}{|\vec A| |\vec B|}$.
First,calculate the dot product: $\vec A \cdot \vec B = (2)(1) + (1)(0) + (-1)(-1) = 2 + 0 + 1 = 3$.
Next,calculate the magnitudes: $|\vec A| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec B| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$.
Now,substitute these values into the formula: $\cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{2}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Since $\cos \theta = \frac{\sqrt{3}}{2}$,the angle $\theta = 30^o$.

(A) Let the unit vectors be $\hat{i}$ (East) and $\hat{j}$ (North). So,$A = A\hat{i}$ and $B = B\hat{j}$.
$(A) (A+B) = A\hat{i} + B\hat{j}$,which points in the North-east direction. Thus,$(A) \rightarrow (p)$.
$(B) (A-B) = A\hat{i} - B\hat{j}$,which points in the South-east direction. This is not listed in Column $II$,so $(B) \rightarrow (s)$.
$(C) (A \times B) = (A\hat{i}) \times (B\hat{j}) = AB(\hat{i} \times \hat{j}) = AB\hat{k}$,which points Vertically upwards. Thus,$(C) \rightarrow (q)$.
$(D) (A \times B) \times (A \times B) = 0$,because the cross product of any vector with itself is zero. This is not listed in Column $II$,so $(D) \rightarrow (s)$.
Therefore,the correct matching is $(A \rightarrow p, B \rightarrow s, C \rightarrow q, D \rightarrow s)$.

(B) The cross product of any vector $\vec{A}$ with a zero vector $\vec{0}$ is defined as $\vec{A} \times \vec{0} = \vec{0}$.
Since the cross product of two vectors results in another vector,the result of $\vec{A} \times 0$ is a zero vector,which is denoted as $\vec{0}$.
Therefore,the correct option is the zero vector.

(N/A) The dot product of $F$ and $d$ is given by $F \cdot d = F_x d_x + F_y d_y + F_z d_z$.
$F \cdot d = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16$ units.
The magnitudes are $F = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ units and $d = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50}$ units.
Using $F \cdot d = F d \cos \theta$,we have $16 = \sqrt{50} \cdot \sqrt{50} \cdot \cos \theta$.
$16 = 50 \cos \theta \implies \cos \theta = \frac{16}{50} = 0.32$.
Therefore,$\theta = \cos^{-1}(0.32)$.
The projection of $F$ on $d$ is given by $\frac{F \cdot d}{|d|} = \frac{16}{\sqrt{50}} = \frac{16}{5\sqrt{2}} = \frac{16\sqrt{2}}{10} = 1.6\sqrt{2}$ units.

(N/A) There are two kinds of multiplication of vectors:
$(i)$ Scalar product (Dot product):
If the product of two vector quantities results in a scalar,then the product is called a scalar product. This product is also known as the dot product.
The scalar product of two vectors $\vec{A}$ and $\vec{B}$ is denoted by $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta = AB \cos \theta$,where $A$ and $B$ are the magnitudes of $\vec{A}$ and $\vec{B}$ respectively,and $\theta$ is the angle between them.
$(ii)$ Vector product (Cross product):
If the product of two vector quantities results in a vector,then the product is called a vector product.
$A$ vector product is represented by placing a cross sign $(\times)$ between two vectors; hence,it is also called the cross product of vectors.
If $\theta$ is the angle between $\vec{A}$ and $\vec{B}$,then its vector product is $\vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin \theta \hat{n} = AB \sin \theta \hat{n}$,where $\hat{n}$ is the unit vector perpendicular to the plane formed by $\vec{A}$ and $\vec{B}$.

(N/A) The scalar product of two vectors $\vec{A}$ and $\vec{B}$ is defined as:
$\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta = AB \cos \theta$ ... $(1)$
where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
Geometrically,this product represents the product of the magnitude of one vector and the projection of the other vector onto the first.
Consider the projection of $\vec{B}$ onto $\vec{A}$:
$1$. Draw a perpendicular from the head of $\vec{B}$ onto the line containing $\vec{A}$,meeting at point $M$ as shown in the figure.
$2$. The length $OM$ represents the projection of $\vec{B}$ onto $\vec{A}$,which is given by $B \cos \theta$.
$3$. Substituting this into the scalar product formula:
$\vec{A} \cdot \vec{B} = A (B \cos \theta) = A (OM)$
Thus,the scalar product is the magnitude of $\vec{A}$ multiplied by the component (projection) of $\vec{B}$ along $\vec{A}$.
Similarly,it can be interpreted as the magnitude of $\vec{B}$ multiplied by the component of $\vec{A}$ along $\vec{B}$ $(B \times A \cos \theta)$.

(N/A) Let $\vec{A}$ and $\vec{B}$ be two vectors with an angle $\theta$ between them.
The scalar product (dot product) of $\vec{A}$ and $\vec{B}$ is defined as:
$\vec{A} \cdot \vec{B} = AB \cos \theta$
Since the product of scalar magnitudes $A$ and $B$ is commutative $(AB = BA)$,we can write:
$AB \cos \theta = BA \cos \theta$
By definition,$BA \cos \theta$ is the scalar product of $\vec{B}$ and $\vec{A}$:
$BA \cos \theta = \vec{B} \cdot \vec{A}$
Therefore,$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
This proves that the scalar product of two vectors obeys the commutative law.

(N/A) According to the figure,let $\overrightarrow{OP} = \vec{A}$,$\overrightarrow{OQ} = \vec{B}$,and $\overrightarrow{QR} = \vec{C}$.
Then,$\overrightarrow{OR} = \overrightarrow{OQ} + \overrightarrow{QR} = \vec{B} + \vec{C}$.
The scalar product $\vec{A} \cdot (\vec{B} + \vec{C})$ is defined as the product of the magnitude of $\vec{A}$ and the projection of $(\vec{B} + \vec{C})$ onto the direction of $\vec{A}$.
From the figure,the projection of $\vec{B}$ on $\vec{A}$ is $OM$,and the projection of $\vec{C}$ on $\vec{A}$ is $MN$.
Therefore,the projection of $(\vec{B} + \vec{C})$ on $\vec{A}$ is $ON = OM + MN$.
Now,$\vec{A} \cdot (\vec{B} + \vec{C}) = |\vec{A}| (ON) = |\vec{A}| (OM + MN)$.
$= |\vec{A}| (OM) + |\vec{A}| (MN)$.
Since $|\vec{A}| (OM) = \vec{A} \cdot \vec{B}$ and $|\vec{A}| (MN) = \vec{A} \cdot \vec{C}$,we have:
$\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}$.
This proves that the scalar product obeys the distributive law.

(N/A) Let $\overrightarrow{A}$ be a vector. The scalar product (dot product) of the vector $\overrightarrow{A}$ with itself is defined as $\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}| |\overrightarrow{A}| \cos \theta$.
Since the angle $\theta$ between a vector and itself is $0^{\circ}$,we have $\cos 0^{\circ} = 1$.
Therefore,$\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}| |\overrightarrow{A}| (1) = |\overrightarrow{A}|^2$.
Taking the square root on both sides,we get $|\overrightarrow{A}| = \sqrt{\overrightarrow{A} \cdot \overrightarrow{A}}$.
Thus,the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.

(A) If two vectors $\vec{A}$ and $\vec{B}$ are mutually perpendicular,the angle $\theta$ between them is $90^{\circ}$.
The scalar product (dot product) of two vectors is given by the formula $\vec{A} \cdot \vec{B} = AB \cos \theta$.
Substituting $\theta = 90^{\circ}$ into the formula,we get $\vec{A} \cdot \vec{B} = AB \cos 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the scalar product becomes $\vec{A} \cdot \vec{B} = AB \times 0 = 0$.
Therefore,the scalar product of two mutually perpendicular vectors is $0$.

The scalar product (dot product) is calculated as:
$\vec{a} \cdot \vec{b} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k}) \cdot (-2 \hat{i} + \hat{j} - 3 \hat{k})$
$= (3)(-2) + (-4)(1) + (5)(-3)$
$= -6 - 4 - 15 = -25$
The vector product (cross product) is calculated using the determinant:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ -2 & 1 & -3 \end{vmatrix}$
$= \hat{i}((-4)(-3) - (5)(1)) - \hat{j}((3)(-3) - (5)(-2)) + \hat{k}((3)(1) - (-4)(-2))$
$= \hat{i}(12 - 5) - \hat{j}(-9 + 10) + \hat{k}(3 - 8)$
$= 7 \hat{i} - \hat{j} - 5 \hat{k}$

(N/A) Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the sides $\vec{OK}$ and $\vec{OM}$ of a triangle,inclined at an angle $\theta$.
In $\Delta OMN$,where $MN$ is the perpendicular from $M$ to $OK$,we have:
$\sin \theta = \frac{MN}{OM} = \frac{MN}{|\vec{b}|}$
$MN = |\vec{b}| \sin \theta$
The area of $\Delta OMK$ is given by:
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OK \times MN$
Area $= \frac{1}{2} |\vec{a}| |\vec{b}| \sin \theta$
Since the magnitude of the cross product is defined as $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,we substitute this into the area formula:
Area $= \frac{1}{2} |\vec{a} \times \vec{b}|$
Thus,the area of the triangle is one-half the magnitude of the cross product of the two vectors.

(N/A) The volume of a parallelepiped is defined as the product of the area of its base and its height.
Let the vectors representing the three adjacent edges of the parallelepiped be $\vec{a}, \vec{b}$,and $\vec{c}$.
The base of the parallelepiped is formed by vectors $\vec{b}$ and $\vec{c}$. The area of this base is given by the magnitude of the cross product: $Area = |\vec{b} \times \vec{c}|$.
The direction of the vector $\vec{b} \times \vec{c}$ is perpendicular to the base,i.e.,normal to the plane containing $\vec{b}$ and $\vec{c}$.
The height $h$ of the parallelepiped is the projection of vector $\vec{a}$ onto the direction of the normal vector $\vec{n} = \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$.
Thus,$h = |\vec{a} \cdot \hat{n}| = \left| \vec{a} \cdot \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} \right|$.
The volume $V$ is given by $V = Area \times h = |\vec{b} \times \vec{c}| \times \left| \vec{a} \cdot \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} \right| = |\vec{a} \cdot (\vec{b} \times \vec{c})|$.
Therefore,the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ represents the volume of the parallelepiped.

(N/A) Multiplying a vector $\vec{A}$ with a positive real number $\lambda$ results in a new vector $\lambda \vec{A}$ whose magnitude is $\lambda$ times the magnitude of $\vec{A}$,and its direction remains the same as that of $\vec{A}$.
$|\lambda \vec{A}| = \lambda |\vec{A}|$ (for $\lambda > 0$)
For example,if $\vec{A}$ is multiplied by $2$,the resultant vector $2\vec{A}$ points in the same direction as $\vec{A}$ and has a magnitude twice that of $|\vec{A}|$,as shown in figure $(a)$.
Multiplying a vector $\vec{A}$ by a negative real number $\lambda$ results in a vector $\lambda \vec{A}$ whose direction is opposite to that of $\vec{A}$ and whose magnitude is $|\lambda|$ times the magnitude of $\vec{A}$.
For example,multiplying $\vec{A}$ by $-1$ and $-1.5$ gives vectors as shown in figure $(b)$.
The factor $\lambda$ can also be a scalar with physical dimensions. In such cases,the dimension of the resulting vector $\lambda \vec{A}$ is the product of the dimensions of $\lambda$ and $\vec{A}$. For instance,multiplying a constant velocity vector $\vec{v}$ by a time interval $t$ gives a displacement vector $\vec{d} = \vec{v}t$.
Dimensions of $\vec{v} = [LT^{-1}] = m/s$
Dimensions of $\vec{v}t = [LT^{-1}] \cdot [T] = [L] = m$

(A) When a vector $\vec{A}$ is multiplied by a scalar $\lambda$,the resulting vector is $\vec{B} = \lambda \vec{A}$.
If $\lambda > 0$,the magnitude of the new vector becomes $|\lambda| |\vec{A}|$ and the direction remains the same as $\vec{A}$.
If $\lambda < 0$,the magnitude of the new vector becomes $|\lambda| |\vec{A}|$ and the direction becomes opposite to that of $\vec{A}$.

(A) The vector product $\vec{V} = \vec{B} \times \vec{C}$ results in a vector that is perpendicular to the plane containing $\vec{B}$ and $\vec{C}$.
Next,the vector product $\vec{A} \times \vec{V}$ results in a vector that is perpendicular to both $\vec{A}$ and $\vec{V}$.
Since $\vec{V}$ is perpendicular to the plane of $\vec{B}$ and $\vec{C}$,any vector perpendicular to $\vec{V}$ must lie within the plane of $\vec{B}$ and $\vec{C}$.
Therefore,$\vec{A} \times (\vec{B} \times \vec{C})$ lies in the plane of $\vec{B}$ and $\vec{C}$ and is perpendicular to $\vec{A}$.

(N/A) In the Cartesian coordinate system,$\hat{i}, \hat{j}$,and $\hat{k}$ are unit vectors along the $X, Y$,and $Z$ axes,respectively.
$(i)$ For parallel unit vectors:
$\hat{i} \cdot \hat{i} = |\hat{i}| |\hat{i}| \cos 0^{\circ} = (1)(1)(1) = 1$
Similarly,$\hat{j} \cdot \hat{j} = 1$ and $\hat{k} \cdot \hat{k} = 1$.
$(ii)$ For orthogonal unit vectors:
$\hat{i} \cdot \hat{j} = |\hat{i}| |\hat{j}| \cos 90^{\circ} = (1)(1)(0) = 0$
Similarly,$\hat{j} \cdot \hat{k} = 0$ and $\hat{k} \cdot \hat{i} = 0$.
Thus,the scalar products are:
$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$
$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

(N/A) Let two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ be represented in Cartesian components as follows:
$\overrightarrow{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$
$\overrightarrow{B} = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}$
Then,the scalar product is given by:
$\overrightarrow{A} \cdot \overrightarrow{B} = (A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}) \cdot (B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k})$
Expanding the dot product using the distributive property:
$= A_{x} B_{x}(\hat{i} \cdot \hat{i}) + A_{x} B_{y}(\hat{i} \cdot \hat{j}) + A_{x} B_{z}(\hat{i} \cdot \hat{k}) + A_{y} B_{x}(\hat{j} \cdot \hat{i}) + A_{y} B_{y}(\hat{j} \cdot \hat{j}) + A_{y} B_{z}(\hat{j} \cdot \hat{k}) + A_{z} B_{x}(\hat{k} \cdot \hat{i}) + A_{z} B_{y}(\hat{k} \cdot \hat{j}) + A_{z} B_{z}(\hat{k} \cdot \hat{k})$
Using the properties of unit vectors: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{i} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{j} = \hat{k} \cdot \hat{i} = \hat{i} \cdot \hat{k} = 0$.
Substituting these values,we get:
$\overrightarrow{A} \cdot \overrightarrow{B} = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$

If $\theta$ is the angle between vectors $\overrightarrow{A}$ and $\overrightarrow{B}$,then by the definition of the scalar product:
$\overrightarrow{A} \cdot \overrightarrow{B} = AB \cos \theta$
Rearranging the formula to solve for $\cos \theta$:
$\cos \theta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{AB}$
Therefore,the angle $\theta$ is given by:
$\theta = \cos^{-1} \left( \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{AB} \right)$
In a Cartesian coordinate system,where $\overrightarrow{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\overrightarrow{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$,the expression becomes:
$\cos \theta = \frac{A_x B_x + A_y B_y + A_z B_z}{\sqrt{A_x^2 + A_y^2 + A_z^2} \sqrt{B_x^2 + B_y^2 + B_z^2}}$

(N/A) The scalar product (or dot product) of two vectors $\vec{A}$ and $\vec{B}$ is defined as the product of the magnitudes of the two vectors and the cosine of the angle $\theta$ between them.
Mathematically,it is expressed as: $\vec{A} \cdot \vec{B} = AB \cos \theta$,where $A$ and $B$ are the magnitudes of vectors $\vec{A}$ and $\vec{B}$ respectively,and $\theta$ is the angle between them $(0 \le \theta \le \pi)$.
The result of the scalar product is a scalar quantity.

(N/A) The scalar product (or dot product) of two vectors $\vec{A}$ and $\vec{B}$ is defined as the product of their magnitudes and the cosine of the angle $\theta$ between them: $\vec{A} \cdot \vec{B} = AB \cos \theta$.
To obtain the magnitude of a vector $\vec{A}$ from the scalar product,we take the dot product of the vector with itself: $\vec{A} \cdot \vec{A} = A A \cos(0^\circ) = A^2$. Thus,the magnitude $A = \sqrt{\vec{A} \cdot \vec{A}}$.
The scalar product is a scalar quantity,meaning it has magnitude but no direction. Therefore,it does not have a direction.

(N/A) The scalar product (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 two vectors.
For two vectors to be mutually perpendicular,the angle between them must be $\theta = 90^{\circ}$.
Substituting this into the formula,we get $\vec{A} \cdot \vec{B} = |A||B| \cos(90^{\circ})$.
Since $\cos(90^{\circ}) = 0$,the scalar product becomes $\vec{A} \cdot \vec{B} = 0$.
Therefore,the necessary condition for two non-zero vectors to be mutually perpendicular is that their scalar product must be zero.

(A) The dot product of two vectors $\overrightarrow{A} = A_x\widehat{i} + A_y\widehat{j} + A_z\widehat{k}$ and $\overrightarrow{B} = B_x\widehat{i} + B_y\widehat{j} + B_z\widehat{k}$ is given by $\overrightarrow{A} \cdot \overrightarrow{B} = A_x B_x + A_y B_y + A_z B_z$.
Given $\overrightarrow{A} = 2\widehat{i} - 2\widehat{j} + 0\widehat{k}$ and $\overrightarrow{B} = 0\widehat{i} + 0\widehat{j} + 2\widehat{k}$.
Substituting the components: $\overrightarrow{A} \cdot \overrightarrow{B} = (2)(0) + (-2)(0) + (0)(2)$.
$\overrightarrow{A} \cdot \overrightarrow{B} = 0 + 0 + 0 = 0$.

(A) Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.
Given $\vec{A} = (2, -3, 1)$ and $\vec{B} = (3, 4, n)$.
Calculating the dot product: $(2)(3) + (-3)(4) + (1)(n) = 0$.
$6 - 12 + n = 0$.
$-6 + n = 0$.
Therefore,$n = 6$.

(B) Given that $\vec{P} \perp \vec{Q}$,the dot product of the two vectors must be zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$.
Substituting the given components: $(k, 2, 3) \cdot (0, 3, k) = 0$.
Calculating the dot product: $(k \times 0) + (2 \times 3) + (3 \times k) = 0$.
This simplifies to: $0 + 6 + 3k = 0$.
Solving for $k$: $3k = -6$,which gives $k = -2$.

(N/A) Definition: The vector product or cross product of two vectors $\vec{a}$ and $\vec{b}$ is another vector $\vec{c}$,whose magnitude is equal to the product of the magnitudes of the two vectors and the sine of the smaller angle between them.
If the product of two vectors results in a vector quantity,then this product is called a vector product. Suppose there are two vectors $\vec{a}$ and $\vec{b}$ and the angle between them is $\theta$.
Therefore,the vector product is defined as $\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \hat{n} = ab \sin \theta \hat{n}$,where $|\vec{a}| = a$ and $|\vec{b}| = b$.
Here,$\hat{n}$ is a unit vector perpendicular to the plane formed by $\vec{a}$ and $\vec{b}$.
This product is also known as the cross product $(\times)$.
If $\vec{a} \times \vec{b}$ is denoted by $\vec{c}$,then $\vec{c} = ab \sin \theta \hat{n}$.
The magnitude of the resulting vector is $c = ab \sin \theta$.
The direction of $\vec{c}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$,and its direction is determined by the right-hand screw rule.

(N/A) $(1)$ The vector product is anticommutative: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. It does not follow the commutative law.
$(2)$ The vector product is distributive over addition: $\vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})$.
$(3)$ For two parallel or anti-parallel vectors,the vector product is zero: $\vec{a} \times \vec{a} = |a||a| \sin(0^{\circ}) \hat{n} = \vec{0}$.
$(4)$ For two perpendicular vectors,the magnitude of the vector product is maximum: $|\vec{a} \times \vec{b}| = |a||b| \sin(90^{\circ}) = |a||b|$.
$(5)$ The vector product of two vectors is a pseudovector (axial vector). Under reflection,the sign of the vector product does not change because both vectors change sign: $(-\vec{a}) \times (-\vec{b}) = \vec{a} \times \vec{b}$.
$(6)$ For unit vectors in a Cartesian coordinate system: $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0}$ and $\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,$\hat{k} \times \hat{i} = \hat{j}$.

(N/A) The vector product (or cross product) of two vectors $\vec{A}$ and $\vec{B}$ is defined as a vector $\vec{C}$ such that $\vec{C} = \vec{A} \times \vec{B} = AB \sin \theta \hat{n}$.
Here,$A$ and $B$ are the magnitudes of vectors $\vec{A}$ and $\vec{B}$ respectively,$\theta$ is the angle between the two vectors $(0^\circ \le \theta \le 180^\circ)$,and $\hat{n}$ is a unit vector perpendicular to the plane containing both $\vec{A}$ and $\vec{B}$.
The direction of $\hat{n}$ is determined by the right-hand rule.

(N/A) The Right-Hand Screw Law is a rule used to determine the direction of the vector product (cross product) of two vectors,$\vec{A}$ and $\vec{B}$.
$1$. Imagine a right-handed screw placed at the point of intersection of vectors $\vec{A}$ and $\vec{B}$,with its axis perpendicular to the plane containing both vectors.
$2$. Rotate the screw from the direction of the first vector $\vec{A}$ towards the second vector $\vec{B}$ through the smaller angle between them.
$3$. The direction in which the screw advances (moves forward) represents the direction of the resultant vector $\vec{C} = \vec{A} \times \vec{B}$.
$4$. If the screw moves forward,the direction is outward from the plane; if it moves backward,the direction is into the plane.

(A) The cross product of two vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{A} \times \vec{B} = AB \sin \theta \hat{n}$,where $\theta$ is the angle between the vectors.
If the vectors are parallel,$\theta = 0^\circ$,so $\sin 0^\circ = 0$.
If the vectors are antiparallel,$\theta = 180^\circ$,so $\sin 180^\circ = 0$.
In both cases,the cross product $\vec{A} \times \vec{B} = 0$.

The product of two vectors can be defined in two ways: the scalar product (dot product) and the vector product (cross product).
$1$. The scalar product is defined as $\vec{A} \cdot \vec{B} = AB \cos \theta$. Since $\cos \theta = \cos(-\theta)$,the scalar product is commutative,i.e.,$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
$2$. The vector product is defined as $\vec{A} \times \vec{B} = AB \sin \theta \hat{n}$,where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ determined by the right-hand rule.
$3$. According to the right-hand rule,the direction of $\vec{A} \times \vec{B}$ is opposite to the direction of $\vec{B} \times \vec{A}$.
$4$. Therefore,$\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
$5$. Because the direction changes,the vector product is anti-commutative,not commutative.

(B) The linear momentum $\vec{p}$ of a particle is defined as $\vec{p} = m\vec{v}$,where $m$ is the mass and $\vec{v}$ is the velocity vector.
Since $m$ is a scalar,$\vec{p}$ is always in the same direction as $\vec{v}$.
Two vectors are parallel if the angle between them is $0^\circ$.
The cross product of two vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{A} \times \vec{B} = |A||B| \sin(\theta) \hat{n}$.
For $\vec{v}$ and $\vec{p}$,the angle $\theta = 0^\circ$,so $\sin(0^\circ) = 0$.
Therefore,$\vec{v} \times \vec{p} = |v||p| \sin(0^\circ) \hat{n} = 0$.
This confirms that for any particle,the cross product of its velocity and momentum is zero because they are collinear.

(A) Given $|A| = 2$ and $|B| = 4$. The dot product is defined as $\vec A \cdot \vec B = |A||B| \cos \theta = 8 \cos \theta$.
$(a) \vec A \cdot \vec B = 0 \implies 8 \cos \theta = 0 \implies \cos \theta = 0 \implies \theta = 90^\circ$. Matches $(ii)$.
$(b) \vec A \cdot \vec B = 8 \implies 8 \cos \theta = 8 \implies \cos \theta = 1 \implies \theta = 0^\circ$. Matches $(i)$.
$(c) \vec A \cdot \vec B = 4 \implies 8 \cos \theta = 4 \implies \cos \theta = 1/2 \implies \theta = 60^\circ$. Matches $(iv)$.
$(d) \vec A \cdot \vec B = -8 \implies 8 \cos \theta = -8 \implies \cos \theta = -1 \implies \theta = 180^\circ$. Matches $(iii)$.

(A-IV, B-III, C-I, D-II) Given $|\vec{A}| = 2$ and $|\vec{B}| = 4$. The magnitude of the cross product is given by $|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta = 8 \sin \theta$.
$(a) |\vec{A} \times \vec{B}| = 8 \sin \theta = 0 \implies \sin \theta = 0 \implies \theta = 0^{\circ}$. Matches with $(iv)$.
$(b) |\vec{A} \times \vec{B}| = 8 \sin \theta = 8 \implies \sin \theta = 1 \implies \theta = 90^{\circ}$. Matches with $(iii)$.
$(c) |\vec{A} \times \vec{B}| = 8 \sin \theta = 4 \implies \sin \theta = 1/2 \implies \theta = 30^{\circ}$. Matches with $(i)$.
$(d) |\vec{A} \times \vec{B}| = 8 \sin \theta = 4\sqrt{2} \implies \sin \theta = 1/\sqrt{2} \implies \theta = 45^{\circ}$. Matches with $(ii)$.

(C) The cross product of two vectors is anti-commutative,meaning $\overrightarrow{ P } \times \overrightarrow{ Q } = -(\overrightarrow{ Q } \times \overrightarrow{ P })$.
Given the condition $\overrightarrow{ P } \times \overrightarrow{ Q } = \overrightarrow{ Q } \times \overrightarrow{ P }$,we can substitute the anti-commutative property:
$-(\overrightarrow{ Q } \times \overrightarrow{ P }) = \overrightarrow{ Q } \times \overrightarrow{ P }$.
This implies $2(\overrightarrow{ Q } \times \overrightarrow{ P }) = 0$,so $\overrightarrow{ Q } \times \overrightarrow{ P } = 0$.
The magnitude of the cross product is given by $|\overrightarrow{ Q } \times \overrightarrow{ P }| = PQ \sin \theta = 0$.
Since $P$ and $Q$ are non-zero vectors,$\sin \theta = 0$.
In the range $0^{\circ} < \theta < 360^{\circ}$,$\sin \theta = 0$ at $\theta = 180^{\circ}$ (since $0^{\circ}$ and $360^{\circ}$ are excluded).
Therefore,the value of $\theta$ is $180^{\circ}$.

(A) Given the relation $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$.
Using the definitions of dot and cross product: $AB \cos \theta = AB \sin \theta$.
Dividing both sides by $AB$ (assuming $A, B \neq 0$),we get $\cos \theta = \sin \theta$,which implies $\tan \theta = 1$,so $\theta = 45^{\circ}$.
The magnitude of the vector difference $|\vec{A} - \vec{B}|$ is given by the formula $\sqrt{A^{2} + B^{2} - 2AB \cos \theta}$.
Substituting $\theta = 45^{\circ}$,we get $\sqrt{A^{2} + B^{2} - 2AB \cos 45^{\circ}}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the expression becomes $\sqrt{A^{2} + B^{2} - 2AB \cdot \frac{1}{\sqrt{2}}} = \sqrt{A^{2} + B^{2} - \sqrt{2}AB}$.

(A) Given vectors are $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$.
Vector $X = \alpha A + \beta B = \alpha(\hat{i} + \hat{j} - 2\hat{k}) + \beta(\hat{i} - \hat{j} + \hat{k})$.
Simplifying $X$,we get $X = (\alpha + \beta)\hat{i} + (\alpha - \beta)\hat{j} + (-2\alpha + \beta)\hat{k}$.
Since $X$ is perpendicular to $C$,their dot product must be zero: $X \cdot C = 0$.
$(\alpha + \beta)(2) + (\alpha - \beta)(-3) + (-2\alpha + \beta)(4) = 0$.
Expanding the terms: $2\alpha + 2\beta - 3\alpha + 3\beta - 8\alpha + 4\beta = 0$.
Combining like terms: $(2 - 3 - 8)\alpha + (2 + 3 + 4)\beta = 0$.
$-9\alpha + 9\beta = 0$.
$-9\alpha = -9\beta$,which implies $\alpha = \beta$.
Therefore,the ratio $\alpha : \beta = 1 : 1$.

(B) The projection of vector $\vec{A}$ on vector $\vec{B}$ is given by the formula: $\text{Proj}_{\vec{B}} \vec{A} = \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} \right) \hat{B}$,where $\hat{B}$ is the unit vector along $\vec{B}$.
First,calculate the dot product: $\vec{A} \cdot \vec{B} = (1)(1) + (1)(1) + (1)(0) = 1 + 1 = 2$.
Next,calculate the magnitude of $\vec{B}$: $|\vec{B}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Then,the unit vector $\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
Finally,substitute these values into the projection formula:
$\text{Proj}_{\vec{B}} \vec{A} = \left( \frac{2}{\sqrt{2}} \right) \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right) = \sqrt{2} \cdot \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \hat{i} + \hat{j}$.

(D) The direction of motion of particle $P$ is normal to the plane containing $\vec{A}$ and $\vec{B}$. Thus,the unit vector $\hat{v}_1$ is given by $\hat{v}_1 = \pm \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}$.
Calculating the cross product: $\vec{A} \times \vec{B} = (\hat{i} + \hat{j}) \times (\hat{j} + \hat{k}) = \hat{i} \times \hat{j} + \hat{i} \times \hat{k} + \hat{j} \times \hat{j} + \hat{j} \times \hat{k} = \hat{k} - \hat{j} + 0 + \hat{i} = \hat{i} - \hat{j} + \hat{k}$.
The magnitude is $|\vec{A} \times \vec{B}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
So,$\hat{v}_1 = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.
The direction of motion of particle $Q$ is normal to the plane containing $\vec{A}$ and $\vec{C}$. Thus,$\hat{v}_2 = \pm \frac{\vec{A} \times \vec{C}}{|\vec{A} \times \vec{C}|}$.
Calculating the cross product: $\vec{A} \times \vec{C} = (\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j}) = \hat{i} \times (-\hat{i}) + \hat{i} \times \hat{j} + \hat{j} \times (-\hat{i}) + \hat{j} \times \hat{j} = 0 + \hat{k} + \hat{k} + 0 = 2\hat{k}$.
The magnitude is $|\vec{A} \times \vec{C}| = 2$.
So,$\hat{v}_2 = \pm \frac{2\hat{k}}{2} = \pm \hat{k}$.
The angle $\theta$ between $\hat{v}_1$ and $\hat{v}_2$ is given by $\cos \theta = |\hat{v}_1 \cdot \hat{v}_2| = |\pm \frac{1}{\sqrt{3}} \hat{k} \cdot \hat{k}| = \frac{1}{\sqrt{3}}$.
Comparing this with $\cos \theta = \frac{1}{\sqrt{x}}$,we get $x = 3$.

(C) Given that $|\vec{A}| = c$,where $c$ is a nonzero constant.
The cross product of any vector with itself is defined as $\vec{A} \times \vec{A} = |\vec{A}| |\vec{A}| \sin(\theta) \hat{n}$,where $\theta$ is the angle between the vector and itself.
Since the angle between a vector and itself is $0^{\circ}$,we have $\theta = 0^{\circ}$.
Substituting this into the formula: $\vec{A} \times \vec{A} = |\vec{A}|^2 \sin(0^{\circ}) \hat{n}$.
Since $\sin(0^{\circ}) = 0$,the expression becomes $\vec{A} \times \vec{A} = 0$.

(A) Given vectors are $\overrightarrow{A} = (2\hat{i} + 3\hat{j} - \hat{k}) \; m$ and $\overrightarrow{B} = (\hat{i} + 2\hat{j} + 2\hat{k}) \; m$.
The component of vector $\overrightarrow{A}$ along vector $\overrightarrow{B}$ is given by the projection formula: $\text{Component} = \overrightarrow{A} \cdot \hat{B} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$.
First,calculate the dot product $\overrightarrow{A} \cdot \overrightarrow{B}$:
$\overrightarrow{A} \cdot \overrightarrow{B} = (2)(1) + (3)(2) + (-1)(2) = 2 + 6 - 2 = 6$.
Next,calculate the magnitude of vector $\overrightarrow{B}$:
$|\overrightarrow{B}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Finally,calculate the component:
$\text{Component} = \frac{6}{3} = 2 \; m$.

(C) The projection of vector $\vec{A}$ on vector $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.
Given that the projection is zero,we have $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 0$,which implies $\vec{A} \cdot \vec{B} = 0$.
Let $\vec{A} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ and $\vec{B} = \hat{i} + 2 \hat{j} + \alpha \hat{k}$.
Taking the dot product: $(2 \hat{i} + 4 \hat{j} - 2 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + \alpha \hat{k}) = 0$.
$(2)(1) + (4)(2) + (-2)(\alpha) = 0$.
$2 + 8 - 2\alpha = 0$.
$10 - 2\alpha = 0$.
$2\alpha = 10$.
$\alpha = 5$.

(D) Given the expression $F = (\hat{n} \cdot F) \hat{n} + G$,we can isolate $G$ as $G = F - (\hat{n} \cdot F) \hat{n}$.
Using the vector triple product identity $A \times (B \times C) = B(A \cdot C) - C(A \cdot B)$,we evaluate $(\hat{n} \times F) \times \hat{n}$.
Note that $(\hat{n} \times F) \times \hat{n} = -\hat{n} \times (\hat{n} \times F)$.
Applying the identity: $-\hat{n} \times (\hat{n} \times F) = -[\hat{n}(\hat{n} \cdot F) - F(\hat{n} \cdot \hat{n})]$.
Since $\hat{n}$ is a unit vector,$\hat{n} \cdot \hat{n} = 1$.
Thus,$-\hat{n}(\hat{n} \cdot F) + F(1) = F - (\hat{n} \cdot F) \hat{n}$.
Comparing this with our expression for $G$,we find $G = (\hat{n} \times F) \times \hat{n}$.

(D) The cross product of two vectors is anticommutative,which means $\vec{M} \times \vec{N} = -(\vec{N} \times \vec{M})$.
This implies that the vector $(\vec{M} \times \vec{N})$ and the vector $(\vec{N} \times \vec{M})$ point in exactly opposite directions.
Since the vectors are antiparallel (pointing in opposite directions),the angle between them is $180^{\circ}$.