Basic , Modulus and Algebra of vectors Questions in English

(B) Given that $P$ divides the line segment $AB$ in the ratio $m : n$.
Using the section formula for vectors,the position vector of point $P$ with respect to the origin $O$ is given by:
$\overrightarrow{OP} = \frac{n\overrightarrow{OA} + m\overrightarrow{OB}}{m + n}$
Substituting the given values $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$:
$\overrightarrow{OP} = \frac{n\vec{a} + m\vec{b}}{m + n}$
Thus,the correct option is $(B)$.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Since $D, E, F$ are the midpoints of $BC, CA, AB$,their position vectors are:
$\vec{d} = \frac{\vec{b} + \vec{c}}{2}$,$\vec{e} = \frac{\vec{c} + \vec{a}}{2}$,$\vec{f} = \frac{\vec{a} + \vec{b}}{2}$.
Now,we calculate the vectors:
$\overrightarrow{AD} = \vec{d} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$
$\overrightarrow{BE} = \vec{e} - \vec{b} = \frac{\vec{c} + \vec{a}}{2} - \vec{b} = \frac{\vec{c} + \vec{a} - 2\vec{b}}{2}$
$\overrightarrow{CF} = \vec{f} - \vec{c} = \frac{\vec{a} + \vec{b}}{2} - \vec{c} = \frac{\vec{a} + \vec{b} - 2\vec{c}}{2}$
Adding these vectors:
$\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = \frac{(\vec{b} + \vec{c} - 2\vec{a}) + (\vec{c} + \vec{a} - 2\vec{b}) + (\vec{a} + \vec{b} - 2\vec{c})}{2}$
$= \frac{(\vec{a} - 2\vec{a} + \vec{a}) + (\vec{b} - 2\vec{b} + \vec{b}) + (\vec{c} - 2\vec{c} + \vec{c})}{2} = \frac{0}{2} = \vec{0}$.
Thus,the sum is a zero vector.

(D) Given that $\overrightarrow{AC} = 3\overrightarrow{AB}$.
Let $\vec{a}$ and $\vec{b}$ be the position vectors of points $A$ and $B$ respectively.
We know that $\overrightarrow{AB} = \vec{b} - \vec{a}$.
Substituting this into the given equation:
$\overrightarrow{AC} = 3(\vec{b} - \vec{a}) = 3\vec{b} - 3\vec{a}$.
Let $\vec{c}$ be the position vector of point $C$. Then $\overrightarrow{AC} = \vec{c} - \vec{a}$.
Equating the two expressions for $\overrightarrow{AC}$:
$\vec{c} - \vec{a} = 3\vec{b} - 3\vec{a}$.
$\vec{c} = 3\vec{b} - 3\vec{a} + \vec{a}$.
$\vec{c} = 3\vec{b} - 2\vec{a}$.
Thus,the position vector of point $C$ is $3\vec{b} - 2\vec{a}$.

(B) The position vector of the midpoint $M$ of a line segment joining points with position vectors $\vec{a}$ and $\vec{b}$ is given by the formula: $\vec{m} = \frac{\vec{a} + \vec{b}}{2}$.
Given $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 3\hat{k}$.
Adding the vectors: $\vec{a} + \vec{b} = (1+3)\hat{i} + (-1-1)\hat{j} + (2+3)\hat{k} = 4\hat{i} - 2\hat{j} + 5\hat{k}$.
Dividing by $2$: $\vec{m} = \frac{4\hat{i} - 2\hat{j} + 5\hat{k}}{2} = 2\hat{i} - \hat{j} + \frac{5}{2}\hat{k}$.
Thus,the correct option is $B$.

(B) In a parallelogram $ABCD$, the vectors representing opposite sides are equal, so $\overrightarrow{AB} = \overrightarrow{DC}$.
Let the position vector of $D$ be $\vec{d} = xi + yj + zk$.
The position vectors are given as $\vec{a} = i + 3j + 5k$, $\vec{b} = i + j + k$, and $\vec{c} = 7i + 7j + 7k$.
We know that $\overrightarrow{AB} = \vec{b} - \vec{a} = (1-1)i + (1-3)j + (1-5)k = 0i - 2j - 4k$.
Similarly, $\overrightarrow{DC} = \vec{c} - \vec{d} = (7-x)i + (7-y)j + (7-z)k$.
Equating $\overrightarrow{AB} = \overrightarrow{DC}$, we get:
$0 = 7 - x \Rightarrow x = 7$
$-2 = 7 - y \Rightarrow y = 9$
$-4 = 7 - z \Rightarrow z = 11$
Thus, the position vector of $D$ is $7i + 9j + 11k$.

(D) We know that in a parallelogram,the diagonals bisect each other. Therefore,$P$ is the midpoint of both diagonals $AC$ and $BD$.
Using the section formula for the midpoint,for diagonal $AC$ with midpoint $P$,we have:
$\overrightarrow{OA} + \overrightarrow{OC} = 2\overrightarrow{OP}$ ......$(i)$
Similarly,for diagonal $BD$ with midpoint $P$,we have:
$\overrightarrow{OB} + \overrightarrow{OD} = 2\overrightarrow{OP}$ ......$(ii)$
Adding equations $(i)$ and $(ii)$,we get:
$(\overrightarrow{OA} + \overrightarrow{OC}) + (\overrightarrow{OB} + \overrightarrow{OD}) = 2\overrightarrow{OP} + 2\overrightarrow{OP}$
$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 4\overrightarrow{OP}$

(A) Let the position vector of point $P$ be $\vec{p} = xi + yj + zk.$
Given that the position vectors of $A, B, C$ are $\vec{a} = i, \vec{b} = j, \vec{c} = k.$
The vector $\overrightarrow{AB} = \vec{b} - \vec{a} = j - i.$
The vector $\overrightarrow{CP} = \vec{p} - \vec{c} = xi + yj + (z - 1)k.$
Given $\overrightarrow{AB} = \overrightarrow{CP},$ we have $j - i = xi + yj + (z - 1)k.$
By comparing the coefficients of $i, j,$ and $k$ on both sides,we get:
$x = -1, y = 1, z - 1 = 0 \Rightarrow z = 1.$
Thus,the position vector of $P$ is $-i + j + k.$

(B) Given position vectors: $\vec{a} = 2i + 3j + 5k$,$\vec{b} = i + 2j + 3k$,$\vec{c} = -5i + 4j - 2k$,$\vec{d} = i + 10j + 10k$.
Calculate vector $\overrightarrow{AB} = \vec{b} - \vec{a} = (1-2)i + (2-3)j + (3-5)k = -i - j - 2k$.
Calculate vector $\overrightarrow{CD} = \vec{d} - \vec{c} = (1 - (-5))i + (10-4)j + (10 - (-2))k = 6i + 6j + 12k$.
Observe that $\overrightarrow{CD} = -6(-i - j - 2k) = -6 \overrightarrow{AB}$.
Since $\overrightarrow{CD}$ is a scalar multiple of $\overrightarrow{AB}$,the vectors are parallel. Thus,$\overrightarrow{AB} \parallel \overrightarrow{CD}$.

(C) Let $\overrightarrow{OA} = 2\hat{i} + 3\hat{j} - \hat{k}$ be the position vector of one end $A$ and $\overrightarrow{OP} = 3\hat{i} + 3\hat{j} + 3\hat{k}$ be the position vector of the midpoint $P$ of the line segment $AB$.
Let $\overrightarrow{OB}$ be the position vector of the other end $B$.
Since $P$ is the midpoint of $AB$,we have the relation:
$\overrightarrow{OP} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$
Multiplying by $2$,we get:
$2\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{OB}$
$\overrightarrow{OB} = 2\overrightarrow{OP} - \overrightarrow{OA}$
Substituting the given vectors:
$\overrightarrow{OB} = 2(3\hat{i} + 3\hat{j} + 3\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k})$
$\overrightarrow{OB} = (6\hat{i} + 6\hat{j} + 6\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k})$
$\overrightarrow{OB} = (6-2)\hat{i} + (6-3)\hat{j} + (6-(-1))\hat{k}$
$\overrightarrow{OB} = 4\hat{i} + 3\hat{j} + 7\hat{k}$
Thus,the position vector of the other end is $4\hat{i} + 3\hat{j} + 7\hat{k}$.

(D) Let the position vectors of the vertices be $\vec{a}, \vec{b}, \vec{c}$ and $\vec{a'}, \vec{b'}, \vec{c'}$.
The centroids $G$ and $G'$ are given by $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$ and $\vec{g'} = \frac{\vec{a'} + \vec{b'} + \vec{c'}}{3}$.
Therefore,$\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$ and $\vec{a'} + \vec{b'} + \vec{c'} = 3\vec{g'}$.
We need to find $\overrightarrow{AA'} + \overrightarrow{BB'} + \overrightarrow{CC'} = (\vec{a'} - \vec{a}) + (\vec{b'} - \vec{b}) + (\vec{c'} - \vec{c})$.
Rearranging the terms,we get $(\vec{a'} + \vec{b'} + \vec{c'}) - (\vec{a} + \vec{b} + \vec{c}) = 3\vec{g'} - 3\vec{g} = 3(\vec{g'} - \vec{g})$.
Since $\vec{g'} - \vec{g} = \overrightarrow{GG'}$,the result is $3\overrightarrow{GG'}$.

(B) Let the origin be at the circumcentre $O$. Then the position vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = R$.
The position vector of the orthocentre $O'$ is given by $\vec{o'} = \vec{a} + \vec{b} + \vec{c}$.
We need to calculate $\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C}$.
Using the definition of position vectors relative to the origin $O$:
$\overrightarrow{O'A} = \vec{a} - \vec{o'} = \vec{a} - (\vec{a} + \vec{b} + \vec{c}) = -(\vec{b} + \vec{c})$
$\overrightarrow{O'B} = \vec{b} - \vec{o'} = \vec{b} - (\vec{a} + \vec{b} + \vec{c}) = -(\vec{a} + \vec{c})$
$\overrightarrow{O'C} = \vec{c} - \vec{o'} = \vec{c} - (\vec{a} + \vec{b} + \vec{c}) = -(\vec{a} + \vec{b})$
Summing these up:
$\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C} = -(\vec{b} + \vec{c} + \vec{a} + \vec{c} + \vec{a} + \vec{b}) = -2(\vec{a} + \vec{b} + \vec{c}) = -2\vec{o'}$.
Since $\vec{o'}$ is the vector $\overrightarrow{OO'}$,we have $\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C} = -2\overrightarrow{OO'} = 2\overrightarrow{O'O}$.

(A) In a regular hexagon $ABCDEF$,let the center be $O$.
Given $\overrightarrow{AB} = a$ and $\overrightarrow{BC} = b$.
In $\triangle ABC$,by triangle law of vector addition,$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = a + b$.
Since $O$ is the center,$\overrightarrow{AO} = \overrightarrow{OC} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}(a + b)$.
Also,$\overrightarrow{AF} = \overrightarrow{OC} = \frac{1}{2}(a + b)$ and $\overrightarrow{FE} = \overrightarrow{AB} = a$.
Now,in $\triangle AFE$,by triangle law of vector addition,$\overrightarrow{AE} = \overrightarrow{AF} + \overrightarrow{FE}$.
Substituting the values,$\overrightarrow{AE} = \frac{1}{2}(a + b) + a = \frac{1}{2}a + \frac{1}{2}b + a = \frac{3}{2}a + \frac{1}{2}b$.
Wait,let us re-evaluate using the property of the hexagon: $\overrightarrow{AE} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE}$.
Alternatively,$\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE}$.
Since $\overrightarrow{AD} = 2\overrightarrow{BC} = 2b$ and $\overrightarrow{DE} = -\overrightarrow{AB} = -a$,
Therefore,$\overrightarrow{AE} = 2b - a$.

(A) Given that the position vector of point $C$ with respect to $B$ is $\overrightarrow{BC} = i + j$.
Similarly,the position vector of $B$ with respect to $A$ is $\overrightarrow{AB} = i - j$.
Using the triangle law of vector addition,the position vector of $C$ with respect to $A$ is given by $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$.
Substituting the given values,we get $\overrightarrow{AC} = (i - j) + (i + j) = 2i$.
Therefore,the correct option is $A$.

(A) Let $\overrightarrow{OA} = 6b - 2a$ and $\overrightarrow{OP} = a - b$. Let $\overrightarrow{OB} = r$.
Since $P$ divides $AB$ in the ratio $1 : 2$,by the section formula,the position vector of $P$ is given by:
$\overrightarrow{OP} = \frac{1(\overrightarrow{OB}) + 2(\overrightarrow{OA})}{1 + 2}$
Substituting the given values:
$a - b = \frac{r + 2(6b - 2a)}{3}$
$3(a - b) = r + 12b - 4a$
$3a - 3b = r + 12b - 4a$
$r = 3a - 3b - 12b + 4a$
$r = 7a - 15b$
Thus,the position vector of $B$ is $7a - 15b$.

(B) The position vector of the midpoint $M$ of the line segment joining points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$ is given by the formula:
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$
Given $\vec{a} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} - 3\hat{k}$.
Substituting the values:
$\vec{m} = \frac{(\hat{i} + 3\hat{j} - \hat{k}) + (3\hat{i} - \hat{j} - 3\hat{k})}{2}$
$\vec{m} = \frac{(1+3)\hat{i} + (3-1)\hat{j} + (-1-3)\hat{k}}{2}$
$\vec{m} = \frac{4\hat{i} + 2\hat{j} - 4\hat{k}}{2}$
$\vec{m} = 2\hat{i} + \hat{j} - 2\hat{k}$
Thus,the correct option is $B$.

(B) Since $C$ is the midpoint of $AB$,we have $\overrightarrow{AC} + \overrightarrow{BC} = 0$,which implies $\overrightarrow{AC} = -\overrightarrow{BC}$.
Using the triangle law of vector addition in $\triangle PAC$ and $\triangle PBC$:
$\overrightarrow{PA} = \overrightarrow{PC} + \overrightarrow{CA}$
$\overrightarrow{PB} = \overrightarrow{PC} + \overrightarrow{CB}$
Adding these two equations:
$\overrightarrow{PA} + \overrightarrow{PB} = (\overrightarrow{PC} + \overrightarrow{CA}) + (\overrightarrow{PC} + \overrightarrow{CB})$
$\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC} + (\overrightarrow{CA} + \overrightarrow{CB})$
Since $C$ is the midpoint of $AB$,$\overrightarrow{CA} = -\overrightarrow{CB}$,so $\overrightarrow{CA} + \overrightarrow{CB} = 0$.
Therefore,$\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC}$.

(D) Given that $D$ is the mid-point of $AB$,so $\overrightarrow{AD} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}\vec{a}$.
Given that $E$ is the mid-point of $AC$,so $\overrightarrow{AE} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}\vec{b}$.
Using the triangle law of vector addition in $\triangle ADE$,we have $\overrightarrow{AD} + \overrightarrow{DE} = \overrightarrow{AE}$.
Therefore,$\overrightarrow{DE} = \overrightarrow{AE} - \overrightarrow{AD}$.
Substituting the values,$\overrightarrow{DE} = \frac{1}{2}\vec{b} - \frac{1}{2}\vec{a} = \frac{\vec{b}}{2} - \frac{\vec{a}}{2}$.

(B) By the triangle law of vector addition,we have $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.
Substituting the given vectors,we get $a + b = c$.
Rearranging the terms,we obtain $a + b - c = 0$.
Therefore,the correct option is $B$.

(C) Given $\overrightarrow{OA} = a$ and $\overrightarrow{OB} = b$.
Point $C$ lies on $OA$ such that $2AC = CO$. This implies $C$ divides $OA$ in the ratio $1:2$ from $O$ to $A$.
Using the section formula,$\overrightarrow{OC} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}a$.
Since $CD$ is parallel to $OB$ and $|\overrightarrow{CD}| = 3|\overrightarrow{OB}|$,we have $\overrightarrow{CD} = 3\overrightarrow{OB} = 3b$.
Now,$\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD} = \frac{2}{3}a + 3b$.
Finally,$\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = (\frac{2}{3}a + 3b) - a = 3b - \frac{1}{3}a$.

(A) Given the relation $2\overrightarrow{AC} = 3\overrightarrow{CB}$.
We express the vectors in terms of the origin $O$ as $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}$ and $\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC}$.
Substituting these into the given equation:
$2(\overrightarrow{OC} - \overrightarrow{OA}) = 3(\overrightarrow{OB} - \overrightarrow{OC})$
$2\overrightarrow{OC} - 2\overrightarrow{OA} = 3\overrightarrow{OB} - 3\overrightarrow{OC}$
Rearranging the terms to isolate $2\overrightarrow{OA} + 3\overrightarrow{OB}$:
$2\overrightarrow{OA} + 3\overrightarrow{OB} = 2\overrightarrow{OC} + 3\overrightarrow{OC}$
$2\overrightarrow{OA} + 3\overrightarrow{OB} = 5\overrightarrow{OC}$

(C) Given the equation: $\overrightarrow {AO} + \overrightarrow {OB} = \overrightarrow {BO} + \overrightarrow {OC}$.
By the triangle law of vector addition,$\overrightarrow {AO} + \overrightarrow {OB} = \overrightarrow {AB}$.
Similarly,$\overrightarrow {BO} + \overrightarrow {OC} = \overrightarrow {BC}$.
Therefore,$\overrightarrow {AB} = \overrightarrow {BC}$.
This implies that the magnitude of vector $\overrightarrow {AB}$ is equal to the magnitude of vector $\overrightarrow {BC}$,i.e.,$|\overrightarrow {AB}| = |\overrightarrow {BC}|$.
Since the lengths of the two sides $AB$ and $BC$ are equal,the points $A, B, C$ form an isosceles triangle.

(A) Let the vertices of the triangle be $A$,$B$,and $C$ with position vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ respectively.
The medians are vectors directed from the vertices to the midpoints of the opposite sides.
Let the midpoints of sides $BC$,$CA$,and $AB$ be $D$,$E$,and $F$ respectively.
The position vectors of the midpoints are $\vec{d} = \frac{\vec{b} + \vec{c}}{2}$,$\vec{e} = \frac{\vec{a} + \vec{c}}{2}$,and $\vec{f} = \frac{\vec{a} + \vec{b}}{2}$.
The vectors representing the medians are $\vec{AD} = \vec{d} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$.
Similarly,$\vec{BE} = \vec{e} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$ and $\vec{CF} = \vec{f} - \vec{c} = \frac{\vec{a} + \vec{b} - 2\vec{c}}{2}$.
The sum of these vectors is $\vec{AD} + \vec{BE} + \vec{CF} = \frac{(\vec{b} + \vec{c} - 2\vec{a}) + (\vec{a} + \vec{c} - 2\vec{b}) + (\vec{a} + \vec{b} - 2\vec{c})}{2}$.
Simplifying the numerator: $(\vec{b} + \vec{c} - 2\vec{a} + \vec{a} + \vec{c} - 2\vec{b} + \vec{a} + \vec{b} - 2\vec{c}) = (\vec{a} - \vec{a}) + (\vec{b} - \vec{b}) + (\vec{c} - \vec{c}) = 0$.
Thus,the sum is $\frac{0}{2} = 0$.

(B) Let the position vectors of the two points be $\vec{p} = 2a - 3b$ and $\vec{q} = 3a - 2b$.
The formula for the position vector of a point dividing the join of two points with position vectors $\vec{p}$ and $\vec{q}$ internally in the ratio $m : n$ is given by $\vec{r} = \frac{m\vec{q} + n\vec{p}}{m + n}$.
Here,$m = 2$,$n = 3$,$\vec{p} = 2a - 3b$,and $\vec{q} = 3a - 2b$.
Substituting these values into the formula:
$\vec{r} = \frac{2(3a - 2b) + 3(2a - 3b)}{2 + 3}$
$\vec{r} = \frac{6a - 4b + 6a - 9b}{5}$
$\vec{r} = \frac{12a - 13b}{5}$
$\vec{r} = \frac{12}{5}a - \frac{13}{5}b$.
Thus,the correct option is $B$.

(A) Given the position vectors of points $A, B, C$ are $\vec{a} = \hat{i}$,$\vec{b} = \hat{j}$,and $\vec{c} = \hat{k}$.
The vector $\vec{AB}$ is given by $\vec{b} - \vec{a} = \hat{j} - \hat{i}$.
Let the position vector of point $X$ be $\vec{x}$. Then $\vec{CX} = \vec{x} - \vec{c} = \vec{x} - \hat{k}$.
Given $\vec{AB} = \vec{CX}$,we have $\hat{j} - \hat{i} = \vec{x} - \hat{k}$.
Therefore,$\vec{x} = -\hat{i} + \hat{j} + \hat{k}$.

(A) The position vector of a point $C$ that divides the line segment joining points with position vectors $\vec{a}$ and $\vec{b}$ in the ratio $m : n$ is given by the section formula: $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$.
Here,the ratio is $m : n = 2 : 1$.
Substituting the values into the formula,we get:
$\vec{OC} = \frac{2(\vec{b}) + 1(\vec{a})}{2 + 1}$
$\vec{OC} = \frac{\vec{a} + 2\vec{b}}{3}$.

(A) The position vectors of the vertices $A, B$,and $C$ of the triangle are $a, b$,and $c$ respectively.
The position vector of the centroid $G$ of the triangle is given by $g = \frac{a + b + c}{3}$.
We need to find the sum of the vectors $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}$.
Using the definition of a vector between two points,$\overrightarrow{GA} = a - g$,$\overrightarrow{GB} = b - g$,and $\overrightarrow{GC} = c - g$.
Summing these,we get:
$\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = (a - g) + (b - g) + (c - g)$
$= (a + b + c) - 3g$
Substituting $g = \frac{a + b + c}{3}$:
$= (a + b + c) - 3 \left( \frac{a + b + c}{3} \right)$
$= (a + b + c) - (a + b + c) = 0$.

(C) The coordinates of the midpoint $C$ of the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ are given by $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Given $A(2, -1)$ and $B(-4, 3)$,the coordinates of $C$ are $\left( \frac{2 + (-4)}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{-2}{2}, \frac{2}{2} \right) = (-1, 1)$.
Since $O$ is the origin $(0, 0)$,the position vector $\overrightarrow{OC}$ is given by $(-1 - 0)i + (1 - 0)j = -i + j$.

(D) In a regular hexagon $ABCDEF$,let $O$ be the center of the hexagon. We can express the vectors in terms of the sides of the hexagon.
We know that $\overrightarrow {AD} = 2\overrightarrow {AO}$,$\overrightarrow {EB} = 2\overrightarrow {EO}$,and $\overrightarrow {FC} = 2\overrightarrow {FO}$.
Alternatively,using the properties of a regular hexagon:
$\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {AF} = 2\overrightarrow {BC} + 2\overrightarrow {AB}$ (since $\overrightarrow {CD} = \overrightarrow {AF} + \overrightarrow {AB}$ is not direct,let's use the center $O$).
Actually,$\overrightarrow {AD} = 2\overrightarrow {BC}$,$\overrightarrow {EB} = 2\overrightarrow {FA}$,and $\overrightarrow {FC} = 2\overrightarrow {AB}$.
Summing these: $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = 2\overrightarrow {BC} + 2\overrightarrow {FA} + 2\overrightarrow {AB}$.
$= 2(\overrightarrow {FA} + \overrightarrow {AB} + \overrightarrow {BC}) = 2(\overrightarrow {FC}) = 2(2\overrightarrow {AB}) = 4\overrightarrow {AB}$.

(C) Let the position vector of point $A$ be $\vec{OA} = a + 2b$.
Let the position vector of point $P$ be $\vec{OP} = a$.
Let the position vector of point $B$ be $\vec{OB} = x$.
Since $P$ divides $AB$ in the ratio $2:3$,by the section formula:
$\vec{OP} = \frac{2(\vec{OB}) + 3(\vec{OA})}{2 + 3}$
$a = \frac{2x + 3(a + 2b)}{5}$
$5a = 2x + 3a + 6b$
$2x = 5a - 3a - 6b$
$2x = 2a - 6b$
$x = a - 3b$
Therefore,the position vector of $B$ is $a - 3b$.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Since $D, E, F$ are midpoints of $AB, AC, BC$ respectively,their position vectors are:
$\vec{d} = \frac{\vec{a} + \vec{b}}{2}$,$\vec{e} = \frac{\vec{a} + \vec{c}}{2}$,$\vec{f} = \frac{\vec{b} + \vec{c}}{2}$.
Now,$\overrightarrow{BE} = \vec{e} - \vec{b} = \frac{\vec{a} + \vec{c}}{2} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$.
And $\overrightarrow{AF} = \vec{f} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$.
Adding these two vectors:
$\overrightarrow{BE} + \overrightarrow{AF} = \frac{\vec{a} + \vec{c} - 2\vec{b} + \vec{b} + \vec{c} - 2\vec{a}}{2} = \frac{2\vec{c} - \vec{a} - \vec{b}}{2} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$.
Since $\vec{d} = \frac{\vec{a} + \vec{b}}{2}$,we have $\overrightarrow{BE} + \overrightarrow{AF} = \vec{c} - \vec{d} = \overrightarrow{DC}$.

(D) The position vectors are $\vec{OA} = i + j$,$\vec{OB} = i - j$,and $\vec{OC} = a i + b j + c k$.
For the points $A, B, C$ to be collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.
Calculate $\vec{AB} = \vec{OB} - \vec{OA} = (i - j) - (i + j) = -2j$.
Calculate $\vec{BC} = \vec{OC} - \vec{OB} = (a i + b j + c k) - (i - j) = (a - 1)i + (b + 1)j + ck$.
Since $\vec{AB}$ and $\vec{BC}$ are collinear,there exists a scalar $k$ such that $\vec{AB} = k \vec{BC}$.
$-2j = k((a - 1)i + (b + 1)j + ck)$.
Comparing the coefficients of $i, j, k$ on both sides:
$k(a - 1) = 0$
$k(b + 1) = -2$
$kc = 0$
From $kc = 0$,since $k \neq 0$ (otherwise $\vec{AB} = 0$,which is not possible),we must have $c = 0$.
From $k(a - 1) = 0$,we get $a = 1$.
From $k(b + 1) = -2$,$b$ can be any arbitrary scalar as long as $k$ is adjusted accordingly.
Therefore,the condition for collinearity is $c = 0, a = 1$ and $b$ is an arbitrary scalar.

(D) Let the points be $P = a + b$,$Q = a - b$,and $R = a + kb$.
For the points to be collinear,the vectors $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ must be parallel.
$\overrightarrow{PQ} = Q - P = (a - b) - (a + b) = -2b$.
$\overrightarrow{QR} = R - Q = (a + kb) - (a - b) = (k + 1)b$.
For these vectors to be parallel,there must exist a scalar $\lambda$ such that $\overrightarrow{PQ} = \lambda \overrightarrow{QR}$.
$-2b = \lambda(k + 1)b$.
Since $b$ is a vector,this implies $-2 = \lambda(k + 1)$.
This equation holds for any $k$ as long as we can choose a suitable $\lambda$. Specifically,if $k = -1$,the points $Q$ and $R$ coincide,making them collinear. For any other $k$,the vectors are proportional. Thus,$k$ can be any real number.

(A) Given the position vectors of points $A, B, C$ are $\vec{OA} = a$,$\vec{OB} = b$,and $\vec{OC} = 3a - 2b$.
First,we find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \vec{OB} - \vec{OA} = b - a$
$\vec{AC} = \vec{OC} - \vec{OA} = (3a - 2b) - a = 2a - 2b = -2(b - a)$
We observe that $\vec{AC} = -2 \vec{AB}$.
Since $\vec{AC}$ is a scalar multiple of $\vec{AB}$,the vectors are parallel.
Because they share a common point $A$,the points $A, B, C$ must be collinear.

(A) By definition,a set of vectors is linearly independent if the only linear combination that results in the zero vector is the one where all scalars are zero.
Since $a, b, c$ are non-collinear vectors (and in the context of three-dimensional space,if they are linearly independent),the equation $xa + yb + zc = 0$ implies that $x = 0, y = 0, z = 0$.
If at least one scalar were non-zero,the vectors would be linearly dependent,which contradicts the property of non-collinear vectors in this context.

(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.
Since the points are collinear,the vector $\overrightarrow{AB}$ must be parallel to the vector $\overrightarrow{BC}$.
$\overrightarrow{AB} = (40 - 60)i + (-8 - 3)j = -20i - 11j$.
$\overrightarrow{BC} = (a - 40)i + (-52 - (-8))j = (a - 40)i - 44j$.
For collinearity,$\overrightarrow{AB} = k\overrightarrow{BC}$ for some scalar $k$.
$-20i - 11j = k((a - 40)i - 44j)$.
Comparing the coefficients of $j$: $-11 = -44k$,which gives $k = \frac{11}{44} = \frac{1}{4}$.
Comparing the coefficients of $i$: $-20 = k(a - 40)$.
Substituting $k = \frac{1}{4}$: $-20 = \frac{1}{4}(a - 40)$.
$-80 = a - 40$.
$a = -80 + 40 = -40$.

(C) The position vector of point $A$ is given by $\overrightarrow{OA} = 4\,i + 5\,j$.
To find the unit vector parallel to $\overrightarrow{OA}$,we use the formula $\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|}$.
First,calculate the magnitude of $\overrightarrow{OA}$:
$|\overrightarrow{OA}| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$.
Now,the unit vector is:
$\hat{OA} = \frac{4\,i + 5\,j}{\sqrt{41}} = \frac{1}{\sqrt{41}}(4\,i + 5\,j)$.
Thus,the correct option is $C$.

(C) Let the points be $A(10, 3)$,$B(12, -5)$,and $C(a, 11)$.
Since the points are collinear,the vector $\overrightarrow{AB}$ must be parallel to the vector $\overrightarrow{BC}$.
$\overrightarrow{AB} = (12 - 10)i + (-5 - 3)j = 2i - 8j$.
$\overrightarrow{BC} = (a - 12)i + (11 - (-5))j = (a - 12)i + 16j$.
For collinearity,$\overrightarrow{AB} = k \overrightarrow{BC}$ for some scalar $k$.
$2i - 8j = k[(a - 12)i + 16j]$.
Comparing the coefficients of $j$: $-8 = 16k \Rightarrow k = -\frac{1}{2}$.
Comparing the coefficients of $i$: $2 = k(a - 12)$.
Substituting $k = -\frac{1}{2}$: $2 = -\frac{1}{2}(a - 12)$.
$-4 = a - 12 \Rightarrow a = 8$.

(D) Let the points be $A$,$B$,and $C$ with position vectors $\vec{OA} = a + b$,$\vec{OB} = a - b$,and $\vec{OC} = a + kb$.
For the points to be collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel,meaning $\vec{AB} = \lambda \vec{BC}$ for some scalar $\lambda$.
Calculate $\vec{AB} = \vec{OB} - \vec{OA} = (a - b) - (a + b) = -2b$.
Calculate $\vec{BC} = \vec{OC} - \vec{OB} = (a + kb) - (a - b) = (k + 1)b$.
For collinearity,$-2b = \lambda(k + 1)b$.
This implies $-2 = \lambda(k + 1)$. Since $\lambda$ can be any non-zero real number,$k$ can take any real value except for the case where the points coincide. However,in the context of vector collinearity,the condition holds for any $k \in \mathbb{R}$ as long as the vectors are linearly dependent.
Thus,the points are collinear for every real number $k$.

(B) Given position vectors are $\vec{A} = 2i + j,$ $\vec{B} = i - 3j,$ $\vec{C} = 3i + 2j,$ and $\vec{D} = i + \lambda j.$
First,calculate the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$:
$\overrightarrow{AB} = \vec{B} - \vec{A} = (i - 3j) - (2i + j) = -i - 4j.$
$\overrightarrow{CD} = \vec{D} - \vec{C} = (i + \lambda j) - (3i + 2j) = -2i + (\lambda - 2)j.$
Since $\overrightarrow{AB} \parallel \overrightarrow{CD},$ their components must be proportional:
$\frac{-1}{-2} = \frac{-4}{\lambda - 2}.$
Solving for $\lambda$:
$\frac{1}{2} = \frac{-4}{\lambda - 2}$
$\lambda - 2 = 2 \times (-4)$
$\lambda - 2 = -8$
$\lambda = -6.$

(A) Two vectors $\vec{a} = a_1\,i + a_2\,j + a_3\,k$ and $\vec{b} = b_1\,i + b_2\,j + b_3\,k$ are parallel if their components are proportional,i.e.,$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$.
Given vectors are $3\,i + 2\,j - k$ and $6\,i - 4xj + yk$.
Comparing the components,we have $\frac{3}{6} = \frac{2}{-4x} = \frac{-1}{y}$.
From $\frac{3}{6} = \frac{2}{-4x}$,we get $\frac{1}{2} = \frac{1}{-2x}$,which implies $-2x = 2$,so $x = -1$.
From $\frac{3}{6} = \frac{-1}{y}$,we get $\frac{1}{2} = \frac{-1}{y}$,which implies $y = -2$.
Therefore,the values are $x = -1$ and $y = -2$.

(B) The vectors $a$,$b$,and $a + b$ are coplanar.
This is because any vector $v = x(a) + y(b)$ is a linear combination of vectors $a$ and $b$,which implies that $v$ lies in the same plane as $a$ and $b$.
Here,$a + b = 1(a) + 1(b)$.
Since $a + b$ can be expressed as a linear combination of $a$ and $b$ with scalars $1$ and $1$,the three vectors are coplanar.

(D) Let the points be $A, B, C$ with position vectors $a, b, c$ respectively.
Since the points are collinear,the vectors $\vec{AB}$ and $\vec{AC}$ are parallel.
Thus,$\vec{AB} = k \vec{AC}$ for some scalar $k$.
This implies $(b - a) = k(c - a)$.
Rearranging the terms,we get $b - a = kc - ka$,which simplifies to $a(k - 1) + b - kc = 0$.
Let $x = k - 1$,$y = 1$,and $z = -k$.
Then $xa + yb + zc = 0$.
Calculating the sum of the scalars: $x + y + z = (k - 1) + 1 + (-k) = 0$.
Therefore,for three collinear points,there exist scalars $x, y, z$ (not all zero) such that $xa + yb + zc = 0$ and $x + y + z = 0$.

(C) Given vectors are $a = (2, 5)$ and $b = (1, 4).$
First,calculate the sum of the vectors: $a + b = (2 + 1, 5 + 4) = (3, 9).$
We can factor out $3$ from the resulting vector: $(3, 9) = 3(1, 3).$
$A$ vector $v$ is parallel to $u$ if $v = k \cdot u$ for some scalar $k \neq 0.$
Here,$(3, 9) = 3 \cdot (1, 3).$
Therefore,the vector $(a + b)$ is parallel to $(1, 3).$

(C) Since vectors $c = (x - 2)a + b$ and $d = (2x + 1)a - b$ are collinear,there exists a scalar $\lambda$ such that $c = \lambda d$.
Substituting the expressions for $c$ and $d$:
$(x - 2)a + b = \lambda ((2x + 1)a - b)$
Rearranging the terms:
$(x - 2)a + b = \lambda(2x + 1)a - \lambda b$
$((x - 2) - \lambda(2x + 1))a + (1 + \lambda)b = 0$
Since $a$ and $b$ are non-collinear,they are linearly independent. Therefore,the coefficients of $a$ and $b$ must be zero:
$1 + \lambda = 0 \Rightarrow \lambda = -1$
$(x - 2) - \lambda(2x + 1) = 0$
Substituting $\lambda = -1$ into the second equation:
$(x - 2) - (-1)(2x + 1) = 0$
$x - 2 + 2x + 1 = 0$
$3x - 1 = 0$
$x = \frac{1}{3}$

(A) Given position vectors are:
$\vec{p} = 2a + 4c$
$\vec{q} = 5a + 3\sqrt{3}b + 4c$
$\vec{r} = -2\sqrt{3}b + c$
$\vec{s} = 2a + c$
Calculate $\overrightarrow{PQ} = \vec{q} - \vec{p} = (5a + 3\sqrt{3}b + 4c) - (2a + 4c) = 3a + 3\sqrt{3}b = 3(a + \sqrt{3}b)$.
Calculate $\overrightarrow{RS} = \vec{s} - \vec{r} = (2a + c) - (-2\sqrt{3}b + c) = 2a + 2\sqrt{3}b = 2(a + \sqrt{3}b)$.
Since $\overrightarrow{PQ} = \frac{3}{2} \overrightarrow{RS}$,the vectors are scalar multiples of each other.
Therefore,$\overrightarrow{PQ}$ is parallel to $\overrightarrow{RS}$.

(C) Two vectors $a = (a_1, a_2)$ and $b = (b_1, b_2)$ are collinear if their components are proportional,i.e.,$\frac{b_1}{a_1} = \frac{b_2}{a_2} = \lambda$.
Given $a = (1, -1)$ and $b = (-2, m)$.
Comparing the components,we have $\frac{-2}{1} = \frac{m}{-1}$.
$-2 = \frac{m}{-1}$.
$m = (-2) \times (-1) = 2$.
Thus,the value of $m$ is $2$.

(B) Let the point $B$ divide the line segment $AC$ in the ratio $\lambda : 1$.
Using the section formula,the position vector of $B$ is given by:
$\vec{B} = \frac{\lambda \vec{C} + 1 \vec{A}}{\lambda + 1}$
Substituting the given position vectors:
$5i - 2k = \frac{\lambda (11i + 3j + 7k) + (i - 2j - 8k)}{\lambda + 1}$
Equating the coefficients of $j$:
$0 = \frac{3\lambda - 2}{\lambda + 1}$
$3\lambda - 2 = 0$
$\lambda = \frac{2}{3}$
Thus,the ratio in which $B$ divides $AC$ is $2:3$.

(C) Given that $a$ and $b$ are two non-collinear vectors.
We are given the equation $xa + yb = 0$.
If $x \neq 0$,then we can write $xa = -yb$,which implies $a = -(\frac{y}{x})b$.
This means that vector $a$ is a scalar multiple of vector $b$,which implies that $a$ and $b$ are collinear.
However,it is given that $a$ and $b$ are non-collinear,which is a contradiction.
Therefore,our assumption that $x \neq 0$ must be false,so $x = 0$.
Substituting $x = 0$ into the equation $xa + yb = 0$,we get $0a + yb = 0$,which simplifies to $yb = 0$.
Since $b$ is a non-zero vector,it follows that $y = 0$.
Thus,for non-collinear vectors,$xa + yb = 0$ implies $x = 0$ and $y = 0$.

(C) By definition,any linear combination of two vectors $a$ and $b$,given by $xa + yb$ where $x$ and $y$ are scalars,lies in the plane spanned by $a$ and $b$.
Since $a$ and $b$ are non-collinear,they define a unique plane.
Therefore,the vector $xa + yb$ is coplanar with $a$ and $b$.

(A) Given that $a + b + c = \alpha d$ and $b + c + d = \beta a$.
Adding $d$ to the first equation,we get $a + b + c + d = (\alpha + 1)d$.
Adding $a$ to the second equation,we get $a + b + c + d = (\beta + 1)a$.
Equating the two expressions for $a + b + c + d$,we have $(\alpha + 1)d = (\beta + 1)a$.
If $\alpha \neq -1$,then $d = \frac{\beta + 1}{\alpha + 1} a$.
Substituting this into the first equation: $a + b + c = \alpha \left( \frac{\beta + 1}{\alpha + 1} \right) a$.
This implies that $a, b, c$ are coplanar,which contradicts the given condition that they are non-coplanar.
Therefore,we must have $\alpha + 1 = 0$,which implies $\alpha = -1$.
Consequently,$a + b + c + d = (\alpha + 1)d = (-1 + 1)d = 0$.