Basic , Modulus and Algebra of vectors Questions in English

(A) Two vectors $\vec{a}$ and $\vec{b}$ are collinear if $\vec{a} = m\vec{b}$ for some non-zero scalar $m$.
Given $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = -2\hat{i} + k\hat{j}$.
Substituting the vectors,we get $\hat{i} - \hat{j} = m(-2\hat{i} + k\hat{j})$.
Equating the coefficients of $\hat{i}$ and $\hat{j}$ on both sides:
For $\hat{i}$: $1 = -2m \implies m = -1/2$.
For $\hat{j}$: $-1 = mk$.
Substituting $m = -1/2$ into the second equation: $-1 = (-1/2)k$.
Solving for $k$,we get $k = 2$.

(A) Let the vector $a$ be represented in terms of its components as $a = x i + y j + z k$.
Then,the dot product of $a$ with the unit vectors $i, j, k$ are:
$a \cdot i = (x i + y j + z k) \cdot i = x$
$a \cdot j = (x i + y j + z k) \cdot j = y$
$a \cdot k = (x i + y j + z k) \cdot k = z$
Substituting these values into the given expression:
$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = x i + y j + z k$
Since $x i + y j + z k = a$,the expression simplifies to $a$.

(D) Let $r = xi + yj + zk$.
Given $r \cdot i = r \cdot j = r \cdot k$, we have $x = y = z$.
Let $x = y = z = a$. Then $r = a(i + j + k)$.
Given $|r| = 3$, we have $\sqrt{a^2 + a^2 + a^2} = 3$.
$\sqrt{3a^2} = 3 \Rightarrow |a|\sqrt{3} = 3$.
$|a| = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Thus, $a = \pm \sqrt{3}$.
Therefore, $r = \pm \sqrt{3}(i + j + k)$.

(D) Let $\vec{v} = \hat{i} + \hat{j} + \hat{k}$. The magnitude is $|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
The angle $\alpha$ between $\vec{v}$ and $\hat{i}$ is given by $\cos \alpha = \frac{\vec{v} \cdot \hat{i}}{|\vec{v}| |\hat{i}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i}}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$. Thus,$\alpha = \cos^{-1}(\frac{1}{\sqrt{3}})$.
The angle $\beta$ between $\vec{v}$ and $\hat{j}$ is given by $\cos \beta = \frac{\vec{v} \cdot \hat{j}}{|\vec{v}| |\hat{j}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{j}}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$. Thus,$\beta = \cos^{-1}(\frac{1}{\sqrt{3}})$.
The angle $\gamma$ between $\vec{v}$ and $\hat{k}$ is given by $\cos \gamma = \frac{\vec{v} \cdot \hat{k}}{|\vec{v}| |\hat{k}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{k}}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$. Thus,$\gamma = \cos^{-1}(\frac{1}{\sqrt{3}})$.
Therefore,$\alpha = \beta = \gamma$.

(B) Let the vector $r$ be represented as $r = xi + yj + zk$.
Then, the dot products are:
$r \cdot i = x$
$r \cdot j = y$
$r \cdot k = z$
Substituting these into the given expression:
$(r \cdot i)^2 + (r \cdot j)^2 + (r \cdot k)^2 = x^2 + y^2 + z^2$
Since the magnitude of vector $r$ is $|r| = \sqrt{x^2 + y^2 + z^2}$, it follows that $|r|^2 = x^2 + y^2 + z^2 = r^2$.
Therefore, $(r \cdot i)^2 + (r \cdot j)^2 + (r \cdot k)^2 = r^2$.

(B) Let the unit vector be $\vec{r} = y\hat{j} + z\hat{k}$. Since it is a unit vector,$|\vec{r}| = \sqrt{y^2 + z^2} = 1$.
The angle $\alpha$ between $\vec{r}$ and the $y$-axis $(\hat{j})$ is given by $\cos(30^\circ) = \frac{\vec{r} \cdot \hat{j}}{|\vec{r}| |\hat{j}|} = \frac{y}{1 \times 1} = y$.
Thus,$y = \cos(30^\circ) = \frac{\sqrt{3}}{2}$.
The angle $\beta$ between $\vec{r}$ and the $z$-axis $(\hat{k})$ is given by $\cos(60^\circ) = \frac{\vec{r} \cdot \hat{k}}{|\vec{r}| |\hat{k}|} = \frac{z}{1 \times 1} = z$.
Thus,$z = \cos(60^\circ) = \frac{1}{2}$.
Since the vector lies in the $yz$-plane,the $x$-component is $0$.
Therefore,the components are $(0, \frac{\sqrt{3}}{2}, \frac{1}{2})$.

(A) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.
Then,the dot products are:
$\vec{a} \cdot \hat{i} = a_1$
$\vec{a} \cdot \hat{j} = a_2$
$\vec{a} \cdot \hat{k} = a_3$
Substituting these values back into the expression for $\vec{a}$,we get:
$\vec{a} = (\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k}$.

(C) In a rhombus,all sides are equal in length. Therefore,the magnitudes of the adjacent side vectors $a$ and $b$ are equal,i.e.,$|a| = |b|$.
For any vector $v$,the dot product $v \cdot v = |v||v| \cos(0^\circ) = |v|^2$.
Applying this to vectors $a$ and $b$:
$a \cdot a = |a|^2$
$b \cdot b = |b|^2$
Since $|a| = |b|$,it follows that $|a|^2 = |b|^2$.
Therefore,$a \cdot a = b \cdot b$.

(B) By definition,the dot product of a unit vector with itself is given by $\hat{a} \cdot \hat{a} = |\hat{a}| |\hat{a}| \cos(0^\circ) = (1)(1)(1) = 1$.
Since $\hat{i}$ is a unit vector,$\hat{i} \cdot \hat{i} = 1$.
Option $(a)$ is incorrect because $\hat{i} \cdot \hat{j} = 0$ as they are orthogonal.
Option $(c)$ is incorrect because the cross product of two vectors results in a vector,not a scalar.
Option $(d)$ is incorrect because $\hat{i} \times (\hat{j} \times \hat{k}) = \hat{i} \times \hat{i} = \vec{0}$.

(D) Given position vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 5\hat{j}$,$\vec{c} = 3\hat{i} + 2\hat{j} - 3\hat{k}$,$\vec{d} = \hat{i} - 6\hat{j} - \hat{k}$.
Calculate $\overrightarrow{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (5-1)\hat{j} + (0-1)\hat{k} = \hat{i} + 4\hat{j} - \hat{k}$.
Calculate $\overrightarrow{CD} = \vec{d} - \vec{c} = (1-3)\hat{i} + (-6-2)\hat{j} + (-1 - (-3))\hat{k} = -2\hat{i} - 8\hat{j} + 2\hat{k}$.
Observe that $\overrightarrow{CD} = -2(\hat{i} + 4\hat{j} - \hat{k}) = -2\overrightarrow{AB}$.
Since $\overrightarrow{CD}$ is a negative scalar multiple of $\overrightarrow{AB}$,the vectors are anti-parallel.
The angle $\theta$ between two anti-parallel vectors is $\pi$ radians.
Alternatively,using the dot product formula: $\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|} = \frac{(1)(-2) + (4)(-8) + (-1)(2)}{\sqrt{1^2+4^2+(-1)^2} \sqrt{(-2)^2+(-8)^2+2^2}} = \frac{-2 - 32 - 2}{\sqrt{18} \sqrt{72}} = \frac{-36}{\sqrt{18} \cdot 2\sqrt{18}} = \frac{-36}{2 \cdot 18} = -1$.
Since $\cos \theta = -1$,$\theta = \pi$.

(A) Let $\theta$ be the angle between vectors $\vec{a}$ and $\vec{b}$,so $\theta = 30^o$.
We need to find the angle between $3\vec{a}$ and $-4\vec{b}$.
Since $3 > 0$,the vector $3\vec{a}$ is in the same direction as $\vec{a}$.
Since $-4 < 0$,the vector $-4\vec{b}$ is in the opposite direction to $\vec{b}$.
The angle between $\vec{a}$ and $\vec{b}$ is $30^o$.
The angle between $\vec{a}$ and $-\vec{b}$ is $180^o - 30^o = 150^o$.
Since $3\vec{a}$ is in the direction of $\vec{a}$ and $-4\vec{b}$ is in the direction of $-\vec{b}$,the angle between $3\vec{a}$ and $-4\vec{b}$ is $150^o$.

(B) Let the two unit vectors be $\vec{a}$ and $\vec{b}$,such that $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
Given that their sum is a unit vector,we have $|\vec{a} + \vec{b}| = 1$.
Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = 1^2$.
Using the identity $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$,we have $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1$,which implies $2(\vec{a} \cdot \vec{b}) = -1$,so $\vec{a} \cdot \vec{b} = -\frac{1}{2}$.
Now,we need to find the magnitude of their difference,$|\vec{a} - \vec{b}|$.
We know that $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
Substituting the known values,$|\vec{a} - \vec{b}|^2 = 1 + 1 - 2(-\frac{1}{2}) = 1 + 1 + 1 = 3$.
Therefore,$|\vec{a} - \vec{b}| = \sqrt{3}$.

(D) Two vectors $a = a_1 i + a_2 j$ and $b = b_1 i + b_2 j$ are parallel if their components are proportional,i.e.,$\frac{a_1}{b_1} = \frac{a_2}{b_2}$.
Given $a = 1i - 2j$ and $b = 2i + \lambda j$.
Comparing the components,we have $a_1 = 1, a_2 = -2$ and $b_1 = 2, b_2 = \lambda$.
Since the vectors are parallel,$\frac{1}{2} = \frac{-2}{\lambda}$.
Cross-multiplying gives $\lambda = 2 \times (-2) = -4$.

(A) Let the vector be $\vec{A}$ with magnitude $|\vec{A}| = 14$.
Given that the vector makes an angle $\theta = 60^\circ$ with the $x-$ axis.
The component of the vector along the $x-$ axis is given by $A_x = |\vec{A}| \cos \theta = 14 \cos 60^\circ$.
Since $\cos 60^\circ = 1/2$,we have $A_x = 14 \times (1/2) = 7$.
The component of the vector along the $y-$ axis is given by $A_y = |\vec{A}| \sin \theta = 14 \sin 60^\circ$.
Since $\sin 60^\circ = \sqrt{3}/2$,we have $A_y = 14 \times (\sqrt{3}/2) = 7\sqrt{3}$.
Thus,the components are $7$ and $7\sqrt{3}$.

(B) Given that $b = 3j + 4k$ and $b_1 = \frac{3}{2}i + \frac{3}{2}j$ is the component of $b$ parallel to $a$.
Since $b = b_1 + b_2$,where $b_2$ is the component of $b$ perpendicular to $a$,we can write $b_2 = b - b_1$.
Substituting the given vectors:
$b_2 = (0i + 3j + 4k) - (\frac{3}{2}i + \frac{3}{2}j + 0k)$
$b_2 = (0 - \frac{3}{2})i + (3 - \frac{3}{2})j + (4 - 0)k$
$b_2 = -\frac{3}{2}i + \frac{3}{2}j + 4k$.
Thus,the correct option is $B$.

(A) The projection of a vector $\vec{a}$ along a vector $\vec{b}$ is given by the formula $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Let $\vec{a} = i + j + k$ and $\vec{b} = j$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (i + j + k) \cdot j = (1)(0) + (1)(1) + (1)(0) = 1$.
Next,calculate the magnitude of vector $\vec{b}$,which is $|\vec{b}| = |j| = 1$.
Therefore,the projection is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{1}{1} = 1$.

(D) The dot product of vectors follows the distributive property over vector subtraction.
For any vectors $a, b,$ and $c$,the property is given by $a \cdot (b - c) = a \cdot b - a \cdot c$.
Option $A$ is false because the vector triple product is not associative.
Option $B$ is false because the cross product is anticommutative,i.e.,$a \times b = -(b \times a)$.
Option $C$ is mathematically undefined because the dot product results in a scalar,and the cross product of two scalars is not defined.
Therefore,the correct statement is $a \cdot (b - c) = a \cdot b - a \cdot c$.

(C) Given that $a \cdot b = 0$. This implies that either $a = 0$,$b = 0$,or $a \perp b$.
Also,given that $a \times b = 0$. This implies that either $a = 0$,$b = 0$,or $a \parallel b$.
For both conditions to be satisfied simultaneously,$a$ and $b$ cannot be both perpendicular and parallel unless at least one of them is a null vector (zero vector).
Therefore,either $a = 0$ or $b = 0$.

(A) The vector product (cross product) of two vectors $u$ and $v$ is anti-commutative,meaning $u \times v = -(v \times u)$.
Therefore,the property $u \times v = v \times u$ is incorrect.
Option $A$ is not a property of vectors.

(D) Let $l, m, n$ be the direction cosines of vector $r$. Since the vector is equally inclined with the coordinate axes,we have $l = m = n$.
We know that $l^2 + m^2 + n^2 = 1$. Substituting $l = m = n$,we get $3l^2 = 1$,which implies $l^2 = \frac{1}{3}$.
Since the tip of $r$ is in the positive octant,$l, m, n$ must be positive. Thus,$l = m = n = \frac{1}{\sqrt{3}}$.
The vector $r$ is given by $r = |r|(li + mj + nk)$.
Given $|r| = 6$,we have $r = 6 \left( \frac{1}{\sqrt{3}}i + \frac{1}{\sqrt{3}}j + \frac{1}{\sqrt{3}}k \right)$.
Simplifying this,$r = \frac{6}{\sqrt{3}}(i + j + k) = 2\sqrt{3}(i + j + k)$.

(B) The equation of a line passing through the points $A(1, -1, 2)$ and $B(3, 1, 1)$ is given by $r = a + \lambda b$,where $a = i - j + 2k$ and the direction vector $b = (3-1)i + (1-(-1))j + (1-2)k = 2i + 2j - k$.
However,the problem states the line passes through $i - j + 2k$ and $3i + j + k$. The direction vector is $v = (3-1)i + (1-(-1))j + (1-2)k = 2i + 2j - k$. The magnitude $|v| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.
The position vector of any point $P$ on the line is $r = (i - j + 2k) + \lambda (2i + 2j - k)$.
The distance from $A(i - j + 2k)$ is $|\lambda v| = |\lambda| |v| = 3|\lambda|$.
Given the distance is $3\sqrt{11}$,we have $3|\lambda| = 3\sqrt{11}$,so $|\lambda| = \sqrt{11}$.
If $\lambda = \sqrt{11}$,$P = i - j + 2k + \sqrt{11}(2i + 2j - k) = (1+2\sqrt{11})i + (-1+2\sqrt{11})j + (2-\sqrt{11})k$.
Re-evaluating the direction vector based on the provided options: if the direction vector was $2i + 2j - k$ (magnitude $3$),the distance calculation leads to the provided options. Checking the vector $B-A = 2i + 2j - k$. If the line is $r = (i - j + 2k) + \lambda (2i + 2j - k)$,the distance is $3|\lambda|$.
Given the options,let's check point $P = -8i - 4j - k$. Vector $AP = (-8-1)i + (-4-(-1))j + (-1-2)k = -9i - 3j - 3k$. Magnitude $|AP| = \sqrt{81+9+9} = \sqrt{99} = 3\sqrt{11}$. This matches the distance. Thus,$-8i - 4j - k$ is a correct position vector.

(C) Let the cube be placed in the coordinate system such that its edges lie along the $x, y,$ and $z$ axes. The diagonals of the three adjacent faces meeting at the origin are represented by the vectors $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c} = \hat{k} + \hat{i}$.
The unit vectors along these directions are $\hat{u}_1 = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$,$\hat{u}_2 = \frac{\hat{j} + \hat{k}}{\sqrt{2}}$,and $\hat{u}_3 = \frac{\hat{k} + \hat{i}}{\sqrt{2}}$.
The forces are given as $\vec{F}_1 = 1 \cdot \hat{u}_1$,$\vec{F}_2 = 2 \cdot \hat{u}_2$,and $\vec{F}_3 = 3 \cdot \hat{u}_3$.
The resultant force $\vec{R} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \frac{1(\hat{i} + \hat{j}) + 2(\hat{j} + \hat{k}) + 3(\hat{k} + \hat{i})}{\sqrt{2}} = \frac{(1+3)\hat{i} + (1+2)\hat{j} + (2+3)\hat{k}}{\sqrt{2}} = \frac{4\hat{i} + 3\hat{j} + 5\hat{k}}{\sqrt{2}}$.
The magnitude of the resultant force is $|\vec{R}| = \frac{\sqrt{4^2 + 3^2 + 5^2}}{\sqrt{2}} = \frac{\sqrt{16 + 9 + 25}}{\sqrt{2}} = \frac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5 \text{ dynes}$.

(B) Given that vector $b$ is in the north-east direction and vector $c$ is in the north-west direction.
Since the angle between north-east and north-west is $90^\circ$,the vectors $b$ and $c$ are perpendicular,so $b \cdot c = 0$.
Given $|b| = |c| = 4$.
We need to find the magnitude and direction of $d = c - b$.
The magnitude is given by $|d|^2 = |c - b|^2 = |c|^2 + |b|^2 - 2(b \cdot c)$.
Substituting the values: $|d|^2 = 4^2 + 4^2 - 0 = 16 + 16 = 32$.
Thus,$|d| = \sqrt{32} = 4\sqrt{2}$.
Regarding the direction,since $c$ is at $135^\circ$ and $b$ is at $45^\circ$ from the positive $x$-axis (east),the vector $d = c - b$ points in the direction of the vector difference,which is towards the west.

(D) Let the two given vertices be $\vec{A} = \hat{i} - \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$.
For any point $\vec{C}$ to be the third vertex of a triangle,the points $\vec{A}$,$\vec{B}$,and $\vec{C}$ must not be collinear.
Three points are collinear if the vectors $\vec{AB}$ and $\vec{AC}$ are parallel.
Here,$\vec{AB} = \vec{B} - \vec{A} = (\hat{j} + \hat{k}) - (\hat{i} - \hat{j}) = -\hat{i} + 2\hat{j} + \hat{k}$.
Checking option $(a)$: $\vec{C} = \hat{i} + \hat{k}$,then $\vec{AC} = (\hat{i} + \hat{k}) - (\hat{i} - \hat{j}) = \hat{j} + \hat{k}$. Since $\vec{AC}$ is not a scalar multiple of $\vec{AB}$,they are not collinear.
Checking option $(b)$: $\vec{C} = \hat{i} - 2\hat{j} - \hat{k}$,then $\vec{AC} = (\hat{i} - 2\hat{j} - \hat{k}) - (\hat{i} - \hat{j}) = -\hat{j} - \hat{k}$. Since $\vec{AC}$ is not a scalar multiple of $\vec{AB}$,they are not collinear.
Checking option $(c)$: $\vec{C} = \hat{i} - \hat{k}$,then $\vec{AC} = (\hat{i} - \hat{k}) - (\hat{i} - \hat{j}) = \hat{j} - \hat{k}$. Since $\vec{AC}$ is not a scalar multiple of $\vec{AB}$,they are not collinear.
Since none of the given vectors are collinear with the side $\vec{AB}$,all of them can form a triangle with the given vertices.

(B) Let $a = xi + yj + zk$. Given $|a| = 50$ and $b = 6i - 8j - \frac{15}{2}k$.
Since $a$ and $b$ are collinear, $a = \lambda b$ for some scalar $\lambda$.
Thus, $a = \lambda(6i - 8j - \frac{15}{2}k) = 6\lambda i - 8\lambda j - \frac{15}{2}\lambda k$.
Given $|a| = 50$, we have $|a|^2 = 2500$.
$|a|^2 = (6\lambda)^2 + (-8\lambda)^2 + (-\frac{15}{2}\lambda)^2 = 36\lambda^2 + 64\lambda^2 + \frac{225}{4}\lambda^2 = 100\lambda^2 + \frac{225}{4}\lambda^2 = \frac{400 + 225}{4}\lambda^2 = \frac{625}{4}\lambda^2$.
Setting $\frac{625}{4}\lambda^2 = 2500$, we get $\lambda^2 = \frac{2500 \times 4}{625} = 4 \times 4 = 16$, so $\lambda = \pm 4$.
Since $a$ makes an acute angle with the positive $z$-axis, the $z$-component of $a$ must be positive.
The $z$-component is $-\frac{15}{2}\lambda$. For this to be positive, $\lambda$ must be negative, so $\lambda = -4$.
Substituting $\lambda = -4$ into $a = \lambda b$, we get $a = -4(6i - 8j - \frac{15}{2}k) = -24i + 32j + 30k$.

(B) The magnitude of a vector remains invariant under rotation of the coordinate axes.
Given the original components are $(2p, 1)$ and the new components are $(p+1, 1)$.
The square of the magnitude is given by $x^2 + y^2$.
Therefore,$(2p)^2 + 1^2 = (p+1)^2 + 1^2$.
$4p^2 + 1 = p^2 + 2p + 1 + 1$.
$3p^2 - 2p - 1 = 0$.
Solving the quadratic equation: $3p^2 - 3p + p - 1 = 0$.
$3p(p-1) + 1(p-1) = 0$.
$(3p+1)(p-1) = 0$.
Thus,$p = 1$ or $p = -\frac{1}{3}$.

(B) Let the direction ratios be $a = 2, b = -3, c = 6$.
The magnitude of the vector formed by these ratios is $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}$.
Substituting the values,we get $l = \frac{2}{7}, m = \frac{-3}{7}, n = \frac{6}{7}$.
Thus,the direction cosines are $\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$.

(C) Given $a \cdot b = 0$,this implies that either $a = 0$,$b = 0$,or $a \perp b$.
Given $a \times b = 0$,this implies that either $a = 0$,$b = 0$,or $a \parallel b$.
Since a vector cannot be both perpendicular and parallel to another non-zero vector,the only condition that satisfies both equations simultaneously is that at least one of the vectors must be a zero vector,i.e.,$a = 0$ or $b = 0$.

(D) Given that $\bar{a} + \bar{b}$ is parallel to $\bar{c}$,there exists a scalar $k \in \mathbb{R}$ such that $\bar{a} + \bar{b} = k\bar{c}$ $(1)$
Also,$\bar{b} + \bar{c}$ is parallel to $\bar{a}$,so there exists a scalar $m \in \mathbb{R}$ such that $\bar{b} + \bar{c} = m\bar{a}$ $(2)$
From $(2)$,$\bar{c} = m\bar{a} - \bar{b}$. Substituting this into $(1)$:
$\bar{a} + \bar{b} = k(m\bar{a} - \bar{b})$
$\bar{a} + \bar{b} = km\bar{a} - k\bar{b}$
$(1 - km)\bar{a} + (1 + k)\bar{b} = \bar{0}$
Since $\bar{a}$ and $\bar{b}$ are non-zero and not parallel,their coefficients must be zero:
$1 - km = 0 \implies km = 1$
$1 + k = 0 \implies k = -1$
Substituting $k = -1$ into $km = 1$,we get $m = -1$.
Now,substituting $k = -1$ into $(1)$:
$\bar{a} + \bar{b} = -1\bar{c} \implies \bar{a} + \bar{b} + \bar{c} = \bar{0}$

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

(D) Let the vertices of the triangle be $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively.
The position vector of the centroid $G$ is given by $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.
The vectors from the vertices to the centroid are $\vec{GA} = \vec{a} - \vec{g}$,$\vec{GB} = \vec{b} - \vec{g}$,and $\vec{GC} = \vec{c} - \vec{g}$.
The sum of these vectors is $\vec{GA} + \vec{GB} + \vec{GC} = (\vec{a} - \vec{g}) + (\vec{b} - \vec{g}) + (\vec{c} - \vec{g})$.
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{g}$.
Substituting $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$,we get:
$= (\vec{a} + \vec{b} + \vec{c}) - 3 \left( \frac{\vec{a} + \vec{b} + \vec{c}}{3} \right) = (\vec{a} + \vec{b} + \vec{c}) - (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$.

(D) Let the position vectors of vertices $A, B, C$ be $\vec{a} = i + j$,$\vec{b} = j + k$,and $\vec{c} = k + i$.
The position vectors of the midpoints $D, E, F$ are:
$\vec{d} = \frac{\vec{b} + \vec{c}}{2} = \frac{(j + k) + (k + i)}{2} = \frac{i + j + 2k}{2}$
$\vec{e} = \frac{\vec{a} + \vec{c}}{2} = \frac{(i + j) + (k + i)}{2} = \frac{2i + j + k}{2}$
$\vec{f} = \frac{\vec{a} + \vec{b}}{2} = \frac{(i + j) + (j + k)}{2} = \frac{i + 2j + k}{2}$
The centroid $G$ of $\Delta DEF$ is given by $\vec{g} = \frac{\vec{d} + \vec{e} + \vec{f}}{3}$.
$\vec{g} = \frac{1}{3} \left( \frac{i + j + 2k + 2i + j + k + i + 2j + k}{2} \right)$
$\vec{g} = \frac{1}{3} \left( \frac{4i + 4j + 4k}{2} \right) = \frac{1}{3} (2i + 2j + 2k) = \frac{2}{3}(i + j + k)$.

(D) Two vectors $\vec{a} = a_1i + a_2j + a_3k$ and $\vec{b} = b_1i + b_2j + b_3k$ are collinear if their components are proportional,i.e.,$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = \lambda$.
Given vectors are $\vec{a} = 3i - 2j + 5k$ and $\vec{b} = -2i + pj - qk$.
Equating the ratios of the components:
$\frac{3}{-2} = \frac{-2}{p} = \frac{5}{-q}$.
From $\frac{3}{-2} = \frac{-2}{p}$,we get $3p = 4$,so $p = 4/3$.
From $\frac{3}{-2} = \frac{5}{-q}$,we get $3(-q) = 5(-2)$,so $-3q = -10$,which means $q = 10/3$.
Therefore,$(p, q) = (4/3, 10/3)$.

(A) The unit vector along the internal angle bisector of $\vec{a}$ and $\vec{b}$ is given by $\hat{u} = \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}$.
First,calculate the magnitudes: $|\vec{a}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$.
$|\vec{b}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
Now,$\frac{\vec{a}}{|\vec{a}|} = \frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k})$ and $\frac{\vec{b}}{|\vec{b}|} = \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k}) = \frac{1}{9}(-6\hat{i} - 3\hat{j} + 6\hat{k})$.
Adding these: $\hat{u} = \frac{1}{9}(7-6)\hat{i} + \frac{1}{9}(-4-3)\hat{j} + \frac{1}{9}(-4+6)\hat{k} = \frac{1}{9}(\hat{i} - 7\hat{j} + 2\hat{k})$.
The magnitude of this vector is $\sqrt{(\frac{1}{9})^2 + (-\frac{7}{9})^2 + (\frac{2}{9})^2} = \sqrt{\frac{1+49+4}{81}} = \sqrt{\frac{54}{81}} = \frac{3\sqrt{6}}{9} = \frac{\sqrt{6}}{3}$.
Since $\vec{c} = k\hat{u}$ and $|\vec{c}| = 5\sqrt{6}$,we have $|k| \cdot \frac{\sqrt{6}}{3} = 5\sqrt{6}$,which gives $|k| = 15$.
Thus,$\vec{c} = 15 \cdot \frac{1}{9}(\hat{i} - 7\hat{j} + 2\hat{k}) = \frac{5}{3}(\hat{i} - 7\hat{j} + 2\hat{k})$.

(D) Given that $\vec{a} + 3\vec{b}$ is collinear with $\vec{c}$,there exists a scalar $k$ such that $\vec{a} + 3\vec{b} = k\vec{c}$.
Also,$\vec{b} + 2\vec{c}$ is collinear with $\vec{a}$,so there exists a scalar $m$ such that $\vec{b} + 2\vec{c} = m\vec{a}$.
From the first equation,$3\vec{b} = k\vec{c} - \vec{a}$,so $\vec{b} = \frac{k}{3}\vec{c} - \frac{1}{3}\vec{a}$.
Substitute this into the second equation: $(\frac{k}{3}\vec{c} - \frac{1}{3}\vec{a}) + 2\vec{c} = m\vec{a}$.
Rearranging terms: $(\frac{k}{3} + 2)\vec{c} = (m + \frac{1}{3})\vec{a}$.
Since $\vec{a}$ and $\vec{c}$ are non-zero and non-collinear,the coefficients must be zero: $\frac{k}{3} + 2 = 0 \implies k = -6$ and $m + \frac{1}{3} = 0 \implies m = -\frac{1}{3}$.
Substituting $k = -6$ into the first equation: $\vec{a} + 3\vec{b} = -6\vec{c}$.
Therefore,$\vec{a} + 3\vec{b} + 6\vec{c} = \vec{0}$.

(A) Let the three points be $A$,$B$,and $C$ with position vectors $\vec{OA} = \vec{a}$,$\vec{OB} = \vec{b}$,and $\vec{OC} = 3\vec{a} - 2\vec{b}$.
To check if the points are collinear,we examine the vectors $\vec{AB}$ and $\vec{BC}$.
$\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$.
$\vec{BC} = \vec{OC} - \vec{OB} = (3\vec{a} - 2\vec{b}) - \vec{b} = 3\vec{a} - 3\vec{b} = 3(\vec{a} - \vec{b})$.
Since $\vec{BC} = -3(\vec{b} - \vec{a}) = -3\vec{AB}$,the vectors $\vec{AB}$ and $\vec{BC}$ are parallel.
Because they share a common point $B$,the points $A$,$B$,and $C$ must be collinear.

(B) Two vectors $\vec{a}$ and $\vec{b}$ are said to be equal if and only if they have the same magnitude and the same direction.
Therefore,if $\vec{a} = \vec{b}$,then both vectors must have equal magnitude and the same direction.

(D) The position vectors are $\vec{p} = -2\hat{i} - \hat{j}$,$\vec{q} = 4\hat{i}$,$\vec{r} = 3\hat{i} + 3\hat{j}$,and $\vec{s} = -3\hat{i} + 2\hat{j}$.
Calculate the vectors representing the sides:
$\vec{PQ} = \vec{q} - \vec{p} = (4 - (-2))\hat{i} + (0 - (-1))\hat{j} = 6\hat{i} + \hat{j}$.
$\vec{QR} = \vec{r} - \vec{q} = (3 - 4)\hat{i} + (3 - 0)\hat{j} = -\hat{i} + 3\hat{j}$.
$\vec{RS} = \vec{s} - \vec{r} = (-3 - 3)\hat{i} + (2 - 3)\hat{j} = -6\hat{i} - \hat{j}$.
$\vec{SP} = \vec{p} - \vec{s} = (-2 - (-3))\hat{i} + (-1 - 2)\hat{j} = \hat{i} - 3\hat{j}$.
Since $\vec{PQ} = -\vec{RS}$ and $\vec{QR} = -\vec{SP}$,the opposite sides are parallel and equal in magnitude,so $PQRS$ is a parallelogram.
Check for perpendicularity: $\vec{PQ} \cdot \vec{QR} = (6)(-1) + (1)(3) = -6 + 3 = -3 \neq 0$. Thus,it is not a rectangle.
Check side lengths: $|\vec{PQ}| = \sqrt{6^2 + 1^2} = \sqrt{37}$ and $|\vec{QR}| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$.
Since the adjacent sides are not equal and the angle is not $90^{\circ}$,it is a parallelogram but not a rhombus or square.

(C) Given vectors are $p = 7i - 2j + 3k$ and $q = 3i + j + 5k$.
First,calculate $2q$:
$2q = 2(3i + j + 5k) = 6i + 2j + 10k$.
Now,find $p - 2q$:
$p - 2q = (7i - 2j + 3k) - (6i + 2j + 10k) = (7-6)i + (-2-2)j + (3-10)k = i - 4j - 7k$.
Finally,calculate the magnitude $|p - 2q|$:
$|p - 2q| = \sqrt{(1)^2 + (-4)^2 + (-7)^2} = \sqrt{1 + 16 + 49} = \sqrt{66}$.

(B) Let the resultant of the given forces be $\vec{R}$.
$\vec{R} = \overline{AB} + \overline{AE} + \overline{DC} + \overline{ED}$
Using the triangle law of vector addition:
$\overline{AE} + \overline{ED} = \overline{AD}$
So,$\vec{R} = \overline{AB} + \overline{AD} + \overline{DC}$
Since $\overline{AD} + \overline{DC} = \overline{AC}$,we have:
$\vec{R} = \overline{AB} + \overline{AC}$
We want the final resultant to be $2\overline{AC}$.
Let the force to be added be $\vec{F}$.
$\vec{R} + \vec{F} = 2\overline{AC}$
$(\overline{AB} + \overline{AC}) + \vec{F} = 2\overline{AC}$
$\vec{F} = 2\overline{AC} - \overline{AC} - \overline{AB}$
$\vec{F} = \overline{AC} - \overline{AB}$
Using the triangle law,$\overline{AB} + \overline{BC} = \overline{AC}$,which implies $\overline{BC} = \overline{AC} - \overline{AB}$.
Therefore,$\vec{F} = \overline{BC}$.

(A) Let the vertices of the regular hexagon be $O, A, B, C, D, E$ in order. Let $\vec{OA} = a$ and $\vec{AB} = b$.
In a regular hexagon,the sides are equal in magnitude and parallel to opposite sides.
The vectors representing the sides in order are:
$1$. $\vec{OA} = a$
$2$. $\vec{AB} = b$
$3$. $\vec{BC} = \vec{OA} - \vec{AB} = a - b$
$4$. $\vec{CD} = -\vec{OA} = -a$
$5$. $\vec{DE} = -\vec{AB} = -b$
$6$. $\vec{EO} = -\vec{BC} = -(a - b) = b - a$
Thus,the remaining four sides are $a - b, -a, -b, b - a$.

(A) For three points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ to be collinear,there must exist scalars $x, y, z$ (not all zero) such that $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$ and $x + y + z = 0$.
This is because the condition for collinearity of three points is that the vector $\vec{AB}$ is a scalar multiple of $\vec{AC}$,i.e.,$(\vec{b} - \vec{a}) = k(\vec{c} - \vec{a})$.
Rearranging this gives $\vec{b} - \vec{a} = k\vec{c} - k\vec{a}$,which implies $(k-1)\vec{a} + \vec{b} - k\vec{c} = \vec{0}$.
Comparing this with $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$,we have $x = k-1$,$y = 1$,and $z = -k$.
Summing these coefficients: $x + y + z = (k-1) + 1 + (-k) = 0$.

(A) Given that the position vector of $C$ relative to $B$ is $\vec{BC} = \hat{i} + \hat{j}$.
The position vector of $B$ relative to $A$ is $\vec{AB} = \hat{i} - \hat{j}$.
We need to find the position vector of $C$ relative to $A$,which is $\vec{AC}$.
Using the triangle law of vector addition,we have:
$\vec{AC} = \vec{AB} + \vec{BC}$.
Substituting the given values:
$\vec{AC} = (\hat{i} - \hat{j}) + (\hat{i} + \hat{j})$.
Simplifying the expression:
$\vec{AC} = \hat{i} + \hat{i} - \hat{j} + \hat{j} = 2\hat{i}$.
Thus,the position vector of $C$ relative to $A$ is $2\hat{i}$.

(A) According to the triangle inequality for vectors,for any two vectors $\vec{a}$ and $\vec{b}$,the magnitude of their sum is always less than or equal to the sum of their magnitudes,i.e.,$|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$.
Also,the reverse triangle inequality states that $|\vec{a} + \vec{b}| \ge ||\vec{a}| - |\vec{b}||$.
Since $|\vec{a}| - |\vec{b}| \le ||\vec{a}| - |\vec{b}||$,it follows that $|\vec{a} + \vec{b}| \ge |\vec{a}| - |\vec{b}|$.
Therefore,the statement $|\vec{a} + \vec{b}| \ge |\vec{a}| - |\vec{b}|$ is always true.

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
We know that $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.
Expanding the expression:
$|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a})$
$= 2(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$
$= 2(1 + 1 + 1) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
We also know that $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$.
So,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = |\vec{a} + \vec{b} + \vec{c}|^2 - 3$.
Substituting this back:
$6 - (|\vec{a} + \vec{b} + \vec{c}|^2 - 3) = 9 - |\vec{a} + \vec{b} + \vec{c}|^2$.
Since $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$,the maximum value is $9$ when $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

(C) Given points are $A(2, 3)$, $B(p, 9)$, and $C(1, -1)$.
For points $A, B$, and $C$ to be collinear, the vector $\vec{AB}$ must be parallel to the vector $\vec{AC}$.
$\vec{AB} = (p - 2)i + (9 - 3)j = (p - 2)i + 6j$.
$\vec{AC} = (1 - 2)i + (-1 - 3)j = -i - 4j$.
Since $\vec{AB} \parallel \vec{AC}$, the ratio of their components must be equal:
$\frac{p - 2}{-1} = \frac{6}{-4}$.
$\frac{p - 2}{-1} = -\frac{3}{2}$.
$p - 2 = \frac{3}{2}$.
$p = 2 + \frac{3}{2} = \frac{4 + 3}{2} = \frac{7}{2}$.

(A) Given vectors are $\vec{a} = -4\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{b} = 12\hat{i} - 6\hat{j} + 15\hat{k}$.
Two vectors $\vec{a}$ and $\vec{b}$ are parallel if $\vec{b} = k\vec{a}$ for some scalar $k$.
Let us check the ratio of the components:
$\frac{b_x}{a_x} = \frac{12}{-4} = -3$
$\frac{b_y}{a_y} = \frac{-6}{2} = -3$
$\frac{b_z}{a_z} = \frac{15}{-5} = -3$
Since the ratios are equal,$\vec{b} = -3\vec{a}$.
Therefore,the vectors are parallel (specifically,anti-parallel).

(D) Two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ 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 $3\hat{i} - 2\hat{j} + 5\hat{k}$ and $-2\hat{i} + p\hat{j} - q\hat{k}$.
Equating the ratios of the components:
$\frac{3}{-2} = \frac{-2}{p} = \frac{5}{-q}$.
First,solve for $p$:
$\frac{3}{-2} = \frac{-2}{p} \Rightarrow 3p = 4 \Rightarrow p = 4/3$.
Next,solve for $q$:
$\frac{3}{-2} = \frac{5}{-q} \Rightarrow -3q = -10 \Rightarrow q = 10/3$.
Thus,$p = 4/3$ and $q = 10/3$.

(B) If $a$ and $b$ are two non-collinear vectors,they form a basis for the plane containing them.
Any vector $r$ that lies in the same plane as $a$ and $b$ (i.e.,is coplanar with $a$ and $b$) can be expressed as a linear combination of $a$ and $b$.
Therefore,there exist unique scalars $m$ and $n$ such that $r = ma + nb$.