Basic , Modulus and Algebra of vectors Questions in English

(A) The position vector of a point $R$ that divides the line segment joining points $P$ and $Q$ with position vectors $\vec{p}$ and $\vec{q}$ in the ratio $m:n$ internally is given by the section formula: $\overrightarrow{OR} = \frac{m\vec{q} + n\vec{p}}{m + n}$.
Given $\vec{p} = 3\vec{a} - 2\vec{b}$,$\vec{q} = \vec{a} + \vec{b}$,$m = 2$,and $n = 1$.
Substituting these values into the formula:
$\overrightarrow{OR} = \frac{2(\vec{a} + \vec{b}) + 1(3\vec{a} - 2\vec{b})}{2 + 1}$
$\overrightarrow{OR} = \frac{2\vec{a} + 2\vec{b} + 3\vec{a} - 2\vec{b}}{3}$
$\overrightarrow{OR} = \frac{5\vec{a}}{3}$.

(A) The position vector of a point $R$ that divides the line segment joining points $P$ and $Q$ with position vectors $\vec{p}$ and $\vec{q}$ externally in the ratio $m:n$ is given by the formula $\overrightarrow{OR} = \frac{m\vec{q} - n\vec{p}}{m - n}$.
Given $\vec{p} = 3\vec{a} - 2\vec{b}$,$\vec{q} = \vec{a} + \vec{b}$,$m = 2$,and $n = 1$.
Substituting these values into the formula:
$\overrightarrow{OR} = \frac{2(\vec{a} + \vec{b}) - 1(3\vec{a} - 2\vec{b})}{2 - 1}$
$\overrightarrow{OR} = \frac{2\vec{a} + 2\vec{b} - 3\vec{a} + 2\vec{b}}{1}$
$\overrightarrow{OR} = (2\vec{a} - 3\vec{a}) + (2\vec{b} + 2\vec{b})$
$\overrightarrow{OR} = 4\vec{b} - \vec{a}$.

(N/A) The magnitude of a vector $\vec{v} = p\hat{i} + q\hat{j} + r\hat{k}$ is given by the formula $|\vec{v}| = \sqrt{p^2 + q^2 + r^2}$.
For vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$:
$|\vec{a}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$.
For vector $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$:
$|\vec{b}| = \sqrt{(2)^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}$.
For vector $\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}$:
$|\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(-\frac{1}{\sqrt{3}}\right)^2} = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = 1$.

Consider $\vec{a} = (\hat{i} + 2\hat{j} + 3\hat{k})$ and $\vec{b} = (2\hat{i} - \hat{j} - 3\hat{k})$.
It can be observed that the magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{1^{2} + 2^{2} + 3^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{2^{2} + (-1)^{2} + (-3)^{2}} = \sqrt{4 + 1 + 9} = \sqrt{14}$.
Since $|\vec{a}| = |\vec{b}| = \sqrt{14}$ and the components are different,$\vec{a}$ and $\vec{b}$ are two different vectors having the same magnitude.

(N/A) Consider $\vec{p} = (\hat{i} + \hat{j} + \hat{k})$ and $\vec{q} = (2\hat{i} + 2\hat{j} + 2\hat{k})$.
The direction cosines of $\vec{p}$ are given by:
$l = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$,$m = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$,$n = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$.
The direction cosines of $\vec{q}$ are given by:
$l = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$m = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$n = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Since the direction cosines of $\vec{p}$ and $\vec{q}$ are the same,the two vectors have the same direction.

(A) Two vectors are equal if and only if their corresponding components are equal.
Given the vectors $2 \hat{i} + 3 \hat{j}$ and $x \hat{i} + y \hat{j}$.
By comparing the coefficients of $\hat{i}$ and $\hat{j}$,we get:
$x = 2$
$y = 3$
Therefore,the values are $x = 2$ and $y = 3$.

(A) The vector with the initial point $P(2,1)$ and terminal point $Q(-5,7)$ is given by the displacement vector $\overrightarrow{PQ}$.
$\overrightarrow{PQ} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j}$
Substituting the coordinates of $P$ and $Q$:
$\overrightarrow{PQ} = (-5 - 2) \hat{i} + (7 - 1) \hat{j}$
$\overrightarrow{PQ} = -7 \hat{i} + 6 \hat{j}$
Thus,the scalar components are $-7$ and $6$.
The vector components are $-7 \hat{i}$ and $6 \hat{j}$.

(A) Given vectors are $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$,and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.
The sum of the vectors is given by $\vec{a} + \vec{b} + \vec{c} = (\vec{a}_x + \vec{b}_x + \vec{c}_x)\hat{i} + (\vec{a}_y + \vec{b}_y + \vec{c}_y)\hat{j} + (\vec{a}_z + \vec{b}_z + \vec{c}_z)\hat{k}$.
Substituting the components:
$\vec{a} + \vec{b} + \vec{c} = (1 - 2 + 1)\hat{i} + (-2 + 4 - 6)\hat{j} + (1 + 5 - 7)\hat{k}$.
Calculating the coefficients:
$x$-component: $1 - 2 + 1 = 0$.
$y$-component: $-2 + 4 - 6 = -4$.
$z$-component: $1 + 5 - 7 = -1$.
Thus,the sum is $0\hat{i} - 4\hat{j} - 1\hat{k} = -4\hat{j} - \hat{k}$.

(A) The unit vector $\hat{a}$ in the direction of vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ is given by $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
First,calculate the magnitude of vector $\vec{a}$:
$|\vec{a}| = \sqrt{1^{2} + 1^{2} + 2^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
Now,divide the vector $\vec{a}$ by its magnitude:
$\hat{a} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$.

(A) The given points are $P(1, 2, 3)$ and $Q(4, 5, 6).$
The vector $\overrightarrow{PQ}$ is given by $\overrightarrow{PQ} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}.$
The magnitude of $\overrightarrow{PQ}$ is $|\overrightarrow{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}.$
The unit vector in the direction of $\overrightarrow{PQ}$ is $\hat{u} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}.$
$\hat{u} = \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \frac{3}{3\sqrt{3}}\hat{i} + \frac{3}{3\sqrt{3}}\hat{j} + \frac{3}{3\sqrt{3}}\hat{k} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}.$

(A) Given vectors are $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$.
First,calculate the sum $\vec{a} + \vec{b}$:
$\vec{a} + \vec{b} = (2 - 1)\hat{i} + (-1 + 1)\hat{j} + (2 - 1)\hat{k} = \hat{i} + 0\hat{j} + \hat{k} = \hat{i} + \hat{k}$.
Next,find the magnitude of the resultant vector $\vec{a} + \vec{b}$:
$|\vec{a} + \vec{b}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.
The unit vector in the direction of $\vec{a} + \vec{b}$ is given by $\frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}$:
$\frac{\hat{i} + \hat{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$.

(A) Let $\vec{a} = 5\hat{i} - \hat{j} + 2\hat{k}$.
The magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{5^2 + (-1)^2 + 2^2} = \sqrt{25 + 1 + 4} = \sqrt{30}$.
The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}}$.
$A$ vector of magnitude $8$ units in the direction of $\vec{a}$ is $8\hat{a} = 8 \left( \frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}} \right)$.
Thus,the required vector is $\frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$.

(N/A) Let $\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{b} = -4 \hat{i} + 6 \hat{j} - 8 \hat{k}$.
We observe that $\vec{b} = -4 \hat{i} + 6 \hat{j} - 8 \hat{k}$.
Taking $-2$ as a common factor,we get $\vec{b} = -2(2 \hat{i} - 3 \hat{j} + 4 \hat{k})$.
This can be written as $\vec{b} = -2 \vec{a}$.
Since $\vec{b} = \lambda \vec{a}$,where $\lambda = -2$,the two vectors are scalar multiples of each other.
Therefore,the given vectors are collinear.

(A) Let the vector be $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The magnitude of the vector is given by $|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The direction cosines $(l, m, n)$ of a vector $a\hat{i} + b\hat{j} + c\hat{k}$ are given by $\left(\frac{a}{|\vec{a}|}, \frac{b}{|\vec{a}|}, \frac{c}{|\vec{a}|}\right)$.
Substituting the values,we get the direction cosines as $\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)$.

(A) Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.
The magnitude of the vector $\vec{a}$ is given by $|\vec{a}| = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}$.
The direction cosines $(l, m, n)$ of the vector $\vec{a}$ are given by the components of the unit vector $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
Thus,$l = \frac{1}{\sqrt{3}}$,$m = \frac{1}{\sqrt{3}}$,and $n = \frac{1}{\sqrt{3}}$.
Let $\alpha, \beta,$ and $\gamma$ be the angles that the vector $\vec{a}$ makes with the positive directions of the $OX, OY,$ and $OZ$ axes respectively.
Then,$\cos \alpha = l = \frac{1}{\sqrt{3}}$,$\cos \beta = m = \frac{1}{\sqrt{3}}$,and $\cos \gamma = n = \frac{1}{\sqrt{3}}$.
Since $\cos \alpha = \cos \beta = \cos \gamma$,it follows that $\alpha = \beta = \gamma$.
Therefore,the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.

(A) The position vector of a point $R$ dividing the line segment joining two points $P$ and $Q$ with position vectors $\vec{a}$ and $\vec{b}$ in the ratio $m:n$ internally is given by the formula: $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$.
Here,$\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$,$\vec{b} = -\hat{i} + \hat{j} + \hat{k}$,$m = 2$,and $n = 1$.
Substituting these values into the formula:
$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2+1}$
$\vec{r} = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} - \hat{k})}{3}$
$\vec{r} = \frac{(-2+1)\hat{i} + (2+2)\hat{j} + (2-1)\hat{k}}{3}$
$\vec{r} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$.

(A) The position vector of point $R$ dividing the line segment joining two points $P$ and $Q$ with position vectors $\vec{a}$ and $\vec{b}$ in the ratio $m: n$ externally is given by: $\vec{r} = \frac{m \vec{b} - n \vec{a}}{m - n}$.
Given position vectors are $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$ with ratio $m: n = 2: 1$.
Substituting the values into the formula:
$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2 \hat{j} - \hat{k})}{2 - 1}$
$= \frac{-2 \hat{i} + 2 \hat{j} + 2 \hat{k} - \hat{i} - 2 \hat{j} + \hat{k}}{1}$
$= -3 \hat{i} + 3 \hat{k}$.

(A) The position vector of the midpoint $R$ of the line segment joining points $P(2, 3, 4)$ and $Q(4, 1, -2)$ is calculated using the midpoint formula: $\overrightarrow{OR} = \frac{\overrightarrow{OP} + \overrightarrow{OQ}}{2}$.
Given $\overrightarrow{OP} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\overrightarrow{OQ} = 4\hat{i} + \hat{j} - 2\hat{k}$.
Substituting these values:
$\overrightarrow{OR} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} - 2\hat{k})}{2}$
Grouping the components:
$\overrightarrow{OR} = \frac{(2 + 4)\hat{i} + (3 + 1)\hat{j} + (4 - 2)\hat{k}}{2}$
$\overrightarrow{OR} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2}$
$\overrightarrow{OR} = 3\hat{i} + 2\hat{j} + \hat{k}$.

(B) According to the triangle law of vector addition,in triangle $ABC$,we have:
$\overrightarrow{ AB } + \overrightarrow{ BC } = \overrightarrow{ AC }$
Now let us evaluate each option:
$(A) \overrightarrow{ AB } + \overrightarrow{ BC } + \overrightarrow{ CA } = \overrightarrow{ AC } + \overrightarrow{ CA } = \overrightarrow{ AC } - \overrightarrow{ AC } = \overrightarrow{0}$. This is true.
$(B) \overrightarrow{ AB } + \overrightarrow{ BC } - \overrightarrow{ CA } = \overrightarrow{ AC } - \overrightarrow{ CA } = \overrightarrow{ AC } + \overrightarrow{ AC } = 2\overrightarrow{ AC } \neq \overrightarrow{0}$. This is not true.
$(C) \overrightarrow{ AB } + \overrightarrow{ BC } - \overrightarrow{ AC } = \overrightarrow{ AC } - \overrightarrow{ AC } = \overrightarrow{0}$. This is true.
$(D) \overrightarrow{ AB } - \overrightarrow{ CB } + \overrightarrow{ CA } = \overrightarrow{ AB } + \overrightarrow{ BC } + \overrightarrow{ CA } = \overrightarrow{ AC } + \overrightarrow{ CA } = \overrightarrow{0}$. This is true.
Therefore,the statement that is not true is $(B)$.

(C) If $\vec{a}$ and $\vec{b}$ are two collinear vectors,they are parallel to the same line.
$1$. By definition,$\vec{b}=\lambda \vec{a}$ for some scalar $\lambda \neq 0$. Thus,option $A$ is correct.
$2$. If $\lambda = \pm 1$,then $\vec{a} = \pm \vec{b}$. This is a specific case of collinearity,so option $B$ is a possible property.
$3$. If $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$,then $\vec{b} = \lambda \vec{a}$ implies $b_1 = \lambda a_1, b_2 = \lambda a_2, b_3 = \lambda a_3$. This means $\frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} = \lambda$. Thus,the components are proportional. Option $C$ states they are not proportional,which is incorrect.
$4$. Collinear vectors can have the same or opposite directions. Option $D$ claims they must have the same direction,which is not necessarily true. However,in the context of multiple-choice questions where only one is 'incorrect' and $C$ is mathematically false,$C$ is the primary incorrect statement regarding properties. Note: $D$ is also technically incorrect as they can have opposite directions. Given the standard interpretation,$C$ is the most fundamentally incorrect statement regarding the definition of collinearity.

(A) We know that $|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$.
Expanding the dot product,we get:
$|\vec{a} - \vec{b}|^2 = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$.
Using the properties $\vec{a} \cdot \vec{a} = |\vec{a}|^2$,$\vec{b} \cdot \vec{b} = |\vec{b}|^2$,and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,we have:
$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$.
Substituting the given values $|\vec{a}| = 2$,$|\vec{b}| = 3$,and $\vec{a} \cdot \vec{b} = 4$:
$|\vec{a} - \vec{b}|^2 = (2)^2 - 2(4) + (3)^2$.
$|\vec{a} - \vec{b}|^2 = 4 - 8 + 9$.
$|\vec{a} - \vec{b}|^2 = 5$.
Therefore,$|\vec{a} - \vec{b}| = \sqrt{5}$.

(N/A) The inequality holds trivially in case either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}.$ So,let $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0.$ Then,
$|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})$
$= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$
$= |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$ (since the scalar product is commutative)
$\leq |\vec{a}|^2 + 2|\vec{a} \cdot \vec{b}| + |\vec{b}|^2$ (since $x \leq |x|$ for all $x \in \mathbb{R}$)
$\leq |\vec{a}|^2 + 2|\vec{a}||\vec{b}| + |\vec{b}|^2$ (by Cauchy-Schwarz inequality,$|\vec{a} \cdot \vec{b}| \leq |\vec{a}||\vec{b}|$)
$= (|\vec{a}| + |\vec{b}|)^2$
Taking the square root on both sides,we get:
$|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$

(N/A) We have the position vectors of points $A$,$B$,and $C$ as $\vec{a} = -2\hat{i} + 3\hat{j} + 5\hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$,and $\vec{c} = 7\hat{i} - \hat{k}$.
First,we calculate the vectors $\vec{AB}$ and $\vec{BC}$:
$\vec{AB} = \vec{b} - \vec{a} = (1 - (-2))\hat{i} + (2 - 3)\hat{j} + (3 - 5)\hat{k} = 3\hat{i} - \hat{j} - 2\hat{k}$
$\vec{BC} = \vec{c} - \vec{b} = (7 - 1)\hat{i} + (0 - 2)\hat{j} + (-1 - 3)\hat{k} = 6\hat{i} - 2\hat{j} - 4\hat{k}$
Observe that $\vec{BC} = 2(3\hat{i} - \hat{j} - 2\hat{k}) = 2\vec{AB}$.
Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ and they share a common point $B$,the vectors are parallel and the points $A$,$B$,and $C$ must be collinear.

Let $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})=\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}$
$\vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}$
$\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})=\frac{6}{7} \hat{i}+\frac{2}{7} \hat{j}-\frac{3}{7} \hat{k}$
$|\vec{a}|=\sqrt{(\frac{2}{7})^{2}+(\frac{3}{7})^{2}+(\frac{6}{7})^{2}}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1$
$|\vec{b}|=\sqrt{(\frac{3}{7})^{2}+(-\frac{6}{7})^{2}+(\frac{2}{7})^{2}}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=\sqrt{\frac{49}{49}}=1$
$|\vec{c}|=\sqrt{(\frac{6}{7})^{2}+(\frac{2}{7})^{2}+(-\frac{3}{7})^{2}}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=\sqrt{\frac{49}{49}}=1$
Thus,each of the given three vectors is a unit vector.
$\vec{a} \cdot \vec{b}=\frac{2}{7} \times \frac{3}{7}+\frac{3}{7} \times(-\frac{6}{7})+\frac{6}{7} \times \frac{2}{7}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=0$
$\vec{b} \cdot \vec{c}=\frac{3}{7} \times \frac{6}{7}+(-\frac{6}{7}) \times \frac{2}{7}+\frac{2}{7} \times(-\frac{3}{7})=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=0$
$\vec{c} \cdot \vec{a}=\frac{6}{7} \times \frac{2}{7}+\frac{2}{7} \times \frac{3}{7}+(-\frac{3}{7}) \times \frac{6}{7}=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0$
Since the dot product of each pair is $0$,the given three vectors are mutually perpendicular to each other.

(C) Given that $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0$.
From the property of the dot product,$\vec{a} \cdot \vec{a} = |\vec{a}|^2$.
Since $\vec{a} \cdot \vec{a} = 0$,we have $|\vec{a}|^2 = 0$,which implies $|\vec{a}| = 0$.
Therefore,$\vec{a}$ is a zero vector $(\vec{a} = \vec{0})$.
Now,consider the second condition $\vec{a} \cdot \vec{b} = 0$. Since $\vec{a} = \vec{0}$,the equation becomes $\vec{0} \cdot \vec{b} = 0$.
This equation holds true for any vector $\vec{b}$ in the space.
Thus,$\vec{b}$ can be any vector.

(B) vector $\lambda \vec{a}$ is a unit vector if its magnitude is equal to $1$,i.e.,$|\lambda \vec{a}| = 1$.
Using the property of the modulus of a vector,$|k \vec{v}| = |k| |\vec{v}|$,we have:
$|\lambda| |\vec{a}| = 1$
Given that the magnitude of $\vec{a}$ is $a$,we substitute $|\vec{a}| = a$:
$|\lambda| a = 1$
Solving for $a$,we get:
$a = \frac{1}{|\lambda|}$
Thus,$\lambda \vec{a}$ is a unit vector if $a = \frac{1}{|\lambda|}$.
The correct answer is $B$.

(A) Let the unit vector $\vec{a}$ be $\vec{a} = a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}$.
Since $\vec{a}$ is a unit vector,$|\vec{a}| = 1$.
The direction cosines of $\vec{a}$ are $\cos \alpha, \cos \beta, \cos \gamma$,where $\alpha = \frac{\pi}{3}, \beta = \frac{\pi}{4}$,and $\gamma = \theta$.
Thus,$a_{1} = \cos \frac{\pi}{3} = \frac{1}{2}$,$a_{2} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $a_{3} = \cos \theta$.
We know that $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.
Substituting the values: $\left(\frac{1}{2}\right)^{2} + \left(\frac{1}{\sqrt{2}}\right)^{2} + \cos^{2} \theta = 1$.
$\frac{1}{4} + \frac{1}{2} + \cos^{2} \theta = 1$.
$\frac{3}{4} + \cos^{2} \theta = 1 \Rightarrow \cos^{2} \theta = \frac{1}{4}$.
Since $\theta$ is acute,$\cos \theta = \frac{1}{2}$,which implies $\theta = \frac{\pi}{3}$.
Therefore,the components of $\vec{a}$ are $\left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right)$.

(A) Given $\vec{a} \cdot \vec{b} = 0$. This implies that either $|\vec{a}| = 0$ or $|\vec{b}| = 0$,or $\vec{a} \perp \vec{b}$ (if both are non-zero vectors).
Given $\vec{a} \times \vec{b} = \vec{0}$. This implies that either $|\vec{a}| = 0$ or $|\vec{b}| = 0$,or $\vec{a} \parallel \vec{b}$ (if both are non-zero vectors).
Since two non-zero vectors cannot be both perpendicular and parallel to each other simultaneously,the only possibility is that at least one of the vectors must be a zero vector.
Therefore,we conclude that either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$.

(N/A) Let $\vec{r} = x \hat{i} + y \hat{j}$ be a unit vector in the $XY$-plane.
Since $\vec{r}$ is a unit vector,its magnitude is $|\vec{r}| = 1$,which implies $\sqrt{x^2 + y^2} = 1$,or $x^2 + y^2 = 1$.
We can represent any point on the unit circle in the $XY$-plane using the parameter $\theta$,where $\theta \in [0, 2\pi)$.
Thus,we can set $x = \cos \theta$ and $y = \sin \theta$.
Substituting these into the expression for $\vec{r}$,we get:
$\vec{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$
where $\theta$ is the angle the vector makes with the positive $X$-axis.
As $\theta$ varies from $0$ to $2\pi$,this expression generates all possible unit vectors in the $XY$-plane.

(A) unit vector $\vec{r}$ in the $XY$-plane making an angle $\theta$ with the positive $x$-axis is given by $\vec{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$.
Given $\theta = 30^{\circ}$.
Substituting the value of $\theta$:
$\vec{r} = \cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j}$
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\sin 30^{\circ} = \frac{1}{2}$,we get:
$\vec{r} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}$.

The vector joining the points $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ is given by the displacement vector $\overrightarrow{PQ}$.
$\overrightarrow{PQ} = \text{Position vector of } Q - \text{Position vector of } P$
$= (x_{2} - x_{1})\hat{i} + (y_{2} - y_{1})\hat{j} + (z_{2} - z_{1})\hat{k}$
The scalar components of the vector $\overrightarrow{PQ}$ are $(x_{2} - x_{1})$,$(y_{2} - y_{1})$,and $(z_{2} - z_{1})$.
The magnitude of the vector $\overrightarrow{PQ}$ is given by $|\overrightarrow{PQ}| = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}$.

(A) Let $O$ be the origin $(0,0)$ representing the initial position of the girl.
The girl walks $4 \, km$ towards the west. So,the position of point $A$ is $(-4, 0)$,which can be represented as $\overrightarrow{OA} = -4 \hat{i}$.
From point $A$,she walks $3 \, km$ in a direction $30^{\circ}$ east of north. This means the angle with the positive $y$-axis (North) is $30^{\circ}$ towards the east. The angle with the positive $x$-axis (East) is $90^{\circ} - 30^{\circ} = 60^{\circ}$.
The displacement vector $\overrightarrow{AB}$ can be written as:
$\overrightarrow{AB} = 3(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j})$
$= 3(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = \frac{3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$.
The total displacement $\overrightarrow{OB}$ is given by the vector sum:
$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB}$
$= -4 \hat{i} + (\frac{3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j})$
$= (-4 + \frac{3}{2}) \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$
$= \frac{-8+3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$
$= \frac{-5}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$.
Thus,the girl's displacement from her initial point is $\frac{-5}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$.

(N/A) In $\Delta ABC,$ let $\overrightarrow{CB}=\vec{a}, \overrightarrow{CA}=\vec{b},$ and $\overrightarrow{AB}=\vec{c}$ (as shown in the figure).
Now,by the triangle law of vector addition,we have $\vec{a}=\vec{b}+\vec{c}.$
It is clearly known that $|\vec{a}|, |\vec{b}|,$ and $|\vec{c}|$ represent the lengths of the sides of $\Delta ABC.$
Also,it is a known geometric property that the sum of the lengths of any two sides of a triangle is always greater than the third side.
Therefore,$|\vec{a}| < |\vec{b}| + |\vec{c}|.$
Hence,it is not true that $|\vec{a}| = |\vec{b}| + |\vec{c}|.$

(C) vector $\vec{v} = x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector if its magnitude is $1$,i.e.,$|\vec{v}| = 1$.
The magnitude of the vector is $|x(\hat{i}+\hat{j}+\hat{k})| = |x| \sqrt{1^2 + 1^2 + 1^2} = |x| \sqrt{3}$.
Setting the magnitude equal to $1$,we get $|x| \sqrt{3} = 1$.
This implies $|x| = \frac{1}{\sqrt{3}}$.
Therefore,$x = \pm \frac{1}{\sqrt{3}}$.

(A) Given vectors are $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$.
Let $\vec{c}$ be the resultant vector,so $\vec{c} = \vec{a} + \vec{b}$.
$\vec{c} = (2+1) \hat{i} + (3-2) \hat{j} + (-1+1) \hat{k} = 3 \hat{i} + \hat{j}$.
The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}$.
The unit vector in the direction of $\vec{c}$ is $\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{3 \hat{i} + \hat{j}}{\sqrt{10}}$.
$A$ vector of magnitude $5$ units parallel to $\vec{c}$ is given by $\pm 5 \hat{c}$.
$\pm 5 \left( \frac{3 \hat{i} + \hat{j}}{\sqrt{10}} \right) = \pm \frac{5}{\sqrt{10}} (3 \hat{i} + \hat{j}) = \pm \frac{5 \sqrt{10}}{10} (3 \hat{i} + \hat{j}) = \pm \frac{\sqrt{10}}{2} (3 \hat{i} + \hat{j}) = \pm \frac{3 \sqrt{10}}{2} \hat{i} \pm \frac{\sqrt{10}}{2} \hat{j}$.

(A) Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-2 \hat{j}+\hat{k}.$
Let $\vec{v} = 2 \vec{a}-\vec{b}+3 \vec{c}.$
Substituting the values,we get $\vec{v} = 2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k}).$
Expanding this,$\vec{v} = (2\hat{i}+2\hat{j}+2\hat{k}) - (2\hat{i}-\hat{j}+3\hat{k}) + (3\hat{i}-6\hat{j}+3\hat{k}).$
Grouping the components,$\vec{v} = (2-2+3)\hat{i} + (2+1-6)\hat{j} + (2-3+3)\hat{k} = 3\hat{i}-3\hat{j}+2\hat{k}.$
The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{3^{2}+(-3)^{2}+2^{2}} = \sqrt{9+9+4} = \sqrt{22}.$
The unit vector parallel to $\vec{v}$ is $\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{22}} = \frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}.$

(N/A) Given position vectors are $\overrightarrow{OP} = 2\vec{a} + \vec{b}$ and $\overrightarrow{OQ} = \vec{a} - 3\vec{b}$.
Point $R$ divides the line segment $PQ$ externally in the ratio $m:n = 1:2$. The section formula for external division is $\overrightarrow{OR} = \frac{m\overrightarrow{OQ} - n\overrightarrow{OP}}{m-n}$.
Substituting the values,we get:
$\overrightarrow{OR} = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1 - 2}$
$= \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1}$
$= \frac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b}$.
To show $P$ is the mid-point of $RQ$,we calculate the mid-point of $RQ$:
Mid-point $= \frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2}$
$= \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} = \overrightarrow{OP}$.
Since the mid-point of $RQ$ is $P$,$P$ is the mid-point of the line segment $RQ$.

(A) Let $\vec{c}$ be the sum of vectors $\vec{a}$ and $\vec{b}$.
$\vec{c} = \vec{a} + \vec{b} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + (-\hat{i} + \hat{j} + 3 \hat{k})$
$\vec{c} = (2-1) \hat{i} + (-1+1) \hat{j} + (2+3) \hat{k} = \hat{i} + 0 \hat{j} + 5 \hat{k} = \hat{i} + 5 \hat{k}$
Now,calculate the magnitude of $\vec{c}$:
$|\vec{c}| = \sqrt{1^2 + 0^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$
The unit vector in the direction of $\vec{c}$ is given by $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.
$\hat{c} = \frac{\hat{i} + 5 \hat{k}}{\sqrt{26}} = \frac{1}{\sqrt{26}} \hat{i} + \frac{5}{\sqrt{26}} \hat{k}$.

(A) The vector $\overrightarrow{PQ}$ with initial point $P(1,3,2)$ and terminal point $Q(-1,0,8)$ is given by $\overrightarrow{PQ} = (-1-1)\hat{i} + (0-3)\hat{j} + (8-2)\hat{k} = -2\hat{i} - 3\hat{j} + 6\hat{k}$.
The direction opposite to $\overrightarrow{PQ}$ is the direction of $\overrightarrow{QP}$.
$\overrightarrow{QP} = -\overrightarrow{PQ} = 2\hat{i} + 3\hat{j} - 6\hat{k}$.
The magnitude of $\overrightarrow{QP}$ is $|\overrightarrow{QP}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
The unit vector in the direction of $\overrightarrow{QP}$ is $\widehat{QP} = \frac{\overrightarrow{QP}}{|\overrightarrow{QP}|} = \frac{2\hat{i} + 3\hat{j} - 6\hat{k}}{7}$.
The required vector of magnitude $11$ in the direction of $\overrightarrow{QP}$ is $11 \times \widehat{QP} = 11 \left( \frac{2\hat{i} + 3\hat{j} - 6\hat{k}}{7} \right) = \frac{22}{7}\hat{i} + \frac{33}{7}\hat{j} - \frac{66}{7}\hat{k}$.

(A) $(i)$ The position vector of the point $R$ dividing the line segment joining $P$ and $Q$ internally in the ratio $m:n = 1:2$ is given by the section formula: $\overrightarrow{OR} = \frac{m\overrightarrow{OQ} + n\overrightarrow{OP}}{m+n}$.
Substituting the values: $\overrightarrow{OR} = \frac{1(\vec{a} - 2\vec{b}) + 2(2\vec{a} + \vec{b})}{1+2} = \frac{\vec{a} - 2\vec{b} + 4\vec{a} + 2\vec{b}}{3} = \frac{5\vec{a}}{3}$.
(ii) The position vector of the point $R$ dividing the line segment joining $P$ and $Q$ externally in the ratio $m:n = 1:2$ is given by the section formula: $\overrightarrow{OR} = \frac{m\overrightarrow{OQ} - n\overrightarrow{OP}}{m-n}$.
Substituting the values: $\overrightarrow{OR} = \frac{1(\vec{a} - 2\vec{b}) - 2(2\vec{a} + \vec{b})}{1-2} = \frac{\vec{a} - 2\vec{b} - 4\vec{a} - 2\vec{b}}{-1} = \frac{-3\vec{a} - 4\vec{b}}{-1} = 3\vec{a} + 4\vec{b}$.

(A) Let $\vec{c}$ be the sum of vectors $\vec{a}$ and $\vec{b}$.
$\vec{c} = \vec{a} + \vec{b} = (2\hat{i} - \hat{j} + \hat{k}) + (2\hat{j} + \hat{k})$
$\vec{c} = 2\hat{i} + (2 - 1)\hat{j} + (1 + 1)\hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$
The magnitude of $\vec{c}$ is given by $|\vec{c}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
The unit vector in the direction of $\vec{c}$ is $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.
$\hat{c} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} = \frac{1}{3}(2\hat{i} + \hat{j} + 2\hat{k})$.

Given $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}.$
$(i)$ $6 \vec{b} = 6(2 \hat{i}+\hat{j}+2 \hat{k}) = 12 \hat{i}+6 \hat{j}+12 \hat{k}.$
Magnitude $|6 \vec{b}| = \sqrt{12^2+6^2+12^2} = \sqrt{144+36+144} = \sqrt{324} = 18.$
Unit vector = $\frac{6 \vec{b}}{|6 \vec{b}|} = \frac{12 \hat{i}+6 \hat{j}+12 \hat{k}}{18} = \frac{2}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k}.$
(ii) $2 \vec{a}-\vec{b} = 2(\hat{i}+\hat{j}+2 \hat{k}) - (2 \hat{i}+\hat{j}+2 \hat{k}) = (2 \hat{i}+2 \hat{j}+4 \hat{k}) - (2 \hat{i}+\hat{j}+2 \hat{k}) = 0 \hat{i} + 1 \hat{j} + 2 \hat{k} = \hat{j} + 2 \hat{k}.$
Magnitude $|2 \vec{a}-\vec{b}| = \sqrt{0^2+1^2+2^2} = \sqrt{5}.$
Unit vector = $\frac{2 \vec{a}-\vec{b}}{|2 \vec{a}-\vec{b}|} = \frac{\hat{j}+2 \hat{k}}{\sqrt{5}} = \frac{1}{\sqrt{5}} \hat{j} + \frac{2}{\sqrt{5}} \hat{k}.$

(A) Given the coordinates of $P$ and $Q$ are $(5, 0, 8)$ and $(3, 3, 2)$ respectively.
$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$
$= (3\hat{i} + 3\hat{j} + 2\hat{k}) - (5\hat{i} + 0\hat{j} + 8\hat{k})$
$= (3-5)\hat{i} + (3-0)\hat{j} + (2-8)\hat{k} = -2\hat{i} + 3\hat{j} - 6\hat{k}$
The magnitude of $\overrightarrow{PQ}$ is $|\overrightarrow{PQ}| = \sqrt{(-2)^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.$
The unit vector in the direction of $\overrightarrow{PQ}$ is given by $\hat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} = \frac{-2\hat{i} + 3\hat{j} - 6\hat{k}}{7} = -\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}.$

(N/A) Given that $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$.
We know that $\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \vec{a} - \vec{b}$.
According to the problem,the point $C$ is on the line $BA$ produced such that $\overrightarrow{BC} = 1.5 \overrightarrow{BA}$.
Therefore,$\overrightarrow{BC} = 1.5(\vec{a} - \vec{b})$.
Since $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}$,we have:
$\overrightarrow{OC} - \overrightarrow{OB} = 1.5\vec{a} - 1.5\vec{b}$.
Substituting $\overrightarrow{OB} = \vec{b}$,we get:
$\overrightarrow{OC} = 1.5\vec{a} - 1.5\vec{b} + \vec{b}$.
$\overrightarrow{OC} = 1.5\vec{a} - 0.5\vec{b}$.
This can be written as $\overrightarrow{OC} = \frac{3\vec{a} - \vec{b}}{2}$.

(D) Let the points be $A(k,-10,3), B(1,-1,3)$ and $C(3,5,3)$.
For points $A, B, C$ to be collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel,i.e.,$\overrightarrow{AB} = \lambda \overrightarrow{BC}$ for some scalar $\lambda$.
$\overrightarrow{AB} = (1-k)\hat{i} + (-1 - (-10))\hat{j} + (3-3)\hat{k} = (1-k)\hat{i} + 9\hat{j} + 0\hat{k}$.
$\overrightarrow{BC} = (3-1)\hat{i} + (5 - (-1))\hat{j} + (3-3)\hat{k} = 2\hat{i} + 6\hat{j} + 0\hat{k}$.
Since $\overrightarrow{AB} = \lambda \overrightarrow{BC}$,we have:
$(1-k)\hat{i} + 9\hat{j} = \lambda(2\hat{i} + 6\hat{j})$.
Comparing the components:
$1-k = 2\lambda$ --- $(1)$
$9 = 6\lambda$ --- $(2)$
From $(2)$,$\lambda = \frac{9}{6} = \frac{3}{2}$.
Substituting $\lambda = \frac{3}{2}$ in $(1)$:
$1-k = 2 \times \frac{3}{2} = 3$.
$1-k = 3 \implies k = 1-3 = -2$.
Thus,the value of $k$ is $-2$.

Given that the magnitude of the vector is $|\vec{r}| = 2 \sqrt{3}$.
Since $\vec{r}$ is equally inclined to the three axes,its direction cosines $l, m, n$ are equal,i.e.,$l = m = n$.
We know that the sum of the squares of direction cosines is $l^2 + m^2 + n^2 = 1$.
Substituting $l = m = n$,we get $l^2 + l^2 + l^2 = 1$,which implies $3l^2 = 1$.
Thus,$l^2 = \frac{1}{3}$,so $l = \pm \frac{1}{\sqrt{3}}$.
Since $l = m = n$,the unit vector $\hat{r}$ is given by $\hat{r} = \pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k}$.
Since $\vec{r} = |\vec{r}| \hat{r}$,we have $\vec{r} = 2 \sqrt{3} \left( \pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k} \right)$.
Therefore,$\vec{r} = \pm 2 \hat{i} \pm 2 \hat{j} \pm 2 \hat{k} = \pm 2(\hat{i} + \hat{j} + \hat{k})$.

(A) Let the direction ratios be $a=2k, b=3k, c=-6k$ for some constant $k$.
The magnitude of the vector is given by $|\vec{r}| = \sqrt{a^2 + b^2 + c^2} = 14$.
$\sqrt{(2k)^2 + (3k)^2 + (-6k)^2} = 14$
$\sqrt{4k^2 + 9k^2 + 36k^2} = 14$
$\sqrt{49k^2} = 14$
$7|k| = 14 \Rightarrow |k| = 2$.
Since $\vec{r}$ makes an acute angle with the $x$-axis,the direction cosine $l = \frac{a}{|\vec{r}|} = \frac{2k}{14} = \frac{k}{7}$ must be positive. Thus,$k=2$.
The direction cosines are $l = \frac{2(2)}{14} = \frac{4}{14} = \frac{2}{7}$,$m = \frac{3(2)}{14} = \frac{6}{14} = \frac{3}{7}$,and $n = \frac{-6(2)}{14} = \frac{-12}{14} = -\frac{6}{7}$.
The components of $\vec{r}$ are $a = 2(2) = 4$,$b = 3(2) = 6$,and $c = -6(2) = -12$.
Thus,$\vec{r} = 4\hat{i} + 6\hat{j} - 12\hat{k}$.

(A) The initial vector is $\vec{v} = \sqrt{3} \hat{i} + \hat{j}$. Its magnitude is $|\vec{v}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.
The angle of the initial vector with the positive $x$-axis is $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$.
Rotating this vector by $45^{\circ}$ counterclockwise gives a new vector with angle $\theta' = 30^{\circ} + 45^{\circ} = 75^{\circ}$.
The new vector is $\vec{v}' = |\vec{v}|(\cos 75^{\circ} \hat{i} + \sin 75^{\circ} \hat{j}) = 2(\cos 75^{\circ} \hat{i} + \sin 75^{\circ} \hat{j})$.
Thus,$\alpha = 2 \cos 75^{\circ}$ and $\beta = 2 \sin 75^{\circ}$.
The vertices of the triangle are $(\alpha, \beta), (0, \beta)$,and $(0,0)$. This is a right-angled triangle with base length $|\alpha| = \alpha$ and height $|\beta| = \beta$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \alpha \beta = \frac{1}{2} (2 \cos 75^{\circ}) (2 \sin 75^{\circ}) = 2 \sin 75^{\circ} \cos 75^{\circ} = \sin(2 \times 75^{\circ}) = \sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}$.

(D) Since $\overrightarrow{a}_{1}$ and $\overrightarrow{a}_{2}$ are collinear,their components must be proportional:
$\frac{x}{1} = \frac{-1}{y} = \frac{1}{z} = k$ (where $k$ is a constant).
From this,we get $x = k$,$y = -\frac{1}{k}$,and $z = \frac{1}{k}$.
Substituting these into the vector $x \hat{i} + y \hat{j} + z \hat{k}$,we get $k \hat{i} - \frac{1}{k} \hat{j} + \frac{1}{k} \hat{k}$.
For the vectors to be collinear,the ratio must hold. If we consider the simplest case where $k=1$,then $x=1, y=-1, z=1$.
The vector becomes $\hat{i} - \hat{j} + \hat{k}$.
The magnitude of this vector is $\sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The unit vector is $\frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.

(A) The points $A, B, C$ are collinear if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.
$\overrightarrow{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (\alpha-4)\hat{j} + (4-3)\hat{k} = \hat{i} + (\alpha-4)\hat{j} + \hat{k}$.
$\overrightarrow{AC} = \vec{c} - \vec{a} = (3-1)\hat{i} + (-2-4)\hat{j} + (5-3)\hat{k} = 2\hat{i} - 6\hat{j} + 2\hat{k}$.
For collinearity,the ratios of components must be equal: $\frac{1}{2} = \frac{\alpha-4}{-6} = \frac{1}{2}$.
Solving $\frac{\alpha-4}{-6} = \frac{1}{2}$ gives $\alpha-4 = -3$,so $\alpha = 1$.
Since the points are non-collinear for $\alpha \neq 1$,the smallest positive integer $\alpha$ for which they are non-collinear is $\alpha = 2$.
Now,find the midpoint $M$ of $BC$: $M = \frac{\vec{b} + \vec{c}}{2} = \frac{(2+3)\hat{i} + (2-2)\hat{j} + (4+5)\hat{k}}{2} = \frac{5}{2}\hat{i} + 0\hat{j} + \frac{9}{2}\hat{k}$.
The length of the median $AM$ is the magnitude of $\vec{M} - \vec{A} = (\frac{5}{2}-1)\hat{i} + (0-4)\hat{j} + (\frac{9}{2}-3)\hat{k} = \frac{3}{2}\hat{i} - 4\hat{j} + \frac{3}{2}\hat{k}$.
$AM = \sqrt{(\frac{3}{2})^2 + (-4)^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \sqrt{\frac{18}{4} + 16} = \sqrt{4.5 + 16} = \sqrt{20.5} = \sqrt{\frac{82}{4}} = \frac{\sqrt{82}}{2}$.