Scalar or Dot product of two vectors and its applications Questions in English

(NONE) We know that the dot product of two vectors is given by $a \cdot b = |a| |b| \cos \theta$.
Given $|a|=2$,$|b|=3$,and $\theta = 120^{\circ}$.
$a \cdot b = (2)(3) \cos 120^{\circ} = 6 \times (-1/2) = -3$.
Now,we need to find the magnitude $\left|\frac{a}{2} - \frac{b}{3}\right|$.
Let $X = \left|\frac{a}{2} - \frac{b}{3}\right|$. Then $X^2 = \left(\frac{a}{2} - \frac{b}{3}\right) \cdot \left(\frac{a}{2} - \frac{b}{3}\right)$.
$X^2 = \frac{1}{4}|a|^2 + \frac{1}{9}|b|^2 - 2 \left(\frac{1}{2} \cdot \frac{1}{3}\right) (a \cdot b)$.
$X^2 = \frac{1}{4}(2)^2 + \frac{1}{9}(3)^2 - \frac{1}{3} (a \cdot b)$.
$X^2 = \frac{1}{4}(4) + \frac{1}{9}(9) - \frac{1}{3}(-3)$.
$X^2 = 1 + 1 + 1 = 3$.
Therefore,$X = \sqrt{3}$.

(D) We are given the identity $|a \times b|^2 + |a \cdot b|^2 = |a|^2 |b|^2$.
Given that $|a \times b|^2 + |a \cdot b|^2 = 36$ and $|a| = 3$.
Substituting these values into the identity:
$|a|^2 |b|^2 = 36$
$(3)^2 |b|^2 = 36$
$9 |b|^2 = 36$
$|b|^2 = \frac{36}{9} = 4$
$|b| = \sqrt{4} = 2$.
Thus,the value of $|b|$ is $2$.

(D) Given,$a \cdot b = 0$ and $(a + b)$ makes a $60^{\circ}$ angle with $a$.
Using the formula for the angle between two vectors:
$\cos 60^{\circ} = \frac{(a + b) \cdot a}{|a + b||a|}$
$\Rightarrow \frac{1}{2} = \frac{a \cdot a + b \cdot a}{|a + b||a|}$
Since $a \cdot b = 0$,we have $b \cdot a = 0$.
$\Rightarrow \frac{1}{2} = \frac{|a|^2}{|a + b||a|} = \frac{|a|}{|a + b|}$
$\Rightarrow |a + b| = 2|a|$
Squaring both sides:
$|a + b|^2 = 4|a|^2$
$|a|^2 + |b|^2 + 2(a \cdot b) = 4|a|^2$
Since $a \cdot b = 0$:
$|a|^2 + |b|^2 = 4|a|^2$
$|b|^2 = 3|a|^2$
$|b| = \sqrt{3}|a|$

(B) We know that for any two vectors $\vec{a}$ and $\vec{b}$,the Lagrange's identity states that $|\vec{a} \times \vec{b}|^{2} + |\vec{a} \cdot \vec{b}|^{2} = |\vec{a}|^{2} |\vec{b}|^{2}$.
Given $|\vec{a}| = 16$ and $|\vec{b}| = 4$.
Substituting these values into the identity:
$\sqrt{|\vec{a} \times \vec{b}|^{2} + |\vec{a} \cdot \vec{b}|^{2}} = \sqrt{|\vec{a}|^{2} |\vec{b}|^{2}}$
$= |\vec{a}| |\vec{b}|$
$= (16) \times (4)$
$= 64$.

(C) Given that the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is $\frac{2\pi}{3}$.
The projection of $\vec{a}$ in the direction of $\vec{b}$ is given by the formula $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = -2$.
We know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
Substituting this into the projection formula:
$\frac{|\vec{a}| |\vec{b}| \cos \theta}{|\vec{b}|} = -2$
$|\vec{a}| \cos \theta = -2$
Substitute $\theta = \frac{2\pi}{3}$:
$|\vec{a}| \cos \left(\frac{2\pi}{3}\right) = -2$
$|\vec{a}| \left(-\frac{1}{2}\right) = -2$
$|\vec{a}| = (-2) \times (-2)$
$|\vec{a}| = 4$
Thus,the correct option is $C$.

(C) Given that,$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$.
We know the Lagrange's identity for vectors:
$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2} \sin^{2} \theta + |\vec{a}|^{2} |\vec{b}|^{2} \cos^{2} \theta$.
Factoring out $|\vec{a}|^{2} |\vec{b}|^{2}$,we get:
$|\vec{a}|^{2} |\vec{b}|^{2} (\sin^{2} \theta + \cos^{2} \theta) = |\vec{a}|^{2} |\vec{b}|^{2}$.
Therefore,$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2}$.
Substituting the given values:
$144 = (4)^{2} |\vec{b}|^{2}$.
$144 = 16 |\vec{b}|^{2}$.
$|\vec{b}|^{2} = \frac{144}{16} = 9$.
Taking the square root,we get $|\vec{b}| = 3$.

(A) Given that $\bar{a}$ and $\bar{b}$ are mutually perpendicular unit vectors.
This implies $|\bar{a}| = 1$,$|\bar{b}| = 1$,and $\bar{a} \cdot \bar{b} = 0$.
We need to evaluate the dot product:
$(3\bar{a} + 2\bar{b}) \cdot (5\bar{a} - 6\bar{b}) = 3\bar{a} \cdot (5\bar{a} - 6\bar{b}) + 2\bar{b} \cdot (5\bar{a} - 6\bar{b})$
$= 15(\bar{a} \cdot \bar{a}) - 18(\bar{a} \cdot \bar{b}) + 10(\bar{b} \cdot \bar{a}) - 12(\bar{b} \cdot \bar{b})$
$= 15|\bar{a}|^2 - 18(0) + 10(0) - 12|\bar{b}|^2$
$= 15(1)^2 - 12(1)^2$
$= 15 - 12 = 3$.

(D) Given that,$ 2 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| $.
We know that the dot product of two vectors is defined as $ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta $,where $ \theta $ is the angle between the vectors.
Substituting this into the given equation:
$ 2 (|\vec{a}| |\vec{b}| \cos \theta) = |\vec{a}| |\vec{b}| $.
Dividing both sides by $ |\vec{a}| |\vec{b}| $ (assuming non-zero vectors):
$ 2 \cos \theta = 1 $.
$ \cos \theta = \frac{1}{2} $.
Since $ \cos 60^{\circ} = \frac{1}{2} $,the angle $ \theta = 60^{\circ} $.

(C) Given that $a \perp b$,therefore the dot product $a \cdot b = 0$.
Since $(a+b) \perp (a+mb)$,their dot product must be zero:
$(a+b) \cdot (a+mb) = 0$
Expanding the dot product:
$a \cdot a + m(a \cdot b) + (b \cdot a) + m(b \cdot b) = 0$
Substituting $a \cdot b = 0$ and $b \cdot a = 0$:
$|a|^{2} + m(0) + 0 + m|b|^{2} = 0$
$|a|^{2} + m|b|^{2} = 0$
Solving for $m$:
$m|b|^{2} = -|a|^{2}$
$m = -\frac{|a|^{2}}{|b|^{2}}$

(B) Given that $a, b$ and $c$ are unit vectors.
Therefore,$|a| = |b| = |c| = 1 \dots (i) $
We are given the equation $a+b+c = 0$.
Squaring both sides,we get:
$|a+b+c|^2 = |0|^2$
$|a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
Substituting the values from $(i)$:
$1 + 1 + 1 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
$3 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
$2(a \cdot b + b \cdot c + c \cdot a) = -3$
$a \cdot b + b \cdot c + c \cdot a = -\frac{3}{2}$

(C) We know that for any two vectors $a$ and $b$,$(a \times b)^{2} + (a \cdot b)^{2} = |a|^{2} |b|^{2} \sin^{2} \theta + |a|^{2} |b|^{2} \cos^{2} \theta$.
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,this simplifies to $|a|^{2} |b|^{2}$.
Given the equation $(a \times b)^{2} + (a \cdot b)^{2} = 144$,we have $|a|^{2} |b|^{2} = 144$.
Given $|a| = 4$,we substitute $|a|^{2} = 16$ into the equation:
$16 |b|^{2} = 144$.
Dividing both sides by $16$,we get $|b|^{2} = 9$.
Taking the square root,we find $|b| = 3$.

(C) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are unit vectors,so $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$.
Given $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.
Squaring both sides:
$(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) = 0$.
$|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0$.
$1+1+1+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0$.
$3+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0 \implies \overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a} = -3/2$.
Since $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,the vectors form an equilateral triangle,so the angle between any two vectors is $120^{\circ}$.
Thus,$\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a} = 1 \times 1 \times \cos(120^{\circ}) = -1/2$.
Substituting these values:
$3(-1/2)+2(-1/2)+(-1/2) = -3/2 - 2/2 - 1/2 = -6/2 = -3$.

(C) Given that $\hat{i}, \hat{j}, \hat{k}$ are unit vectors along the positive $x, y, z$-axes respectively.
$(a)$ $\sum \hat{i} \times(\hat{j}+\hat{k}) = \hat{i} \times(\hat{j}+\hat{k}) + \hat{j} \times(\hat{k}+\hat{i}) + \hat{k} \times(\hat{i}+\hat{j})$
$= (\hat{k} - \hat{j}) + (\hat{i} - \hat{k}) + (\hat{j} - \hat{i}) = \vec{0}$. This is true.
$(b)$ $\sum \hat{i} \times(\hat{j} \times \hat{k}) = \hat{i} \times \hat{i} + \hat{j} \times \hat{j} + \hat{k} \times \hat{k} = \vec{0} + \vec{0} + \vec{0} = \vec{0}$. This is true.
$(c)$ $\sum \hat{i} \cdot(\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{k} = 1 + 1 + 1 = 3$. This is false as it is not equal to $\vec{0}$.
$(d)$ $\sum \hat{i} \cdot(\hat{j}+\hat{k}) = (\hat{i} \cdot \hat{j} + \hat{i} \cdot \hat{k}) + (\hat{j} \cdot \hat{k} + \hat{j} \cdot \hat{i}) + (\hat{k} \cdot \hat{i} + \hat{k} \cdot \hat{j}) = 0 + 0 + 0 = 0$. This is true.

(C) Let $\overrightarrow{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$.
Given that $\overrightarrow{a} \cdot \hat{i} = 1$,we have $a_{1} = 1$.
Given that $\overrightarrow{a} \cdot (\hat{i} + \hat{j}) = 1$,we have $a_{1} + a_{2} = 1$. Substituting $a_{1} = 1$,we get $1 + a_{2} = 1$,which implies $a_{2} = 0$.
Given that $\overrightarrow{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$,we have $a_{1} + a_{2} + a_{3} = 1$. Substituting $a_{1} = 1$ and $a_{2} = 0$,we get $1 + 0 + a_{3} = 1$,which implies $a_{3} = 0$.
Therefore,$\overrightarrow{a} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{i}$.

(C) Given vectors are $\overrightarrow{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ and $\overrightarrow{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$.
The formula for the projection of vector $\overrightarrow{a}$ on vector $\overrightarrow{b}$ is given by $\frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|}$.
First,calculate the dot product $\overrightarrow{a} \cdot \overrightarrow{b}$:
$\overrightarrow{a} \cdot \overrightarrow{b} = (3)(2) + (-1)(3) + (5)(1) = 6 - 3 + 5 = 8$.
Next,calculate the magnitude of vector $\overrightarrow{b}$:
$|\overrightarrow{b}| = \sqrt{2^{2} + 3^{2} + 1^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
Therefore,the projection is $\frac{8}{\sqrt{14}}$.

(B) Given the magnitudes of vectors $OA$ and $OB$ are $|OA| = 5$ and $|OB| = 6$ respectively.
The angle between the two vectors is $\theta = \angle BOA = 60^{\circ}$.
The dot product of two vectors is given by the formula: $OA \cdot OB = |OA| |OB| \cos \theta$.
Substituting the given values:
$OA \cdot OB = 5 \times 6 \times \cos 60^{\circ}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have:
$OA \cdot OB = 30 \times \frac{1}{2} = 15$.
Thus,the value of $OA \cdot OB$ is $15$.

(C) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$.
We know that $|a-b|^2 = (a-b) \cdot (a-b) = |a|^2 - 2(a \cdot b) + |b|^2$.
Since $a \cdot b = |a||b| \cos \theta$,we have $|a-b|^2 = 1^2 - 2(1)(1) \cos \theta + 1^2 = 2 - 2 \cos \theta$.
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get $|a-b|^2 = 2(2 \sin^2 \frac{\theta}{2}) = 4 \sin^2 \frac{\theta}{2}$.
Taking the square root on both sides,we get $|a-b| = 2 \sin \frac{\theta}{2}$.
Therefore,$\sin \frac{\theta}{2} = \frac{|a-b|}{2}$.

(C) Given that $|a+b|^{2}+|a \cdot b|^{2}=144$ and $|a|=6$.
We know the identity $|a+b|^{2} + |a-b|^{2} = 2(|a|^{2} + |b|^{2})$,but here we use the property related to vectors.
Actually,the expression $|a+b|^{2} + |a \cdot b|^{2}$ is not a standard identity.
Assuming the intended expression is $|a+b|^{2} + |a-b|^{2} = 2(|a|^{2} + |b|^{2})$ or similar,let us re-evaluate the given equation $|a+b|^{2} + |a \cdot b|^{2} = 144$.
If we assume the question implies $|a|^{2}|b|^{2} = 144$,then $|6|^{2}|b|^{2} = 144$.
$36|b|^{2} = 144$.
$|b|^{2} = \frac{144}{36} = 4$.
Therefore,$|b| = 2$.

(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
We are given that $\sqrt{3}\vec{a}-\vec{b}$ is also a unit vector,so $|\sqrt{3}\vec{a}-\vec{b}| = 1$.
Squaring both sides,we get $|\sqrt{3}\vec{a}-\vec{b}|^2 = 1^2$.
Using the property $|\vec{u}-\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$,we have:
$3|\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{3}(\vec{a} \cdot \vec{b}) = 1$.
Substituting $|\vec{a}| = 1$ and $|\vec{b}| = 1$:
$3(1)^2 + (1)^2 - 2\sqrt{3}(|\vec{a}||\vec{b}|\cos \theta) = 1$.
$3 + 1 - 2\sqrt{3}(1)(1)\cos \theta = 1$.
$4 - 2\sqrt{3}\cos \theta = 1$.
$2\sqrt{3}\cos \theta = 3$.
$\cos \theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Since $\cos \theta = \frac{\sqrt{3}}{2}$,the angle $\theta = 30^{\circ}$.

(C) Given that for a quadrilateral $ABCD$,$AB=a$,$BC=b$,$AD=b-a$.
Since $M$ is the midpoint of $BC$,$BM = \frac{b}{2}$.
Using the section formula for point $N$ on $DM$ where $DN = \frac{4}{5} DM$,we have $DN:NM = 4:1$.
Using the vector representation or coordinate geometry approach,the position vector of $N$ is given by $\vec{N} = \frac{1 \cdot \vec{D} + 4 \cdot \vec{M}}{5}$.
Multiplying by $5$,we get $5\vec{N} = \vec{D} + 4\vec{M}$.
Since $\vec{M} = \vec{B} + \frac{1}{2}\vec{BC} = \vec{B} + \frac{b}{2}$ and $\vec{D} = \vec{A} + (b-a)$,we substitute these values.
$5\vec{AN} = 4\vec{AM} + \vec{AD} = 4(a + \frac{b}{2}) + (b-a) = 4a + 2b + b - a = 3(a+b) = 3AC$.
Thus,$5 AN = 3 AC$.

(D) The angle bisector theorem states that the bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the lengths of the adjacent sides $AB$ and $AC$,i.e.,in the ratio $c:b$,where $c = |AB|$ and $b = |AC|$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = |(2-4)\hat{i} + (3-7)\hat{j} + (4-8)\hat{k}| = |-2\hat{i} - 4\hat{j} - 4\hat{k}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
$AC = |(2-4)\hat{i} + (5-7)\hat{j} + (7-8)\hat{k}| = |-2\hat{i} - 2\hat{j} - 1\hat{k}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,the ratio is $6:3 = 2:1$.
The position vector of the point $D$ dividing $BC$ in ratio $2:1$ is given by the section formula:
$\vec{D} = \frac{2\vec{C} + 1\vec{B}}{2+1} = \frac{2(2\hat{i} + 5\hat{j} + 7\hat{k}) + 1(2\hat{i} + 3\hat{j} + 4\hat{k})}{3}$.
$\vec{D} = \frac{(4+2)\hat{i} + (10+3)\hat{j} + (14+4)\hat{k}}{3} = \frac{6\hat{i} + 13\hat{j} + 18\hat{k}}{3} = 2\hat{i} + \frac{13}{3}\hat{j} + 6\hat{k}$.
Therefore,the correct option is $D$.

(B) Given vectors $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.
First,calculate the magnitudes:
$|\vec{OA}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$.
$|\vec{OB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = 6$.
Let $\vec{a} = \frac{\vec{OA}}{|\vec{OA}|} = \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}$ and $\vec{b} = \frac{\vec{OB}}{|\vec{OB}|} = \frac{2}{6}\hat{i} + \frac{4}{6}\hat{j} + \frac{4}{6}\hat{k} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.
The internal angle bisector vector $\vec{OP}$ is along the direction of $(\vec{a} + \vec{b})$.
$\vec{a} + \vec{b} = (\frac{2}{3} + \frac{1}{3})\hat{i} + (\frac{2}{3} + \frac{2}{3})\hat{j} + (\frac{1}{3} + \frac{2}{3})\hat{k} = 1\hat{i} + \frac{4}{3}\hat{j} + 1\hat{k}$.
The length of the angle bisector $k$ is the magnitude of the vector $\vec{OP} = \lambda(\vec{a} + \vec{b})$. Using the section formula,the bisector vector is $\vec{OP} = \frac{|\vec{OB}|\vec{OA} + |\vec{OA}|\vec{OB}}{|\vec{OA}| + |\vec{OB}|} = \frac{6\vec{OA} + 3\vec{OB}}{9} = \frac{2\vec{OA} + \vec{OB}}{3}$.
$\vec{OP} = \frac{2(2\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$.
$k^2 = |\vec{OP}|^2 = 2^2 + (\frac{8}{3})^2 + 2^2 = 4 + \frac{64}{9} + 4 = 8 + \frac{64}{9} = \frac{72 + 64}{9} = \frac{136}{9}$.
Therefore,$9k^2 = 136$.

(D) Given the equations for the position vector $\bar{x}$ of point $P$:
$1$) $\bar{x} \cdot (\bar{c} - \bar{b}) = \bar{a} \cdot \bar{c} - \bar{a} \cdot \bar{b} \implies (\bar{x} - \bar{a}) \cdot (\bar{c} - \bar{b}) = 0$.
This implies that the vector $\vec{AP}$ is perpendicular to the side $BC$.
$2$) $\bar{x} \cdot (\bar{a} - \bar{c}) = \bar{a} \cdot \bar{b} - \bar{b} \cdot \bar{c} \implies (\bar{x} - \bar{b}) \cdot (\bar{a} - \bar{c}) = 0$.
This implies that the vector $\vec{BP}$ is perpendicular to the side $AC$.
Since $P$ is a point such that $\vec{AP} \perp BC$ and $\vec{BP} \perp AC$,$P$ is the intersection of the altitudes of the triangle $ABC$.
Therefore,$P$ is the orthocentre of the triangle $ABC$.

(B) Let the vertices be $A(2,3,5)$,$B(-1,3,2)$,and $C(3,5,-2)$.
Calculate the vectors representing the sides:
$\vec{AB} = B - A = (-1-2, 3-3, 2-5) = (-3, 0, -3)$.
$\vec{BC} = C - B = (3-(-1), 5-3, -2-2) = (4, 2, -4)$.
$\vec{CA} = A - C = (2-3, 3-5, 5-(-2)) = (-1, -2, 7)$.
Calculate the magnitudes:
$|\vec{AB}| = \sqrt{(-3)^2 + 0^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.
$|\vec{BC}| = \sqrt{4^2 + 2^2 + (-4)^2} = \sqrt{16+4+16} = \sqrt{36} = 6$.
$|\vec{CA}| = \sqrt{(-1)^2 + (-2)^2 + 7^2} = \sqrt{1+4+49} = \sqrt{54} = 3\sqrt{6}$.
Using the Law of Cosines: $\cos \alpha = \frac{|\vec{AB}|^2 + |\vec{AC}|^2 - |\vec{BC}|^2}{2|\vec{AB}||\vec{AC}|}$.
Note that $\vec{AC} = -\vec{CA} = (1, 2, -7)$,so $|\vec{AC}| = 3\sqrt{6}$.
$\cos \alpha = \frac{18 + 54 - 36}{2(3\sqrt{2})(3\sqrt{6})} = \frac{36}{18\sqrt{12}} = \frac{36}{36\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Then $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{3} = \frac{2}{3}$.
Similarly,$\cos \beta = \frac{|\vec{BA}|^2 + |\vec{BC}|^2 - |\vec{AC}|^2}{2|\vec{BA}||\vec{BC}|} = \frac{18 + 36 - 54}{2(3\sqrt{2})(6)} = 0$.
So $\beta = 90^\circ$ and $\sin^2 \beta = 1$.
Finally,$\sin^2 \gamma = 1 - \sin^2 \alpha - \sin^2 \beta$ is not correct,but $\gamma = 180^\circ - (\alpha + \beta)$.
Since $\cos \beta = 0$,$\beta = 90^\circ$,then $\alpha + \gamma = 90^\circ$,so $\gamma = 90^\circ - \alpha$.
Thus $\sin^2 \gamma = \sin^2(90^\circ - \alpha) = \cos^2 \alpha = \frac{1}{3}$.
Sum $= \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = \frac{2}{3} + 1 + \frac{1}{3} = 2$.

(B) Let the position vector of the orthocentre be $\bar{h}$.
We know that for any triangle,the centroid $G$ divides the line segment joining the orthocentre $H$ and the circumcentre $P$ in the ratio $2:1$.
The position vector of the centroid $G$ is given by $\bar{g} = \frac{\bar{a}+\bar{b}+\bar{c}}{3}$.
Using the section formula,$\bar{g} = \frac{1(\bar{h}) + 2(\bar{p})}{1+2} = \frac{\bar{h} + 2\bar{p}}{3}$.
Equating the two expressions for $\bar{g}$:
$\frac{\bar{a}+\bar{b}+\bar{c}}{3} = \frac{\bar{h} + 2\bar{p}}{3}$.
$\bar{h} + 2\bar{p} = \bar{a}+\bar{b}+\bar{c}$.
Given $\bar{p} = \frac{\bar{a}+\bar{b}+\bar{c}}{4}$,substitute this into the equation:
$\bar{h} + 2\left(\frac{\bar{a}+\bar{b}+\bar{c}}{4}\right) = \bar{a}+\bar{b}+\bar{c}$.
$\bar{h} + \frac{\bar{a}+\bar{b}+\bar{c}}{2} = \bar{a}+\bar{b}+\bar{c}$.
$\bar{h} = (\bar{a}+\bar{b}+\bar{c}) - \frac{\bar{a}+\bar{b}+\bar{c}}{2} = \frac{\bar{a}+\bar{b}+\bar{c}}{2}$.

(B) First,find the unit vectors along $\vec{a}$ and $\vec{b}$.
$|\vec{a}| = \sqrt{4^2 + 7^2 + (-4)^2} = \sqrt{16 + 49 + 16} = \sqrt{81} = 9$.
$|\vec{b}| = \sqrt{12^2 + (-3)^2 + 4^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = 13$.
Unit vectors are $\hat{a} = \frac{4 \hat{i} + 7 \hat{j} - 4 \hat{k}}{9}$ and $\hat{b} = \frac{12 \hat{i} - 3 \hat{j} + 4 \hat{k}}{13}$.
The vector along the internal angle bisector is given by $\vec{v} = \lambda(\hat{a} + \hat{b})$.
$\hat{a} + \hat{b} = \frac{13(4 \hat{i} + 7 \hat{j} - 4 \hat{k}) + 9(12 \hat{i} - 3 \hat{j} + 4 \hat{k})}{117} = \frac{(52 + 108) \hat{i} + (91 - 27) \hat{j} + (-52 + 36) \hat{k}}{117} = \frac{160 \hat{i} + 64 \hat{j} - 16 \hat{k}}{117} = \frac{16}{117}(10 \hat{i} + 4 \hat{j} - \hat{k})$.
Let $\vec{c} = k(10 \hat{i} + 4 \hat{j} - \hat{k})$. The magnitude is $|\vec{c}| = |k| \sqrt{10^2 + 4^2 + (-1)^2} = |k| \sqrt{100 + 16 + 1} = |k| \sqrt{117} = |k| \sqrt{9 \times 13} = 3|k| \sqrt{13}$.
Given $|\vec{c}| = 3 \sqrt{13}$,we have $3|k| \sqrt{13} = 3 \sqrt{13}$,so $k = 1$.
Thus,$\vec{c} = 10 \hat{i} + 4 \hat{j} - \hat{k}$.

(C) Given $|\vec{a}|=2, |\vec{b}|=3$ and $\theta = \frac{\pi}{3}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{3}) = 2 \times 3 \times \frac{1}{2} = 3$.
Let the adjacent sides be $\vec{p} = 2\vec{a} + 3\vec{b}$ and $\vec{q} = \vec{a} - \vec{b}$.
The diagonals are $\vec{d_1} = \vec{p} + \vec{q} = (2\vec{a} + 3\vec{b}) + (\vec{a} - \vec{b}) = 3\vec{a} + 2\vec{b}$ and $\vec{d_2} = \vec{p} - \vec{q} = (2\vec{a} + 3\vec{b}) - (\vec{a} - \vec{b}) = \vec{a} + 4\vec{b}$.
Calculate the squared lengths:
$|\vec{d_1}|^2 = |3\vec{a} + 2\vec{b}|^2 = 9|\vec{a}|^2 + 4|\vec{b}|^2 + 12(\vec{a} \cdot \vec{b}) = 9(4) + 4(9) + 12(3) = 36 + 36 + 36 = 108$.
$|\vec{d_1}| = \sqrt{108} = 6\sqrt{3}$.
$|\vec{d_2}|^2 = |\vec{a} + 4\vec{b}|^2 = |\vec{a}|^2 + 16|\vec{b}|^2 + 8(\vec{a} \cdot \vec{b}) = 4 + 16(9) + 8(3) = 4 + 144 + 24 = 172$.
$|\vec{d_2}| = \sqrt{172} = 2\sqrt{43}$.
Comparing the lengths,$6\sqrt{3} = \sqrt{108} \approx 10.39$ and $2\sqrt{43} = \sqrt{172} \approx 13.11$.
The shorter diagonal is $6\sqrt{3}$.

(C) Let $\vec{A} = 7 \hat{i}-4 \hat{j}+5 \hat{k}$ be the position vector of vertex $A$.
Let $\vec{G}_{BCD} = \frac{\vec{B}+\vec{C}+\vec{D}}{3} = -\hat{i}+4 \hat{j}-3 \hat{k}$ be the position vector of the centroid of triangle $BCD$.
The centroid $\vec{G}$ of the tetrahedron $ABCD$ is given by the formula:
$\vec{G} = \frac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4}$
We can rewrite this as:
$\vec{G} = \frac{1}{4} [\vec{A} + 3(\frac{\vec{B}+\vec{C}+\vec{D}}{3})] = \frac{1}{4} [\vec{A} + 3 \vec{G}_{BCD}]$
Substituting the given values:
$\vec{G} = \frac{1}{4} [(7 \hat{i}-4 \hat{j}+5 \hat{k}) + 3(-\hat{i}+4 \hat{j}-3 \hat{k})]$
$\vec{G} = \frac{1}{4} [7 \hat{i}-4 \hat{j}+5 \hat{k} - 3 \hat{i} + 12 \hat{j} - 9 \hat{k}]$
$\vec{G} = \frac{1}{4} [4 \hat{i} + 8 \hat{j} - 4 \hat{k}]$
$\vec{G} = \hat{i} + 2 \hat{j} - \hat{k}$

(C) The component of vector $\vec{a}$ along $\vec{b}$ is given by $\vec{p} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(2) + (3)(-4) + (13)(3) = 2 - 12 + 39 = 29$.
Next,calculate the magnitude squared $|\vec{b}|^2 = (2)^2 + (-4)^2 + (3)^2 = 4 + 16 + 9 = 29$.
So,the component of $\vec{a}$ along $\vec{b}$ is $\vec{p} = \left( \frac{29}{29} \right) \vec{b} = \vec{b} = 2\hat{i} - 4\hat{j} + 3\hat{k}$.
The component of $\vec{a}$ perpendicular to $\vec{b}$ is $\vec{x} = \vec{a} - \vec{p} = \vec{a} - \vec{b}$.
$\vec{x} = (\hat{i} + 3\hat{j} + 13\hat{k}) - (2\hat{i} - 4\hat{j} + 3\hat{k}) = (1-2)\hat{i} + (3+4)\hat{j} + (13-3)\hat{k} = -\hat{i} + 7\hat{j} + 10\hat{k}$.

(B) Given that $\vec{a} \cdot \vec{b} = 0$,$\vec{b} \cdot \vec{c} = 0$,$|\vec{a}| = 2$,$|\vec{b}| = 3$,$|\vec{c}| = 5$ and $|\vec{a} + \vec{b} + \vec{c}| = 4 \sqrt{3}$.
We know that $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
Substituting the given values:
$(4 \sqrt{3})^2 = 2^2 + 3^2 + 5^2 + 2(0 + 0 + |\vec{c}| |\vec{a}| \cos \theta)$.
$48 = 4 + 9 + 25 + 2(5)(2) \cos \theta$.
$48 = 38 + 20 \cos \theta$.
$10 = 20 \cos \theta$.
$\cos \theta = \frac{10}{20} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{3}$.

(B) In $\triangle ABC$,$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b} \dots (i)$
In $\triangle ADC$,$\overrightarrow{AD} + \overrightarrow{DC} = \overrightarrow{AC}$. Given $\overrightarrow{DA} = \vec{a} - \vec{b}$,so $\overrightarrow{AD} = -(\vec{a} - \vec{b}) = \vec{b} - \vec{a}$.
Thus,$\overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD} = (\vec{a} + \vec{b}) - (\vec{b} - \vec{a}) = 2\vec{a}$.
$M$ is the midpoint of $BC$,so $\overrightarrow{BM} = \frac{1}{2}\overrightarrow{BC} = \frac{\vec{b}}{2}$.
In $\triangle BDM$,$\overrightarrow{DM} = \overrightarrow{BM} - \overrightarrow{BD}$. Since $\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = \vec{b} - 2\vec{a}$,we have $\overrightarrow{DM} = \frac{\vec{b}}{2} - (\vec{b} - 2\vec{a}) = 2\vec{a} - \frac{\vec{b}}{2} = \frac{4\vec{a} - \vec{b}}{2}$.
Given $\overrightarrow{DX} = \frac{4}{5}\overrightarrow{DM} = \frac{4}{5} \left( \frac{4\vec{a} - \vec{b}}{2} \right) = \frac{8\vec{a} - 2\vec{b}}{5}$.
Now,$\overrightarrow{AX} = \overrightarrow{AD} + \overrightarrow{DX} = (\vec{b} - \vec{a}) + \frac{8\vec{a} - 2\vec{b}}{5} = \frac{5\vec{b} - 5\vec{a} + 8\vec{a} - 2\vec{b}}{5} = \frac{3\vec{a} + 3\vec{b}}{5} = \frac{3}{5}(\vec{a} + \vec{b})$.
Since $\overrightarrow{AX} = \frac{3}{5}\overrightarrow{AC}$,the vectors $\overrightarrow{AX}$ and $\overrightarrow{AC}$ are parallel and share a common point $A$.
Therefore,the points $A, X$ and $C$ are collinear.

(B) Let the vectors of magnitude $a, 2a, 3a$ be along the diagonals of three adjacent faces of a cube meeting at the origin $O$.
Let the cube edges be along the coordinate axes. The diagonals of the three adjacent faces can be represented as vectors along $(\hat{i}+\hat{j})$,$(\hat{j}+\hat{k})$,and $(\hat{k}+\hat{i})$.
Normalizing these directions,the vectors are:
$\vec{v_1} = a \frac{\hat{i}+\hat{j}}{\sqrt{2}}$,$\vec{v_2} = 2a \frac{\hat{j}+\hat{k}}{\sqrt{2}}$,$\vec{v_3} = 3a \frac{\hat{k}+\hat{i}}{\sqrt{2}}$.
The resultant vector $\vec{R} = \vec{v_1} + \vec{v_2} + \vec{v_3}$ is:
$\vec{R} = \frac{a}{\sqrt{2}} [(\hat{i}+\hat{j}) + 2(\hat{j}+\hat{k}) + 3(\hat{k}+\hat{i})] = \frac{a}{\sqrt{2}} (4\hat{i} + 3\hat{j} + 5\hat{k})$.
The magnitude $|\vec{R}|$ is:
$|\vec{R}| = \frac{a}{\sqrt{2}} \sqrt{4^2 + 3^2 + 5^2} = \frac{a}{\sqrt{2}} \sqrt{16 + 9 + 25} = \frac{a}{\sqrt{2}} \sqrt{50} = \frac{a}{\sqrt{2}} (5\sqrt{2}) = 5a$.

(D) Let the position vectors of $A, B, C, D$ be $\vec{0}, \vec{b}, \vec{c}, \vec{d}$ respectively.
Since $ABCD$ is a parallelogram,we have $\vec{b} + \vec{d} = \vec{c}$.
$M$ is the midpoint of $BC$,so $\vec{M} = \frac{\vec{b} + \vec{c}}{2}$.
$N$ is the midpoint of $CD$,so $\vec{N} = \frac{\vec{c} + \vec{d}}{2}$.
Now,$\overline{AM} + \overline{AN} = \vec{M} + \vec{N} = \frac{\vec{b} + \vec{c}}{2} + \frac{\vec{c} + \vec{d}}{2}$.
$= \frac{\vec{b} + \vec{d} + 2\vec{c}}{2}$.
Substituting $\vec{b} + \vec{d} = \vec{c}$,we get:
$= \frac{\vec{c} + 2\vec{c}}{2} = \frac{3\vec{c}}{2} = \frac{3}{2} \overline{AC}$.

(B) Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ respectively.
Given $\vec{BC} = \lambda \vec{AD}$,we have $\vec{c} - \vec{b} = \lambda(\vec{d} - \vec{a}) \dots (i)$.
We are given $\vec{x} = \vec{AC} + \vec{BD}$.
Substituting the position vectors,$\vec{x} = (\vec{c} - \vec{a}) + (\vec{d} - \vec{b})$.
Rearranging the terms,$\vec{x} = (\vec{c} - \vec{b}) + (\vec{d} - \vec{a})$.
Using equation $(i)$,$\vec{x} = \lambda(\vec{d} - \vec{a}) + 1(\vec{d} - \vec{a})$.
Thus,$\vec{x} = (\lambda + 1)(\vec{d} - \vec{a})$.
Since $\vec{AD} = \vec{d} - \vec{a}$,we have $\vec{x} = (\lambda + 1) \vec{AD}$.
Comparing this with $\vec{x} = p \vec{AD}$,we get $p = \lambda + 1$.

(B) Since $D$ is the midpoint of $AB$,we have $\vec{AD} = \frac{1}{2} \vec{AB}$.
Using the triangle law of vector addition in $\triangle CAD$,we have $\vec{CD} = \vec{CA} + \vec{AD} = \vec{CA} + \frac{1}{2} \vec{AB}$.
Now,we calculate the dot product $\vec{CA} \cdot \vec{CD}$:
$\vec{CA} \cdot \vec{CD} = \vec{CA} \cdot (\vec{CA} + \frac{1}{2} \vec{AB})$
$= |\vec{CA}|^2 + \frac{1}{2} (\vec{CA} \cdot \vec{AB})$
$= b^2 + \frac{1}{2} |\vec{CA}| |\vec{AB}| \cos(\pi - A)$
$= b^2 - \frac{1}{2} bc \cos A$
Using the cosine rule in $\triangle ABC$,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting this value:
$= b^2 - \frac{1}{2} bc \left( \frac{b^2 + c^2 - a^2}{2bc} \right)$
$= b^2 - \frac{b^2 + c^2 - a^2}{4}$
$= \frac{4b^2 - b^2 - c^2 + a^2}{4}$
$= \frac{1}{4}(a^2 + 3b^2 - c^2)$.

(B) Let the adjacent sides of the parallelogram be $p = 5a + 2b$ and $q = a - 3b$.
The diagonals of the parallelogram are given by $d_1 = p + q$ and $d_2 = p - q$.
$d_1 = (5a + 2b) + (a - 3b) = 6a - b$
$d_2 = (5a + 2b) - (a - 3b) = 4a + 5b$
Given $|a| = 2\sqrt{2}$,$|b| = 3$,and the angle $\theta = 45^{\circ}$ between $a$ and $b$,we have $a \cdot b = |a||b| \cos 45^{\circ} = (2\sqrt{2})(3)(\frac{1}{\sqrt{2}}) = 6$.
Now,the length of the first diagonal is:
$|d_1| = |6a - b| = \sqrt{(6a - b) \cdot (6a - b)} = \sqrt{36|a|^2 + |b|^2 - 12(a \cdot b)}$
$|d_1| = \sqrt{36(8) + 9 - 12(6)} = \sqrt{288 + 9 - 72} = \sqrt{225} = 15$.
Now,the length of the second diagonal is:
$|d_2| = |4a + 5b| = \sqrt{(4a + 5b) \cdot (4a + 5b)} = \sqrt{16|a|^2 + 25|b|^2 + 40(a \cdot b)}$
$|d_2| = \sqrt{16(8) + 25(9) + 40(6)} = \sqrt{128 + 225 + 240} = \sqrt{593}$.
Thus,the lengths of the diagonals are $15$ and $\sqrt{593}$.

(B) Given that $\overline{e_1}$ and $\overline{e_2}$ are unit vectors,we have $|\overline{e_1}| = 1$ and $|\overline{e_2}| = 1$.
Given $|\overline{e_1} + \overline{e_2}| = \sqrt{3}$.
Squaring both sides,we get $|\overline{e_1} + \overline{e_2}|^2 = 3$.
$(\overline{e_1} + \overline{e_2}) \cdot (\overline{e_1} + \overline{e_2}) = 3$.
$|\overline{e_1}|^2 + |\overline{e_2}|^2 + 2(\overline{e_1} \cdot \overline{e_2}) = 3$.
$1 + 1 + 2(\overline{e_1} \cdot \overline{e_2}) = 3$.
$2(\overline{e_1} \cdot \overline{e_2}) = 1 \implies \overline{e_1} \cdot \overline{e_2} = \frac{1}{2}$.
Now,we calculate the dot product $(2 \overline{e_1} - 5 \overline{e_2}) \cdot (3 \overline{e_1} + \overline{e_2})$:
$= 6(\overline{e_1} \cdot \overline{e_1}) + 2(\overline{e_1} \cdot \overline{e_2}) - 15(\overline{e_2} \cdot \overline{e_1}) - 5(\overline{e_2} \cdot \overline{e_2})$.
$= 6|\overline{e_1}|^2 - 13(\overline{e_1} \cdot \overline{e_2}) - 5|\overline{e_2}|^2$.
$= 6(1) - 13(\frac{1}{2}) - 5(1)$.
$= 6 - 6.5 - 5 = -5.5 = -\frac{11}{2}$.

(B) Given that,$\vec{a}=\frac{3}{2} \hat{k}$ and $\vec{b}=\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{2} = \hat{i}+\hat{j}-\frac{1}{2} \hat{k}$.
First,calculate $\vec{a}+\vec{b} = \hat{i}+\hat{j}+(\frac{3}{2}-\frac{1}{2}) \hat{k} = \hat{i}+\hat{j}+\hat{k}$.
Next,calculate $\vec{a}-\vec{b} = -\hat{i}-\hat{j}+(\frac{3}{2}+\frac{1}{2}) \hat{k} = -\hat{i}-\hat{j}+2 \hat{k}$.
Let $\vec{u} = \vec{a}+\vec{b}$ and $\vec{v} = \vec{a}-\vec{b}$.
The dot product $\vec{u} \cdot \vec{v} = (1)(-1) + (1)(-1) + (1)(2) = -1 - 1 + 2 = 0$.
Since the dot product is $0$,the vectors are perpendicular.
Therefore,the angle $\theta$ between them is $90^{\circ}$.

(A) The position vectors of points $u$ and $v$ are given as $(2, 1)$ and $(3, -5)$ respectively.
The equation of the line passing through points $(x_1, y_1) = (2, 1)$ and $(x_2, y_2) = (3, -5)$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
Substituting the values: $y - 1 = \frac{-5 - 1}{3 - 2}(x - 2) \Rightarrow y - 1 = -6(x - 2) \Rightarrow y - 1 = -6x + 12 \Rightarrow 6x + y = 13$.
Now,we check which points satisfy the equation $6x + y = 13$:
For $P\left(\frac{5}{2}, -2\right)$: $6\left(\frac{5}{2}\right) + (-2) = 15 - 2 = 13$. (Satisfies)
For $Q\left(\frac{7}{3}, -1\right)$: $6\left(\frac{7}{3}\right) + (-1) = 14 - 1 = 13$. (Satisfies)
For $R\left(\frac{9}{4}, 0\right)$: $6\left(\frac{9}{4}\right) + 0 = \frac{27}{2} = 13.5 \neq 13$. (Does not satisfy)
Thus,only points $P$ and $Q$ lie on the line.

(D) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
Consider the expression $|a - b|^2$:
$|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)$
Since $a \cdot b = |a||b| \cos \theta = 1 \times 1 \times \cos \theta = \cos \theta$,we have:
$|a - b|^2 = 1^2 + 1^2 - 2 \cos \theta$
$|a - b|^2 = 2 - 2 \cos \theta$
$|a - b|^2 = 2(1 - \cos \theta)$
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$:
$|a - b|^2 = 2(2 \sin^2 \frac{\theta}{2})$
$|a - b|^2 = 4 \sin^2 \frac{\theta}{2}$
Taking the square root on both sides:
$|a - b| = 2 \sin \frac{\theta}{2}$
Therefore,$\sin \frac{\theta}{2} = \frac{1}{2} |a - b|$.

(D) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$.
Also,$a+2b$ is a unit vector,so $|a+2b| = 1$.
Squaring both sides,we get $|a+2b|^2 = 1^2 = 1$.
Expanding the dot product,we have $|a|^2 + 4|b|^2 + 4(a \cdot b) = 1$.
Substituting the values,$1^2 + 4(1)^2 + 4|a||b| \cos \theta = 1$.
$1 + 4 + 4 \cos \theta = 1$.
$5 + 4 \cos \theta = 1$,which implies $4 \cos \theta = -4$,so $\cos \theta = -1$.
This means $\theta = \pi$.
Now,we evaluate the expression $\sin \theta + \cos^3 \theta + \tan^5 \theta$.
Substituting $\theta = \pi$: $\sin \pi + \cos^3 \pi + \tan^5 \pi = 0 + (-1)^3 + 0 = -1$.

(D) Given that $a$,$b$,and $c$ are non-zero vectors such that no two of these are collinear.
Since $a + 2b$ is collinear with $c$,we have $a + 2b = m c$ for some non-zero scalar $m$ $(i)$.
Since $b + 3c$ is collinear with $a$,we have $b + 3c = n a$ for some non-zero scalar $n$ $(ii)$.
From $(ii)$,we have $b = n a - 3c$.
Substitute this into $(i)$:
$a + 2(n a - 3c) = m c$
$a + 2n a - 6c = m c$
$(1 + 2n) a = (m + 6) c$
Since $a$ and $c$ are non-collinear,the coefficients must be zero:
$1 + 2n = 0 \Rightarrow n = -\frac{1}{2}$
$m + 6 = 0 \Rightarrow m = -6$
Substituting $m = -6$ into equation $(i)$:
$a + 2b = -6c$
$a + 2b + 6c = 0$.

(B) Given $|u+v|=|u-v|$.
Squaring both sides,we get:
$|u+v|^2 = |u-v|^2$
$(u+v) \cdot (u+v) = (u-v) \cdot (u-v)$
$|u|^2 + |v|^2 + 2(u \cdot v) = |u|^2 + |v|^2 - 2(u \cdot v)$
$2(u \cdot v) = -2(u \cdot v)$
$4(u \cdot v) = 0$
$u \cdot v = 0$
Since the dot product of two non-zero vectors is zero,the vectors $u$ and $v$ are perpendicular.
Hence,option $B$ is correct.

(D) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$.
Also,$a+b$ is a unit vector,so $|a+b| = 1$.
Squaring both sides,we get $|a+b|^2 = 1^2 = 1$.
Using the property $|a+b|^2 = |a|^2 + |b|^2 + 2(a \cdot b)$,we have:
$1^2 + 1^2 + 2(a \cdot b) = 1$
$1 + 1 + 2(a \cdot b) = 1$
$2 + 2(a \cdot b) = 1$
$2(a \cdot b) = -1$
$a \cdot b = -\frac{1}{2}$.
Since $a \cdot b = |a||b| \cos \theta$,where $\theta$ is the angle between $a$ and $b$:
$(1)(1) \cos \theta = -\frac{1}{2}$
$\cos \theta = -\frac{1}{2}$.
Therefore,$\theta = 120^{\circ}$.
Hence,option $D$ is correct.

(B) Let $p$ be the unit vector along the angular bisector of $a$ and $b$. Since $a$ and $b$ are unit vectors,the vector $a+b$ lies along the angular bisector of $a$ and $b$.
Thus,$p = \lambda(a+b)$ for some scalar $\lambda$.
Since $p$ is a unit vector,$|p| = 1$.
$|p|^2 = \lambda^2 |a+b|^2 = 1$.
$|a+b|^2 = |a|^2 + |b|^2 + 2(a \cdot b) = 1 + 1 + 2 \cos \theta = 2(1 + \cos \theta) = 4 \cos^2 (\theta / 2)$.
So,$\lambda^2 (4 \cos^2 (\theta / 2)) = 1$.
$\lambda = \frac{1}{2 \cos (\theta / 2)}$.
Therefore,$p = \frac{a+b}{2 \cos (\theta / 2)}$.

(B) Given vectors are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$,and $\vec{c} = \lambda \hat{i} + 2 \hat{j} + 3 \hat{k}$.
Let $\vec{v} = \vec{b} + \vec{c} = (\lambda + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}$.
The unit vector along $\vec{v}$ is $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{(\lambda + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{(\lambda + 2)^2 + 6^2 + (-2)^2}} = \frac{(\lambda + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{(\lambda + 2)^2 + 40}}$.
The scalar product of $\vec{a}$ and $\hat{u}$ is $1$,so $\vec{a} \cdot \hat{u} = 1$.
$\frac{(\lambda + 2)(1) + (6)(1) + (-2)(1)}{\sqrt{(\lambda + 2)^2 + 40}} = 1$.
$\frac{\lambda + 2 + 6 - 2}{\sqrt{(\lambda + 2)^2 + 40}} = 1$.
$\frac{\lambda + 6}{\sqrt{(\lambda + 2)^2 + 40}} = 1$.
$\lambda + 6 = \sqrt{(\lambda + 2)^2 + 40}$.
Squaring both sides: $(\lambda + 6)^2 = (\lambda + 2)^2 + 40$.
$\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 4 + 40$.
$12\lambda + 36 = 4\lambda + 44$.
$8\lambda = 8$.
$\lambda = 1$.

(D) Let the origin be at $A$. Then $\vec{A} = 0$,$\vec{B} = u$,and $\vec{C} = v$.
Since $D$ is the midpoint of $BC$,the position vector of $D$ is $\vec{D} = \frac{u+v}{2}$.
In $\triangle ABD$,let $M$ be the midpoint of $AD$. The median through vertex $B$ is the segment $BM$.
The position vector of $M$ is $\vec{M} = \frac{\vec{A} + \vec{D}}{2} = \frac{0 + \frac{u+v}{2}}{2} = \frac{u+v}{4}$.
The vector $\overrightarrow{BM} = \vec{M} - \vec{B} = \frac{u+v}{4} - u = \frac{u+v-4u}{4} = \frac{v-3u}{4}$.
The length of the median is $|\overrightarrow{BM}| = |\frac{v-3u}{4}| = \frac{|v-3u|}{4}$.

(A) Let the required vector $d$ be in the plane of $b$ and $c$. Thus,$d = b + \lambda c$ for some scalar $\lambda$.
Substituting the given vectors:
$d = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} + \hat{j} - 2\hat{k}) = (1 + \lambda)\hat{i} + (2 + \lambda)\hat{j} - (1 + 2\lambda)\hat{k}$.
The projection of $d$ on $a$ is given by $\frac{|d \cdot a|}{|a|} = \sqrt{\frac{2}{3}}$.
First,calculate $a \cdot d$:
$a \cdot d = 2(1 + \lambda) - 1(2 + \lambda) + 1(-1 - 2\lambda) = 2 + 2\lambda - 2 - \lambda - 1 - 2\lambda = -\lambda - 1$.
Next,calculate $|a| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
So,$\frac{|-\lambda - 1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}}$.
Thus,$|-\lambda - 1| = 2$,which implies $-\lambda - 1 = 2$ or $-\lambda - 1 = -2$.
If $-\lambda - 1 = 2$,then $\lambda = -3$.
Substituting $\lambda = -3$ into $d$:
$d = (1 - 3)\hat{i} + (2 - 3)\hat{j} - (1 - 6)\hat{k} = -2\hat{i} - \hat{j} + 5\hat{k}$.
This matches option $A$.

(B) The equation of the line passing through points $A(a+2b-5c)$ and $B(-a-2b-3c)$ is given by $r = A + \lambda_1(B-A)$.
$r = (a+2b-5c) + \lambda_1(-2a-4b+2c) = (a+2b-5c) + \lambda_1'(a+2b-c)$ where $\lambda_1' = -2\lambda_1$.
For simplicity,let $r = (a+2b-5c) + \lambda_1(a+2b-c)$ $(i)$.
The equation of the line passing through points $C(-4c)$ and $D(6a-4b+4c)$ is $r = C + \lambda_2(D-C)$.
$r = -4c + \lambda_2(6a-4b+8c) = -4c + 2\lambda_2(3a-2b+4c)$ (ii).
Equating $(i)$ and (ii) for the intersection point:
$(1+\lambda_1)a + (2+2\lambda_1)b + (-5-\lambda_1)c = (6\lambda_2)a + (-4\lambda_2)b + (-4+8\lambda_2)c$.
Comparing coefficients of $a, b, c$:
$1+\lambda_1 = 6\lambda_2$,$2+2\lambda_1 = -4\lambda_2$,$-5-\lambda_1 = -4+8\lambda_2$.
From the first two equations: $2(1+\lambda_1) = 2+2\lambda_1 = -4\lambda_2$,so $2(6\lambda_2) = -4\lambda_2 \implies 16\lambda_2 = 0 \implies \lambda_2 = 0$.
Then $\lambda_1 = -1$. The intersection point is $r = -4c + 0 = -4c$.
Checking the options,for $\mu = -3$,option $B$ gives $r = (3a+6b-c) - 3(a+2b+c) = 3a+6b-c-3a-6b-3c = -4c$. Thus,the line in option $B$ passes through the intersection point.

(D) Let the position vectors of $P, Q, R$ be $\vec{p}, \vec{q}, \vec{r}$ respectively.
Since $M$ is the mid-point of $QR$,the position vector of $M$ is $\vec{m} = \frac{\vec{q} + \vec{r}}{2}$.
Since $C$ is the mid-point of $PM$,the position vector of $C$ is $\vec{c} = \frac{\vec{p} + \vec{m}}{2} = \frac{\vec{p} + \frac{\vec{q} + \vec{r}}{2}}{2} = \frac{2\vec{p} + \vec{q} + \vec{r}}{4}$.
The line $QC$ passes through $Q$ and $C$. The vector equation of line $QC$ is $\vec{r} = \vec{q} + t(\vec{c} - \vec{q}) = \vec{q} + t(\frac{2\vec{p} + \vec{q} + \vec{r}}{4} - \vec{q}) = \vec{q} + t(\frac{2\vec{p} - 3\vec{q} + \vec{r}}{4})$.
The line $PR$ passes through $P$ and $R$. The vector equation of line $PR$ is $\vec{r} = \vec{p} + s(\vec{r} - \vec{p}) = (1-s)\vec{p} + s\vec{r}$.
Equating the coefficients of $\vec{p}, \vec{q}, \vec{r}$ for the intersection point $N$,we find $t = \frac{4}{3}$ and $s = \frac{1}{3}$.
Thus,$\vec{n} = (1 - \frac{1}{3})\vec{p} + \frac{1}{3}\vec{r} = \frac{2\vec{p} + \vec{r}}{3}$.
Since $\vec{n} = \vec{q} + \frac{4}{3}(\frac{2\vec{p} - 3\vec{q} + \vec{r}}{4}) = \vec{q} + \frac{2\vec{p} - 3\vec{q} + \vec{r}}{3} = \frac{2\vec{p} + \vec{r}}{3}$,the point $N$ lies on $PR$.
Using the section formula,$N$ divides $QC$ in the ratio $t : (1-t) = \frac{4}{3} : (1 - \frac{4}{3}) = \frac{4}{3} : -\frac{1}{3} = 4 : -1$.
Thus,$\overrightarrow{QN} = 4\overrightarrow{QC}$,which implies $\frac{|\overrightarrow{QN}|}{|\overrightarrow{CN}|} = 4$.