Multiplication of Vectors Questions in English

(A) Let the North direction be along the positive $y$-axis $(+\hat{j})$ and the East direction be along the positive $x$-axis $(+\hat{i})$.
Since the vector $\vec{A}$ points towards North,we have $\vec{A} = A\hat{j}$.
The vector $\vec{B}$ points upwards,which is perpendicular to the horizontal plane (North-East plane). Let this be the $z$-axis,so $\vec{B} = B\hat{k}$.
The cross product is $\vec{A} \times \vec{B} = (A\hat{j}) \times (B\hat{k})$.
Using the cross product rules for unit vectors $(\hat{j} \times \hat{k} = \hat{i})$,we get $\vec{A} \times \vec{B} = AB\hat{i}$.
The direction $\hat{i}$ corresponds to the East direction.
Therefore,the vector $\vec{A} \times \vec{B}$ points towards East.

(B) Given the equation $\vec{A} + \vec{B} + \vec{C} = 0$.
We can write $\vec{A} = -(\vec{B} + \vec{C})$.
Taking the cross product of both sides with $\vec{B}$:
$\vec{A} \times \vec{B} = -(\vec{B} + \vec{C}) \times \vec{B}$.
Using the distributive property of the cross product:
$\vec{A} \times \vec{B} = -(\vec{B} \times \vec{B}) - (\vec{C} \times \vec{B})$.
Since the cross product of any vector with itself is zero $(\vec{B} \times \vec{B} = 0)$:
$\vec{A} \times \vec{B} = 0 - (\vec{C} \times \vec{B})$.
Using the property $\vec{C} \times \vec{B} = -(\vec{B} \times \vec{C})$:
$\vec{A} \times \vec{B} = -(-(\vec{B} \times \vec{C})) = \vec{B} \times \vec{C}$.
Thus,$\vec{A} \times \vec{B} = \vec{B} \times \vec{C} = \vec{C} \times \vec{A}$.

(B) Given vectors are $\vec{A} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = 4 \hat{i} + \hat{j} + 2 \hat{k}$.
To find the cross product $\vec{A} \times \vec{B}$,we use the determinant method:
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 4 & 1 & 2 \end{vmatrix}$
$= \hat{i} [(-3)(2) - (4)(1)] - \hat{j} [(2)(2) - (4)(4)] + \hat{k} [(2)(1) - (-3)(4)]$
$= \hat{i} [-6 - 4] - \hat{j} [4 - 16] + \hat{k} [2 + 12]$
$= -10 \hat{i} + 12 \hat{j} + 14 \hat{k}$
Now,calculate the magnitude $|\vec{A} \times \vec{B}| = \sqrt{(-10)^2 + (12)^2 + (14)^2}$
$= \sqrt{100 + 144 + 196}$
$= \sqrt{440}$
$= \sqrt{4 \times 110} = 2 \sqrt{110}$.

(A) Two vectors are perpendicular if their dot product is zero.
$(a \hat{i} + b \hat{j} + \hat{k}) \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 0$
$2a - 3b + 4 = 0 \Rightarrow 2a - 3b = -4$
We are given the second equation: $3a + 2b = 7$.
To solve for $a$ and $b$,multiply the first equation by $2$ and the second by $3$:
$4a - 6b = -8$
$9a + 6b = 21$
Adding these equations: $13a = 13 \Rightarrow a = 1$.
Substituting $a = 1$ into $3a + 2b = 7$: $3(1) + 2b = 7 \Rightarrow 2b = 4 \Rightarrow b = 2$.
The ratio $\frac{a}{b} = \frac{1}{2}$.
Given $\frac{a}{b} = \frac{x}{2}$,we have $\frac{1}{2} = \frac{x}{2}$,which implies $x = 1$.

(D) Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$.
Given $\vec{P} = \hat{i} + 2m \hat{j} + m \hat{k}$ and $\vec{Q} = 4 \hat{i} - 2 \hat{j} + m \hat{k}$.
Taking the dot product:
$(\hat{i} + 2m \hat{j} + m \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} + m \hat{k}) = 0$
$(1)(4) + (2m)(-2) + (m)(m) = 0$
$4 - 4m + m^2 = 0$
This is a quadratic equation in the form $(m - 2)^2 = 0$.
Therefore,$m = 2$.

(D) First,calculate the cross product $\overrightarrow{P} \times \overrightarrow{Q}$ using the determinant method:
$\overrightarrow{P} \times \overrightarrow{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5 \end{vmatrix}$
$= \hat{i}(\sqrt{3} \times 2.5 - 2 \times \sqrt{3}) - \hat{j}(3 \times 2.5 - 2 \times 4) + \hat{k}(3 \times \sqrt{3} - 4 \times \sqrt{3})$
$= \hat{i}(2.5\sqrt{3} - 2\sqrt{3}) - \hat{j}(7.5 - 8) + \hat{k}(3\sqrt{3} - 4\sqrt{3})$
$= 0.5\sqrt{3}\hat{i} + 0.5\hat{j} - \sqrt{3}\hat{k}$
$= \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k}$
Next,find the magnitude $|\overrightarrow{P} \times \overrightarrow{Q}| = \sqrt{(\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 + (-\sqrt{3})^2} = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = \sqrt{1 + 3} = 2$.
The unit vector is $\frac{\overrightarrow{P} \times \overrightarrow{Q}}{|\overrightarrow{P} \times \overrightarrow{Q}|} = \frac{1}{2}(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k}) = \frac{1}{4}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k})$.
Comparing this with $\frac{1}{x}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k})$,we get $x = 4$.

(B) The scalar triple product $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ implies that the three vectors are coplanar.
This can be calculated using the determinant of the components:
$\begin{vmatrix} -x & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$-x(4(3) - 3(-1)) - (-6)((-1)(3) - 3(-8)) + (-2)((-1)(-1) - 4(-8)) = 0$
$-x(12 + 3) + 6(-3 + 24) - 2(1 + 32) = 0$
$-15x + 6(21) - 2(33) = 0$
$-15x + 126 - 66 = 0$
$-15x + 60 = 0$
$15x = 60$
$x = 4$

(A) Given that the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}\right)$,so $\cos \theta = \frac{5}{9}$.
We are given the condition $|\vec{a}+\vec{b}| = \sqrt{2}|\vec{a}-\vec{b}|$.
Squaring both sides,we get $|\vec{a}+\vec{b}|^2 = 2|\vec{a}-\vec{b}|^2$.
Expanding the dot products: $a^2 + b^2 + 2\vec{a}\cdot\vec{b} = 2(a^2 + b^2 - 2\vec{a}\cdot\vec{b})$.
$a^2 + b^2 + 2ab\cos\theta = 2a^2 + 2b^2 - 4ab\cos\theta$.
Rearranging the terms: $6ab\cos\theta = a^2 + b^2$.
Substituting $\cos\theta = \frac{5}{9}$: $6ab\left(\frac{5}{9}\right) = a^2 + b^2$.
$\frac{10}{3}ab = a^2 + b^2$.
Given $|\vec{a}| = n|\vec{b}|$,so $a = nb$. Substituting this into the equation:
$\frac{10}{3}(nb)b = (nb)^2 + b^2$.
$\frac{10}{3}nb^2 = n^2b^2 + b^2$.
Dividing by $b^2$ (assuming $b \neq 0$): $\frac{10}{3}n = n^2 + 1$.
$3n^2 - 10n + 3 = 0$.
Solving the quadratic equation: $(3n - 1)(n - 3) = 0$.
Thus,$n = \frac{1}{3}$ or $n = 3$.
The integer value of $n$ is $3$.

(A) Given: $|\vec{A}|=4$,$|\vec{A}+\vec{B}|=10$,and $\vec{A} \cdot(\vec{A}+\vec{B})=20$.
Expanding the dot product: $\vec{A} \cdot \vec{A} + \vec{A} \cdot \vec{B} = 20$.
Since $\vec{A} \cdot \vec{A} = |\vec{A}|^2 = 4^2 = 16$,we have $16 + \vec{A} \cdot \vec{B} = 20$,which implies $\vec{A} \cdot \vec{B} = 4$.
Now,consider the magnitude of the sum: $|\vec{A}+\vec{B}|^2 = (\vec{A}+\vec{B}) \cdot (\vec{A}+\vec{B}) = |\vec{A}|^2 + |\vec{B}|^2 + 2(\vec{A} \cdot \vec{B})$.
Substituting the known values: $10^2 = 4^2 + |\vec{B}|^2 + 2(4)$.
$100 = 16 + |\vec{B}|^2 + 8$.
$100 = 24 + |\vec{B}|^2$.
$|\vec{B}|^2 = 100 - 24 = 76$.
Therefore,$|\vec{B}| = \sqrt{76}$ units.

(A) Given that $|\vec{A}_1|=2, |\vec{A}_2|=3$ and $|\vec{A}_1+\vec{A}_2|=3$.
Squaring the magnitude of the sum:
$(|\vec{A}_1+\vec{A}_2|)^2 = 3^2$
$A_1^2 + A_2^2 + 2 \vec{A}_1 \cdot \vec{A}_2 = 9$
$4 + 9 + 2 \vec{A}_1 \cdot \vec{A}_2 = 9$
$13 + 2 \vec{A}_1 \cdot \vec{A}_2 = 9$
$2 \vec{A}_1 \cdot \vec{A}_2 = -4$
$\vec{A}_1 \cdot \vec{A}_2 = -2$
Now,expand the expression $(\vec{A}_1+2 \vec{A}_2) \cdot (3 \vec{A}_1-4 \vec{A}_2)$:
$= 3 \vec{A}_1 \cdot \vec{A}_1 - 4 \vec{A}_1 \cdot \vec{A}_2 + 6 \vec{A}_2 \cdot \vec{A}_1 - 8 \vec{A}_2 \cdot \vec{A}_2$
$= 3 |\vec{A}_1|^2 + 2 \vec{A}_1 \cdot \vec{A}_2 - 8 |\vec{A}_2|^2$
Substitute the values:
$= 3(2)^2 + 2(-2) - 8(3)^2$
$= 3(4) - 4 - 8(9)$
$= 12 - 4 - 72$
$= 8 - 72 = -64$

(D) The cross product $\overrightarrow{C} = \overrightarrow{A} \times \overrightarrow{B}$ is a vector that is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$.
Since $\overrightarrow{A} \times \overrightarrow{B}$ is perpendicular to the plane containing $\overrightarrow{A}$ and $\overrightarrow{B}$,it must be perpendicular to any vector lying in that plane.
Both $(\overrightarrow{A} + \overrightarrow{B})$ and $(\overrightarrow{A} - \overrightarrow{B})$ are linear combinations of $\overrightarrow{A}$ and $\overrightarrow{B}$,meaning they lie in the same plane.
Therefore,$(\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} + \overrightarrow{B})$ and $(\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} - \overrightarrow{B})$ are both correct.
Statement $(e)$ is incorrect because $\overrightarrow{A} \cdot \overrightarrow{B}$ is a scalar,not a vector,and the concept of perpendicularity is defined for vectors.

(C) Given that $|\hat{a} \cdot \hat{b}| = \frac{1}{2}$.
Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = 1$ and $|\hat{b}| = 1$.
Thus,$|\cos \theta| = \frac{1}{2}$,which implies $\theta = 60^\circ$ or $\theta = 120^\circ$.
We know that $|\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2(\hat{a} \cdot \hat{b})$.
Case $1$: If $\hat{a} \cdot \hat{b} = \frac{1}{2}$,then $|\hat{a} - \hat{b}|^2 = 1 + 1 - 2(\frac{1}{2}) = 1$,so $|\hat{a} - \hat{b}| = 1$.
Case $2$: If $\hat{a} \cdot \hat{b} = -\frac{1}{2}$,then $|\hat{a} - \hat{b}|^2 = 1 + 1 - 2(-\frac{1}{2}) = 3$,so $|\hat{a} - \hat{b}| = \sqrt{3}$.
Therefore,$|\hat{a} - \hat{b}|$ can be $1$ or $\sqrt{3}$.

(D) Two vectors $\vec{A}$ and $\vec{B}$ are perpendicular if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.
Let $\vec{A} = \hat{i} + \hat{j} + \hat{k}$.
For option $A$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1 \neq 0$.
For option $B$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = (1)(1) + (1)(-1) + (1)(-1) = 1 - 1 - 1 = -1 \neq 0$.
For option $C$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (-\hat{i} - \hat{j} - \hat{k}) = (1)(-1) + (1)(-1) + (1)(-1) = -1 - 1 - 1 = -3 \neq 0$.
For option $D$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (3\hat{i} + 2\hat{j} - 5\hat{k}) = (1)(3) + (1)(2) + (1)(-5) = 3 + 2 - 5 = 0$.
Since the dot product is zero for option $D$,it is the correct answer.

(C) Using 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 $\vec{b} \times \vec{b} = 0$.
Also,using the anticommutative property,$-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}$.
Therefore,the expression becomes $0 + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - 0 = 2(\vec{a} \times \vec{b})$.

(D) Given that $\hat{a}_{1}$ and $\hat{a}_{2}$ are unit vectors,so $|\hat{a}_{1}| = 1$ and $|\hat{a}_{2}| = 1$.
Given $|\hat{a}_{1}-\hat{a}_{2}| = \sqrt{3}$.
Squaring both sides,we get $|\hat{a}_{1}-\hat{a}_{2}|^2 = 3$.
$(\hat{a}_{1}-\hat{a}_{2}) \cdot (\hat{a}_{1}-\hat{a}_{2}) = 3 \implies |\hat{a}_{1}|^2 + |\hat{a}_{2}|^2 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3$.
$1 + 1 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3 \implies 2 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3 \implies \hat{a}_{1} \cdot \hat{a}_{2} = -1/2$.
Now,we need to evaluate $(\hat{a}_{1}-\hat{a}_{2}) \cdot (2\hat{a}_{1}-\hat{a}_{2})$.
$= 2(\hat{a}_{1} \cdot \hat{a}_{1}) - (\hat{a}_{1} \cdot \hat{a}_{2}) - 2(\hat{a}_{2} \cdot \hat{a}_{1}) + (\hat{a}_{2} \cdot \hat{a}_{2})$.
$= 2(1) - 3(\hat{a}_{1} \cdot \hat{a}_{2}) + 1$.
$= 3 - 3(-1/2) = 3 + 3/2 = 9/2 = 4.5$.

(A) Given the condition $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$.
Using the definitions of dot and cross products,we have $AB \cos \theta = AB \sin \theta$.
Dividing both sides by $AB$ (assuming $A, B \neq 0$),we get $\cos \theta = \sin \theta$,which implies $\tan \theta = 1$.
Thus,$\theta = 45^{\circ}$ or $\frac{\pi}{4}$ radians.
The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ is given by $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos \theta}$.
Substituting $\theta = 45^{\circ}$,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos 45^{\circ}}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the expression becomes $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \times \frac{1}{\sqrt{2}}}$.
Simplifying,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + \sqrt{2} AB}$.

(D) If two vectors are perpendicular,their dot product must be equal to zero.
Given vectors are $A = 2i + 2j + 3k$ and $B = 3i + 6j + nk$.
The dot product $A \cdot B = (2)(3) + (2)(6) + (3)(n) = 0$.
Calculating the terms: $6 + 12 + 3n = 0$.
$18 + 3n = 0$.
$3n = -18$.
$n = -6$.
Therefore,the value of $n$ is $-6$.

(D) Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular if their dot product is zero,i.e.,$\overrightarrow{A} \cdot \overrightarrow{B} = 0$.
Given $\overrightarrow{A} = a_{1} \hat{\imath} + a_{2} \hat{\jmath}$ and $\overrightarrow{B} = b_{1} \hat{\imath} + b_{2} \hat{\jmath}$.
Calculating the dot product: $(a_{1} \hat{\imath} + a_{2} \hat{\jmath}) \cdot (b_{1} \hat{\imath} + b_{2} \hat{\jmath}) = a_{1}b_{1} + a_{2}b_{2} = 0$.
This implies $a_{1}b_{1} = -a_{2}b_{2}$.
Rearranging the terms to match the options: $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$ is not correct,but looking at $a_{1}b_{1} = -a_{2}b_{2}$,we can write $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$ or $\frac{b_{2}}{a_{1}} = -\frac{a_{2}}{b_{1}}$ is incorrect. Let's re-examine: $a_{1}b_{1} + a_{2}b_{2} = 0 \implies a_{1}b_{1} = -a_{2}b_{2}$. Dividing by $a_{1}b_{1}$,we get $1 = -\frac{a_{2}b_{2}}{a_{1}b_{1}}$. Rearranging gives $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$. Wait,checking option $A$: $\frac{b_{2}}{a_{1}} = -\frac{a_{2}}{b_{1}}$ is equivalent to $b_{1}b_{2} = -a_{1}a_{2}$,which is not the condition. The condition is $a_{1}b_{1} + a_{2}b_{2} = 0$. Thus,$a_{1}b_{1} = -a_{2}b_{2}$. This can be written as $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$. None of the options perfectly match this standard form,but if we rearrange $a_{1}b_{1} = -a_{2}b_{2}$ as $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$,there might be a typo in the options. However,if we look at $\frac{a_{1}}{b_{2}} = -\frac{a_{2}}{b_{1}}$,this implies $a_{1}b_{1} = -a_{2}b_{2}$,which is exactly the condition for perpendicularity. Therefore,option $D$ is correct.

(C) Let $\overrightarrow{A}$ and $\overrightarrow{B}$ be two vectors in a plane. The sum $\overrightarrow{A} + \overrightarrow{B}$ is a vector that lies in the same plane as $\overrightarrow{A}$ and $\overrightarrow{B}$.
By the definition of the cross product,$\overrightarrow{A} \times \overrightarrow{B}$ is a vector that is perpendicular to the plane containing both $\overrightarrow{A}$ and $\overrightarrow{B}$.
Since the vector $(\overrightarrow{A} + \overrightarrow{B})$ lies in the plane and the vector $(\overrightarrow{A} \times \overrightarrow{B})$ is perpendicular to the plane,the angle between them is $90^{\circ}$ or $\frac{\pi}{2} \text{ rad}$.
Mathematically,the dot product of these two vectors is:
$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} \times \overrightarrow{B}) = \overrightarrow{A} \cdot (\overrightarrow{A} \times \overrightarrow{B}) + \overrightarrow{B} \cdot (\overrightarrow{A} \times \overrightarrow{B})$
Using the property of the scalar triple product,$\overrightarrow{A} \cdot (\overrightarrow{A} \times \overrightarrow{B}) = 0$ and $\overrightarrow{B} \cdot (\overrightarrow{A} \times \overrightarrow{B}) = 0$.
Therefore,$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} \times \overrightarrow{B}) = 0$,which confirms that the vectors are orthogonal,and the angle between them is $90^{\circ}$.

(B) Two vectors $\vec{P}$ and $\vec{Q}$ are perpendicular if their dot product is zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$.
Given $\vec{P} = b \hat{i} + 6 \hat{j} + \hat{k}$ and $\vec{Q} = \hat{i} - a \hat{j} + 4 \hat{k}$.
Calculating the dot product: $(b)(1) + (6)(-a) + (1)(4) = 0$.
This simplifies to $b - 6a + 4 = 0$,or $b - 6a = -4$.
We are also given the equation $3b - a = 5$,which implies $a = 3b - 5$.
Substituting $a$ into the first equation: $b - 6(3b - 5) = -4$.
$b - 18b + 30 = -4$.
$-17b = -34$,so $b = 2$.
Now,find $a$: $a = 3(2) - 5 = 6 - 5 = 1$.
Thus,$a = 1$ and $b = 2$.

(C) The magnitude of the vector product (cross product) of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula:
$|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta$
Given:
$|\vec{A}| = 5 \sqrt{3}$ units
$|\vec{B}| = 10$ units
$\theta = 30^{\circ}$
Substituting these values into the formula:
$|\vec{A} \times \vec{B}| = (5 \sqrt{3}) \times (10) \times \sin 30^{\circ}$
$|\vec{A} \times \vec{B}| = 50 \sqrt{3} \times \frac{1}{2}$
$|\vec{A} \times \vec{B}| = 25 \sqrt{3}$ units
Therefore,the correct option is $C$.

(D) We are given $|\vec{a}| = \sqrt{26}$,$|\vec{b}| = 7$,and $|\vec{a} \times \vec{b}| = 35$.
Using the formula for the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 35$.
Substituting the values: $\sqrt{26} \times 7 \times \sin \theta = 35$.
$\sin \theta = \frac{35}{7 \sqrt{26}} = \frac{5}{\sqrt{26}}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{25}{26} = \frac{1}{26}$.
Thus,$|\cos \theta| = \frac{1}{\sqrt{26}}$.
The dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
$\vec{a} \cdot \vec{b} = \sqrt{26} \times 7 \times \left( \pm \frac{1}{\sqrt{26}} \right) = \pm 7$.
Given the options,the correct value is $7$.

(D) Given: $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}$.
First,calculate the vectors $(\vec{a} + 3\vec{b})$ and $(2\vec{a} - \vec{b})$:
$\vec{a} + 3\vec{b} = (\hat{i} + \hat{j} + 2\hat{k}) + 3(3\hat{i} + 2\hat{j} - \hat{k}) = (1+9)\hat{i} + (1+6)\hat{j} + (2-3)\hat{k} = 10\hat{i} + 7\hat{j} - \hat{k}$.
$2\vec{a} - \vec{b} = 2(\hat{i} + \hat{j} + 2\hat{k}) - (3\hat{i} + 2\hat{j} - \hat{k}) = (2-3)\hat{i} + (2-2)\hat{j} + (4+1)\hat{k} = -\hat{i} + 0\hat{j} + 5\hat{k}$.
Now,calculate the dot product: $[(\vec{a} + 3\vec{b}) \cdot (2\vec{a} - \vec{b})] = (10\hat{i} + 7\hat{j} - \hat{k}) \cdot (-\hat{i} + 0\hat{j} + 5\hat{k})$.
$= (10 \times -1) + (7 \times 0) + (-1 \times 5) = -10 + 0 - 5 = -15$.
The magnitude of $-15$ is $|-15| = 15$.

(C) Two vectors are perpendicular if their dot product is zero.
Let $\vec{A} = a \hat{i} + b \hat{j} + \hat{k}$ and $\vec{B} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$.
$\vec{A} \cdot \vec{B} = (a)(2) + (b)(-3) + (1)(4) = 0$.
$2a - 3b + 4 = 0$,which implies $2a - 3b = -4$.
We are given the system of equations:
$1) 2a - 3b = -4$
$2) 3a + 2b = 7$
Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$4a - 6b = -8$
$9a + 6b = 21$
Adding these equations: $13a = 13$,so $a = 1$.
Substitute $a = 1$ into $3a + 2b = 7$: $3(1) + 2b = 7 \implies 2b = 4 \implies b = 2$.
The ratio $a/b = 1/2$.
We are given $a/b = x/2$,so $1/2 = x/2$,which means $x = 1$.

(B) Given that $\vec{A} \cdot \vec{B} = 0$,this implies that vector $\vec{A}$ is perpendicular to vector $\vec{B}$.
Given that $\vec{A} \cdot \vec{C} = 0$,this implies that vector $\vec{A}$ is perpendicular to vector $\vec{C}$.
Since vector $\vec{A}$ is perpendicular to both $\vec{B}$ and $\vec{C}$,it must be parallel to the direction of the cross product of $\vec{B}$ and $\vec{C}$.
The cross product $\vec{B} \times \vec{C}$ results in a vector that is perpendicular to the plane containing both $\vec{B}$ and $\vec{C}$.
Therefore,$\vec{A}$ is parallel to $\vec{B} \times \vec{C}$.

(B) Given vector $\vec{P} = P \sin \theta \hat{i} - P \cos \theta \hat{j}$.
Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$.
Let us check option $B$: $\vec{Q} = Q \cos \theta \hat{i} + Q \sin \theta \hat{j}$.
$\vec{P} \cdot \vec{Q} = (P \sin \theta \hat{i} - P \cos \theta \hat{j}) \cdot (Q \cos \theta \hat{i} + Q \sin \theta \hat{j})$
$= (P \sin \theta)(Q \cos \theta) + (-P \cos \theta)(Q \sin \theta)$
$= PQ \sin \theta \cos \theta - PQ \sin \theta \cos \theta = 0$.
Since the dot product is $0$,the vectors are perpendicular.

(C) Let two vectors $P$ and $Q$ be represented as shown in the figure.
Here,$Q_P$ is the component of vector $Q$ in the direction of vector $P$.
From the right-angled triangle formed by the projection of $Q$ onto $P$,we have:
$\cos \theta = \frac{Q_P}{Q}$
$\Rightarrow Q_P = Q \cos \theta$
We know that the dot product of two vectors is given by $P \cdot Q = PQ \cos \theta$.
Therefore,$\cos \theta = \frac{P \cdot Q}{PQ}$.
Substituting this into the expression for $Q_P$:
$Q_P = Q \left( \frac{P \cdot Q}{PQ} \right)$
$Q_P = \frac{P \cdot Q}{P}$
Thus,the component of $Q$ in the direction of $P$ is $\frac{P \cdot Q}{P}$.

(B) Two vectors $\vec{A}$ and $\vec{B}$ are perpendicular if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.
Given $\vec{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ and $\vec{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$.
Calculating the dot product: $(2\hat{i} + 3\hat{j} + 8\hat{k}) \cdot (-4\hat{i} + 4\hat{j} + \alpha\hat{k}) = 0$.
$(2)(-4) + (3)(4) + (8)(\alpha) = 0$.
$-8 + 12 + 8\alpha = 0$.
$4 + 8\alpha = 0$.
$8\alpha = -4$.
$\alpha = -4/8 = -1/2$.

(C) Given that vectors $A$ and $B$ are mutually perpendicular,so $A \cdot B = 0$.
We need to find the component of vector $(A-B)$ along the direction of vector $(A+B)$.
The component of a vector $P$ along the direction of vector $Q$ is given by the formula: $\text{Component} = \frac{P \cdot Q}{|Q|}$.
Here,$P = A-B$ and $Q = A+B$.
First,calculate the magnitude of $(A+B)$:
$|A+B| = \sqrt{(A+B) \cdot (A+B)} = \sqrt{A \cdot A + B \cdot B + 2(A \cdot B)} = \sqrt{|A|^2 + |B|^2 + 0} = \sqrt{|A|^2 + |B|^2}$.
Now,calculate the dot product $(A-B) \cdot (A+B)$:
$(A-B) \cdot (A+B) = A \cdot A + A \cdot B - B \cdot A - B \cdot B = |A|^2 + 0 - 0 - |B|^2 = |A|^2 - |B|^2$.
Finally,the component is:
$\text{Component} = \frac{(A-B) \cdot (A+B)}{|A+B|} = \frac{|A|^2 - |B|^2}{\sqrt{|A|^2 + |B|^2}}$.

(C) Two vectors are perpendicular if their dot product is equal to $0$.
Let the given vector be $\vec{A} = 4 \hat{i} - 3 \hat{j}$.
We check the dot product of $\vec{A}$ with each option:
For option $C$: $\vec{B} = 3 \hat{i} + 4 \hat{j}$.
$\vec{A} \cdot \vec{B} = (4 \hat{i} - 3 \hat{j}) \cdot (3 \hat{i} + 4 \hat{j}) = (4 \times 3) + (-3 \times 4) = 12 - 12 = 0$.
Since the dot product is $0$,the vector $(3 \hat{i} + 4 \hat{j})$ is perpendicular to $(4 \hat{i} - 3 \hat{j})$.

(B) The cross product of two vectors $\overrightarrow{r}$ and $\overrightarrow{F}$ is given by the determinant method:
$\overrightarrow{r} \times \overrightarrow{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}$
Given $\overrightarrow{r} = (-5 \hat{i} - 3 \hat{j} + 0 \hat{k}) \text{ m}$ and $\overrightarrow{F} = (4 \hat{i} - 10 \hat{j} + 0 \hat{k}) \text{ N}$.
Substituting the values:
$\overrightarrow{r} \times \overrightarrow{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & -3 & 0 \\ 4 & -10 & 0 \end{vmatrix}$
$= \hat{i}((-3)(0) - (0)(-10)) - \hat{j}((-5)(0) - (0)(4)) + \hat{k}((-5)(-10) - (-3)(4))$
$= \hat{i}(0) - \hat{j}(0) + \hat{k}(50 - (-12))$
$= \hat{k}(50 + 12) = 62 \hat{k} \text{ Nm}$.

(C) The vector product of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is defined as $\overrightarrow{C} = \overrightarrow{A} \times \overrightarrow{B} = AB \sin \theta \hat{n}$.
For two parallel vectors,the angle $\theta$ between them is $0^\circ$.
Substituting this into the formula: $\overrightarrow{A} \times \overrightarrow{B} = AB \sin(0^\circ) \hat{n} = AB(0) \hat{n} = \vec{0}$.
Thus,the vector product of two parallel vectors is a null vector.
Therefore,option $(C)$ is correct.

(C) The dot product (scalar product) of two vectors results in a scalar,while the cross product (vector product) results in a vector.
$1$. For $(\vec{A} \cdot \vec{A})(\vec{B} \cdot \vec{C})$: Both $(\vec{A} \cdot \vec{A})$ and $(\vec{B} \cdot \vec{C})$ are scalars. The product of two scalars is a scalar. This statement is true.
$2$. For $(\vec{A} \times \vec{B}) \cdot(\vec{B} \times \vec{C})$: Both $(\vec{A} \times \vec{B})$ and $(\vec{B} \times \vec{C})$ are vectors. The dot product of two vectors is a scalar. This statement is true.
$3$. For $(\vec{A} \times \vec{C}) \times(\vec{B} \times \vec{C})$: Both $(\vec{A} \times \vec{C})$ and $(\vec{B} \times \vec{C})$ are vectors. The cross product of two vectors is a vector. Therefore,the statement that it is a scalar value is false.
$4$. For $\vec{A} \times(\vec{B} \times \vec{C})$: $(\vec{B} \times \vec{C})$ is a vector. The cross product of $\vec{A}$ with this vector results in another vector. This statement is true.

(A) Let the given vectors be $\vec{A} = 5 \hat{i} + 12 \hat{j}$ and $\vec{B} = 3 \hat{i} + 4 \hat{j}$.
First,we find the magnitudes of these vectors:
$|\vec{A}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
$|\vec{B}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Now,the unit vectors are $\hat{n}_1 = \frac{\vec{A}}{|\vec{A}|} = \frac{1}{13}(5 \hat{i} + 12 \hat{j})$ and $\hat{n}_2 = \frac{\vec{B}}{|\vec{B}|} = \frac{1}{5}(3 \hat{i} + 4 \hat{j})$.
The dot product is $\hat{n}_1 \cdot \hat{n}_2 = \left[ \frac{1}{13}(5 \hat{i} + 12 \hat{j}) \right] \cdot \left[ \frac{1}{5}(3 \hat{i} + 4 \hat{j}) \right]$.
$= \frac{1}{65} (5 \times 3 + 12 \times 4) = \frac{1}{65} (15 + 48) = \frac{63}{65}$.

(A) Given vectors are $\overrightarrow{r_1} = \hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{r_2} = \hat{i} - \hat{j} + \hat{k}$.
First,we calculate the cross product $\overrightarrow{r_1} \times \overrightarrow{r_2}$ using the determinant method:
$\overrightarrow{r_1} \times \overrightarrow{r_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$
$= \hat{i}(1 - (-1)) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$
$= \hat{i}(2) - \hat{j}(0) + \hat{k}(-2) = 2\hat{i} - 2\hat{k}$.
Now,the magnitude of this vector is $|\overrightarrow{r_1} \times \overrightarrow{r_2}| = \sqrt{2^2 + 0^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.
The unit vector $\hat{n}$ in the direction of $\overrightarrow{r_1} \times \overrightarrow{r_2}$ is given by $\frac{\overrightarrow{r_1} \times \overrightarrow{r_2}}{|\overrightarrow{r_1} \times \overrightarrow{r_2}|} = \frac{2\hat{i} - 2\hat{k}}{2\sqrt{2}} = \frac{\hat{i}}{\sqrt{2}} - \frac{\hat{k}}{\sqrt{2}}$.

(D) Given,$A=2 \hat{i}+4 \hat{j}+4 \hat{k}$ and $B=4 \hat{i}+2 \hat{j}-4 \hat{k}$.
Let $\theta$ be the angle between $A$ and $B$.
By using the dot product formula,$A \cdot B = |A| |B| \cos \theta$,where $|A|$ and $|B|$ are the magnitudes of vectors $A$ and $B$ respectively.
First,calculate the magnitudes:
$|A| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
$|B| = \sqrt{4^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
Now,calculate the dot product:
$A \cdot B = (2)(4) + (4)(2) + (4)(-4) = 8 + 8 - 16 = 0$.
Substituting these values into the dot product formula:
$0 = (6)(6) \cos \theta$.
$0 = 36 \cos \theta$.
$\cos \theta = 0$.
Therefore,$\theta = 90^{\circ}$.

(A) The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (3)(5) + (2)(3) + (5)(1) = 15 + 6 + 5 = 26$.
Next,calculate the magnitudes of the vectors:
$|\vec{a}| = \sqrt{3^2 + 2^2 + 5^2} = \sqrt{9 + 4 + 25} = \sqrt{38}$.
$|\vec{b}| = \sqrt{5^2 + 3^2 + 1^2} = \sqrt{25 + 9 + 1} = \sqrt{35}$.
Now,substitute these values into the formula:
$\cos \theta = \frac{26}{\sqrt{38} \cdot \sqrt{35}} = \frac{26}{\sqrt{1330}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{26}{\sqrt{1330}}\right)$.

(B) The angle $\theta$ between two vectors $\vec{F}$ and $\vec{d}$ is given by the formula: $\cos \theta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}| |\vec{d}|}$
First,calculate the dot product: $\vec{F} \cdot \vec{d} = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16$
Next,calculate the magnitudes: $|\vec{F}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$
$|\vec{d}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50}$
Substitute these values into the formula: $\cos \theta = \frac{16}{\sqrt{50} \times \sqrt{50}} = \frac{16}{50}$
Therefore,$\cos \theta = 0.32$,which implies $\theta = \cos^{-1}(0.32)$.

(C) The component of a vector $\vec{P}$ along the direction of another vector $\vec{Q}$ is given by the projection formula: $\text{Component} = \vec{P} \cdot \hat{Q} = \vec{P} \cdot \frac{\vec{Q}}{|\vec{Q}|}$.
Given $\vec{P} = 2 \hat{i} + 3 \hat{j}$ and $\vec{Q} = \hat{i} + \hat{j}$.
First,calculate the magnitude of vector $\vec{Q}$:
$|\vec{Q}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Now,calculate the dot product $\vec{P} \cdot \vec{Q}$:
$\vec{P} \cdot \vec{Q} = (2 \hat{i} + 3 \hat{j}) \cdot (\hat{i} + \hat{j}) = (2 \times 1) + (3 \times 1) = 2 + 3 = 5$.
Finally,the component of $\vec{P}$ along $\vec{Q}$ is:
$\frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|} = \frac{5}{\sqrt{2}}$.

(C) The component of vector $\vec{A}$ along $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.
The component of vector $\vec{B}$ along $\vec{A}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}$.
According to the problem,the component of $\vec{A}$ along $\vec{B}$ is twice the component of $\vec{B}$ along $\vec{A}$:
$\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 2 \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|} \right)$.
Assuming $\vec{A} \cdot \vec{B} \neq 0$,we can divide both sides by $(\vec{A} \cdot \vec{B})$:
$\frac{1}{|\vec{B}|} = \frac{2}{|\vec{A}|}$.
Rearranging the terms to find the ratio of magnitudes $\frac{|\vec{A}|}{|\vec{B}|}$:
$\frac{|\vec{A}|}{|\vec{B}|} = 2$.
Thus,the ratio is $2: 1$.

(A) The ant's displacement vector $\vec{a}$ is given by the movement along the $x$ and $y$ axes:
$\vec{a} = 10 \hat{i} + 20 \hat{j}$
The position vector $\vec{b}$ of the point with magnitude $r = \sqrt{2} \ cm$ at an angle $\theta = 45^{\circ}$ with the $x$-axis is:
$\vec{b} = r \cos \theta \hat{i} + r \sin \theta \hat{j}$
$\vec{b} = \sqrt{2} \cos 45^{\circ} \hat{i} + \sqrt{2} \sin 45^{\circ} \hat{j}$
$\vec{b} = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) \hat{i} + \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) \hat{j} = \hat{i} + \hat{j}$
The dot product $\vec{a} \cdot \vec{b}$ is:
$\vec{a} \cdot \vec{b} = (10 \hat{i} + 20 \hat{j}) \cdot (\hat{i} + \hat{j})$
$= (10 \times 1) + (20 \times 1) = 10 + 20 = 30 \ cm^2$