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

(B) Let the first term of the $GP$ be $u$ and the common ratio be $z$.
Then,$T_p = u z^{p-1} = a$,$T_q = u z^{q-1} = b$,and $T_r = u z^{r-1} = c$.
Taking the logarithm on both sides:
$\log a = \log u + (p-1) \log z$
$\log b = \log u + (q-1) \log z$
$\log c = \log u + (r-1) \log z$
Let $\vec{A} = (\log a^2) i + (\log b^2) j + (\log c^2) k = 2(\log a) i + 2(\log b) j + 2(\log c) k$.
Let $\vec{B} = (q-r) i + (r-p) j + (p-q) k$.
The dot product $\vec{A} \cdot \vec{B} = 2 [(\log a)(q-r) + (\log b)(r-p) + (\log c)(p-q)]$.
Substituting $\log a, \log b, \log c$ in terms of $p, q, r$:
$(\log a)(q-r) + (\log b)(r-p) + (\log c)(p-q) = [\log u + (p-1)\log z](q-r) + [\log u + (q-1)\log z](r-p) + [\log u + (r-1)\log z](p-q)$.
$= \log u (q-r+r-p+p-q) + \log z [(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$.
$= \log u (0) + \log z [pq - pr - q + r + qr - qp - r + p + rp - rq - p + q] = 0$.
Since the dot product is $0$,the vectors are perpendicular.
Therefore,the angle $\theta = \frac{\pi}{2}$.

(D) Given,$\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$.
Let the diagonals be $\vec{d_1}$ and $\vec{d_2}$.
In a parallelogram $ABCD$,the diagonals are $\vec{AC} = \vec{AB} + \vec{BC}$ and $\vec{BD} = \vec{BC} - \vec{AB}$.
$\vec{AC} = (3\hat{i} - 2\hat{j} + 2\hat{k}) + (\hat{i} - 2\hat{k}) = 4\hat{i} - 2\hat{j}$.
$\vec{BD} = (\hat{i} - 2\hat{k}) - (3\hat{i} - 2\hat{j} + 2\hat{k}) = -2\hat{i} + 2\hat{j} - 4\hat{k}$.
The angle $\theta$ between the diagonals is given by $\cos \theta = \frac{|\vec{AC} \cdot \vec{BD}|}{|\vec{AC}| |\vec{BD}|}$.
$\vec{AC} \cdot \vec{BD} = (4)(-2) + (-2)(2) + (0)(-4) = -8 - 4 = -12$.
$|\vec{AC}| = \sqrt{4^2 + (-2)^2 + 0^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.
$|\vec{BD}| = \sqrt{(-2)^2 + 2^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
$\cos \theta = \frac{|-12|}{(2\sqrt{5})(2\sqrt{6})} = \frac{12}{4\sqrt{30}} = \frac{3}{\sqrt{30}} = \sqrt{\frac{9}{30}} = \sqrt{\frac{3}{10}}$.
Thus,$\theta = \cos^{-1}\left(\sqrt{\frac{3}{10}}\right)$.
Since this value is not among the options,the correct answer is $D$.

(A) Two vectors are orthogonal if their dot product is equal to $0$.
Given vectors are $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$.
Taking the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (-2x)(3x) + (-3y)(2y) = 0$.
This simplifies to $1 - 6x^2 - 6y^2 = 0$.
Rearranging the terms,we get $6x^2 + 6y^2 = 1$,which implies $x^2 + y^2 = \frac{1}{6}$.
This is the equation of a circle with center at the origin $(0, 0)$ and radius $\frac{1}{\sqrt{6}}$.
Thus,the locus of the point $(x, y)$ is a circle.

(A) The angle $\theta$ between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$.
Since the angle $\theta > 90^{\circ}$,we have $\cos \theta < 0$,which implies $\overrightarrow{a} \cdot \overrightarrow{b} < 0$.
Calculating the dot product: $\overrightarrow{a} \cdot \overrightarrow{b} = (\lambda)(\lambda) + (-7)(1) + (3)(2\lambda) = \lambda^2 - 7 + 6\lambda$.
Setting the dot product to be less than zero: $\lambda^2 + 6\lambda - 7 < 0$.
Factoring the quadratic expression: $(\lambda + 7)(\lambda - 1) < 0$.
Solving the inequality,we find that $\lambda$ must lie between the roots $-7$ and $1$.
Therefore,$-7 < \lambda < 1$.

(D) Given that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.
We can write this as $\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$.
Squaring both sides,we get $(\overrightarrow{a}+\overrightarrow{b})^2=(-\overrightarrow{c})^2$.
Expanding the dot product,we have $|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2(\overrightarrow{a} \cdot \overrightarrow{b})=|\overrightarrow{c}|^2$.
Since $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \cos \theta$,where $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$,we have $|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}| \cos \theta=|\overrightarrow{c}|^2$.
Substituting the given values $3^2+4^2+2(3)(4) \cos \theta = (\sqrt{37})^2$.
$9+16+24 \cos \theta = 37$.
$25+24 \cos \theta = 37$.
$24 \cos \theta = 37-25 = 12$.
$\cos \theta = \frac{12}{24} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(B) The orthogonal projection of a vector $a$ onto a vector $b$ is defined as the component of $a$ along the direction of $b$.
The formula for the projection of $a$ on $b$ is given by:
$\text{Proj}_{b} a = \left( \frac{a \cdot b}{|b|^2} \right) b$
This represents a vector that is parallel to $b$ and has a magnitude equal to the scalar projection of $a$ on $b$.

(D) Given that,$|\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}|$.
Using the definitions of cross product and dot product:
$|\vec{a}| |\vec{b}| \sin \theta = |\vec{a}| |\vec{b}| \cos \theta$.
Dividing both sides by $|\vec{a}| |\vec{b}| \cos \theta$ (assuming $\vec{a}, \vec{b} \neq 0$ and $\cos \theta \neq 0$):
$\frac{\sin \theta}{\cos \theta} = 1$.
$\tan \theta = 1$.
Since $\theta$ is the angle between two vectors,$0 \leq \theta \leq \pi$.
Therefore,$\theta = \frac{\pi}{4}$.

(D) Let $\vec{u} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{v} = a \hat{i}+4 \hat{j}+b \hat{k}$.
Given $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{2}{3}$.
The dot product $\vec{u} \cdot \vec{v} = (2)(a) + (-1)(4) + (2)(b) = 2a - 4 + 2b = 2(a+b-2)$.
The magnitudes are $|\vec{u}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$ and $|\vec{v}| = \sqrt{a^2 + 4^2 + b^2} = \sqrt{a^2+b^2+16}$.
Substituting these into the formula: $\frac{2}{3} = \frac{2(a+b-2)}{3 \sqrt{a^2+b^2+16}}$.
Simplifying,we get $\sqrt{a^2+b^2+16} = a+b-2$.
Squaring both sides: $a^2+b^2+16 = (a+b-2)^2 = (a+b)^2 - 4(a+b) + 4$.
$a^2+b^2+16 = a^2+b^2+2ab - 4a - 4b + 4$.
$16 = 2ab - 4a - 4b + 4$.
$12 = 2ab - 4(a+b)$.
$6 = ab - 2(a+b)$.
$ab = 2(a+b) + 6 = 2(a+b+3)$.
Thus,$2(a+b+3) = ab$.

(B) The volume of a parallelepiped formed by coterminous edges $\bar{a}, \bar{b}, \bar{c}$ is given by the scalar triple product $|[\bar{a} \bar{b} \bar{c}]|$.
Since $\bar{a}$ and $\bar{b}$ determine the base,the area of the base is given by $|\bar{a} \times \bar{b}|$.
The volume of a parallelepiped is also defined as $\text{Volume} = \text{Area of base} \times \text{Height}$.
Therefore,$\text{Height} = \frac{\text{Volume}}{\text{Area of base}} = \frac{|[\bar{a} \bar{b} \bar{c}]|}{|\bar{a} \times \bar{b}|}$.

(D) Given: $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$.
Since $\vec{a} \parallel \vec{c}$,we can write $\vec{c} = k \vec{a}$ for some scalar $k$.
Given $\vec{b} = \vec{c} - \vec{d}$,we have $\vec{d} = \vec{c} - \vec{b} = k \vec{a} - \vec{b}$.
Since $\vec{a} \perp \vec{d}$,their dot product is zero: $\vec{a} \cdot \vec{d} = 0$.
$\vec{a} \cdot (k \vec{a} - \vec{b}) = 0 \Rightarrow k |\vec{a}|^2 - \vec{a} \cdot \vec{b} = 0$.
Calculate $|\vec{a}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$.
Calculate $\vec{a} \cdot \vec{b} = (2)(0) + (-1)(2) + (1)(-3) = 0 - 2 - 3 = -5$.
Substituting these values: $k(6) - (-5) = 0 \Rightarrow 6k = -5 \Rightarrow k = -\frac{5}{6}$.
Thus,$\vec{c} = -\frac{5}{6} \vec{a}$.
Then $\vec{d} = \vec{c} - \vec{b} = -\frac{5}{6} \vec{a} - \vec{b}$.
Finally,$\vec{c} + \vec{d} = (-\frac{5}{6} \vec{a}) + (-\frac{5}{6} \vec{a} - \vec{b}) = -\frac{10}{6} \vec{a} - \vec{b} = -\frac{5}{3} \vec{a} - \vec{b} = -\frac{1}{3}(5 \vec{a} + 3 \vec{b})$.

(C) The orthogonal projection of $a$ on $b$ is given by $x = \frac{a \cdot b}{|b|^2} b$.
The orthogonal projection of $b$ on $a$ is given by $y = \frac{a \cdot b}{|a|^2} a$.
Given $a = \hat{i} + 2\hat{j} - 2\hat{k}$ and $b = 2\hat{i} - \hat{j} - 2\hat{k}$.
Calculate the dot product: $a \cdot b = (1)(2) + (2)(-1) + (-2)(-2) = 2 - 2 + 4 = 4$.
Calculate the magnitudes squared: $|a|^2 = 1^2 + 2^2 + (-2)^2 = 1 + 4 + 4 = 9$ and $|b|^2 = 2^2 + (-1)^2 + (-2)^2 = 4 + 1 + 4 = 9$.
Thus,$x - y = \frac{a \cdot b}{|b|^2} b - \frac{a \cdot b}{|a|^2} a = \frac{4}{9} b - \frac{4}{9} a = \frac{4}{9} (b - a)$.
Calculate $b - a = (2\hat{i} - \hat{j} - 2\hat{k}) - (\hat{i} + 2\hat{j} - 2\hat{k}) = \hat{i} - 3\hat{j}$.
Then $x - y = \frac{4}{9} (\hat{i} - 3\hat{j})$.
Finally,$|x - y| = \frac{4}{9} \sqrt{1^2 + (-3)^2} = \frac{4}{9} \sqrt{1 + 9} = \frac{4}{9} \sqrt{10}$.

(B) Given that $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors of magnitude $K$,we have $\bar{a} \cdot \bar{b} = \bar{b} \cdot \bar{c} = \bar{c} \cdot \bar{a} = 0$ and $|\bar{a}| = |\bar{b}| = |\bar{c}| = K$,so $\bar{a} \cdot \bar{a} = \bar{b} \cdot \bar{b} = \bar{c} \cdot \bar{c} = K^2$.
Using the vector triple product formula $\bar{u} \times (\bar{v} \times \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}$,the given equation becomes:
$(\bar{a} \cdot \bar{a})(\bar{r}-\bar{b}) - (\bar{a} \cdot (\bar{r}-\bar{b}))\bar{a} + (\bar{b} \cdot \bar{b})(\bar{r}-\bar{c}) - (\bar{b} \cdot (\bar{r}-\bar{c}))\bar{b} + (\bar{c} \cdot \bar{c})(\bar{r}-\bar{a}) - (\bar{c} \cdot (\bar{r}-\bar{a}))\bar{c} = \bar{0}$.
Substituting the dot products:
$K^2(\bar{r}-\bar{b}) - (\bar{a} \cdot \bar{r})\bar{a} + K^2(\bar{r}-\bar{c}) - (\bar{b} \cdot \bar{r})\bar{b} + K^2(\bar{r}-\bar{a}) - (\bar{c} \cdot \bar{r})\bar{c} = \bar{0}$.
$3K^2\bar{r} - K^2(\bar{a}+\bar{b}+\bar{c}) - ((\bar{a} \cdot \bar{r})\bar{a} + (\bar{b} \cdot \bar{r})\bar{b} + (\bar{c} \cdot \bar{r})\bar{c}) = \bar{0}$.
Since $\bar{a}, \bar{b}, \bar{c}$ are orthogonal,any vector $\bar{r}$ can be written as $\bar{r} = \frac{\bar{a} \cdot \bar{r}}{K^2}\bar{a} + \frac{\bar{b} \cdot \bar{r}}{K^2}\bar{b} + \frac{\bar{c} \cdot \bar{r}}{K^2}\bar{c}$.
Thus,$(\bar{a} \cdot \bar{r})\bar{a} + (\bar{b} \cdot \bar{r})\bar{b} + (\bar{c} \cdot \bar{r})\bar{c} = K^2\bar{r}$.
Substituting this back:
$3K^2\bar{r} - K^2(\bar{a}+\bar{b}+\bar{c}) - K^2\bar{r} = \bar{0}$.
$2K^2\bar{r} = K^2(\bar{a}+\bar{b}+\bar{c})$.
$\bar{r} = \frac{\bar{a}+\bar{b}+\bar{c}}{2}$.

(B) Given,$\vec{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\vec{a}=3 \hat{i}+4 \hat{j}-5 \hat{k}$.
First,calculate the dot products:
$\vec{b} \cdot \vec{b} = (2)^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6$.
$\vec{b} \cdot \vec{a} = (2)(3) + (-1)(4) + (-1)(-5) = 6 - 4 + 5 = 7$.
Using the vector triple product formula $\vec{b} \times(\vec{a} \times \vec{b})=(\vec{b} \cdot \vec{b}) \vec{a}-(\vec{b} \cdot \vec{a}) \vec{b}$,we get:
$\vec{b} \times(\vec{a} \times \vec{b}) = 6 \vec{a} - 7 \vec{b}$.
We can rewrite this as $\frac{\vec{a} - (7/6) \vec{b}}{1/6}$.
Comparing this with the given expression $\frac{\vec{a}-k \vec{b}}{l}$,we find $k = 7/6$ and $l = 1/6$.
Now,calculate $|\vec{b}| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6}$.
Finally,$\frac{k}{l|\vec{b}|} = \frac{7/6}{(1/6) \times \sqrt{6}} = \frac{7}{\sqrt{6}}$.
This value represents the scalar projection of $\vec{a}$ on $\vec{b}$,which is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{7}{\sqrt{6}}$.

(C) The component of vector $\bar{b}$ perpendicular to $\bar{a}$ is given by $\bar{b} - \text{proj}_{\bar{a}} \bar{b} = \bar{b} - \left( \frac{\bar{b} \cdot \bar{a}}{|\bar{a}|^2} \right) \bar{a}$.
First,calculate the dot product $\bar{b} \cdot \bar{a} = (1)(1) + (\sqrt{11})(\sqrt{11}) + (-10)(-2) = 1 + 11 + 20 = 32$.
Next,calculate the magnitude squared $|\bar{a}|^2 = (1)^2 + (\sqrt{11})^2 + (-2)^2 = 1 + 11 + 4 = 16$.
Now,find the projection: $\frac{\bar{b} \cdot \bar{a}}{|\bar{a}|^2} \bar{a} = \frac{32}{16} \bar{a} = 2 \bar{a} = 2(\bar{i} + \sqrt{11} \bar{j} - 2 \bar{k}) = 2 \bar{i} + 2\sqrt{11} \bar{j} - 4 \bar{k}$.
Finally,the perpendicular component is $\bar{b} - 2 \bar{a} = (\bar{i} + \sqrt{11} \bar{j} - 10 \bar{k}) - (2 \bar{i} + 2\sqrt{11} \bar{j} - 4 \bar{k}) = (1-2) \bar{i} + (\sqrt{11} - 2\sqrt{11}) \bar{j} + (-10 + 4) \bar{k} = -\bar{i} - \sqrt{11} \bar{j} - 6 \bar{k} = -(\bar{i} + \sqrt{11} \bar{j} + 6 \bar{k})$.
Thus,the correct option is $C$.

(D) Let $\bar{r} = x\bar{i} + y\bar{j} + z\bar{k}$.
Given the dot products:
$x + 2y + 3z = 0$ $(1)$
$2x - 3y + z = -2$ $(2)$
$3x + y - 2z = 6$ $(3)$
Solving this system of linear equations:
From $(1)$,$x = -2y - 3z$.
Substitute into $(2)$: $2(-2y - 3z) - 3y + z = -2 \implies -7y - 5z = -2 \implies 7y + 5z = 2$ $(4)$
Substitute into $(3)$: $3(-2y - 3z) + y - 2z = 6 \implies -5y - 11z = 6$ $(5)$
Multiply $(4)$ by $5$ and $(5)$ by $7$: $35y + 25z = 10$ and $-35y - 77z = 42$.
Adding these: $-52z = 52 \implies z = -1$.
Substitute $z = -1$ into $(4)$: $7y - 5 = 2 \implies 7y = 7 \implies y = 1$.
Substitute $y = 1, z = -1$ into $(1)$: $x + 2(1) + 3(-1) = 0 \implies x - 1 = 0 \implies x = 1$.
Thus,$\bar{r} = \bar{i} + \bar{j} - \bar{k}$.
Finally,calculate $\bar{r} \cdot (3\bar{i} + \bar{j} + \bar{k}) = (1)(3) + (1)(1) + (-1)(1) = 3 + 1 - 1 = 3$.

(B) Given $|\bar{a}| = |\bar{b}| = \sqrt{6}$ and $\bar{a} \cdot \bar{b} = -1$.
We know that $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos(\theta)$.
Substituting the values: $-1 = (\sqrt{6})(\sqrt{6}) \cos(\theta) = 6 \cos(\theta)$.
So,$\cos(\theta) = -\frac{1}{6}$.
Since $\sin^2(\theta) = 1 - \cos^2(\theta)$,we have $\sin^2(\theta) = 1 - (-\frac{1}{6})^2 = 1 - \frac{1}{36} = \frac{35}{36}$.
Thus,$\sin(\theta) = \frac{\sqrt{35}}{6}$.
Now,$|\bar{a} \times \bar{b}| = |\bar{a}| |\bar{b}| \sin(\theta) = (\sqrt{6})(\sqrt{6}) \sin(\theta) = 6 \sin(\theta)$.
Therefore,$|\bar{a} \times \bar{b}| \sin(\theta) = 6 \sin(\theta) \cdot \sin(\theta) = 6 \sin^2(\theta) = 6 \times \frac{35}{36} = \frac{35}{6}$.

(A) Let the position vectors of vertices $A, B, D$ be $\vec{0}, \vec{b}, \vec{d}$ respectively. Then the position vector of $C$ is $\vec{b} + \vec{d}$.
Since $P$ divides $AD$ in the ratio $3:1$,the position vector of $P$ is $\vec{p} = \frac{3\vec{d} + 1\vec{0}}{3+1} = \frac{3}{4}\vec{d}$.
Let $Q$ divide $BP$ in the ratio $k:1$ and $AC$ in the ratio $m:1$.
The position vector of $Q$ on $BP$ is $\vec{q} = \frac{k\vec{p} + 1\vec{b}}{k+1} = \frac{k(\frac{3}{4}\vec{d}) + \vec{b}}{k+1} = \frac{1}{k+1}\vec{b} + \frac{3k}{4(k+1)}\vec{d}$.
The position vector of $Q$ on $AC$ is $\vec{q} = \frac{m\vec{c} + 1\vec{a}}{m+1} = \frac{m(\vec{b} + \vec{d})}{m+1} = \frac{m}{m+1}\vec{b} + \frac{m}{m+1}\vec{d}$.
Comparing the coefficients of $\vec{b}$ and $\vec{d}$:
$\frac{1}{k+1} = \frac{m}{m+1}$ and $\frac{3k}{4(k+1)} = \frac{m}{m+1}$.
Thus,$\frac{1}{k+1} = \frac{3k}{4(k+1)} \Rightarrow 4 = 3k \Rightarrow k = \frac{4}{3}$.
Substituting $k = \frac{4}{3}$ into $\frac{m}{m+1} = \frac{1}{k+1} = \frac{1}{4/3 + 1} = \frac{1}{7/3} = \frac{3}{7}$.
$7m = 3m + 3 \Rightarrow 4m = 3 \Rightarrow m = \frac{3}{4}$.
Therefore,$AQ:QC = m:1 = \frac{3}{4}:1 = 3:4$.

(D) Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Since $\vec{a}$ lies in the plane containing $\vec{b}$ and $\vec{c}$,$\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar. Thus,$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
Calculating $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2+1) - \hat{j}(1-2) + \hat{k}(-1-4) = 3\hat{i} + \hat{j} - 5\hat{k}$.
So,$(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (3\hat{i} + \hat{j} - 5\hat{k}) = 0 \Rightarrow 3a_1 + a_2 - 5a_3 = 0 \dots (i)$.
Since $\vec{a}$ is perpendicular to $\hat{i} + \hat{j} + 3\hat{k}$,we have $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 0 \Rightarrow a_1 + a_2 + 3a_3 = 0 \dots (ii)$.
The projection of $\vec{a}$ on $\vec{b}$ is $3\sqrt{6}$,so $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 3\sqrt{6}$.
$\frac{a_1 + 2a_2 + a_3}{\sqrt{1^2 + 2^2 + 1^2}} = 3\sqrt{6} \Rightarrow a_1 + 2a_2 + a_3 = 3\sqrt{6} \cdot \sqrt{6} = 18 \dots (iii)$.
Solving equations $(i), (ii),$ and $(iii)$:
Subtracting $(ii)$ from $(i)$: $2a_1 - 8a_3 = 0 \Rightarrow a_1 = 4a_3$.
Substituting $a_1 = 4a_3$ into $(ii)$: $4a_3 + a_2 + 3a_3 = 0 \Rightarrow a_2 = -7a_3$.
Substituting into $(iii)$: $4a_3 + 2(-7a_3) + a_3 = 18 \Rightarrow 4a_3 - 14a_3 + a_3 = 18 \Rightarrow -9a_3 = 18 \Rightarrow a_3 = -2$.
Thus,$a_1 = -8$ and $a_2 = 14$.
Therefore,$\vec{a} = -8\hat{i} + 14\hat{j} - 2\hat{k}$.
$|\vec{a}|^2 = (-8)^2 + 14^2 + (-2)^2 = 64 + 196 + 4 = 264$.

(B) The vector $\overrightarrow{AB}$ is given by $B - A = (4-3, 4-4, 4-0) = (1, 0, 4)$.
The vector $\overrightarrow{CD}$ is given by $D - C = (1 - (-6), 1 - 2, 2 - 3) = (7, -1, -1)$.
The dot product $\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(7) + (0)(-1) + (4)(-1) = 7 + 0 - 4 = 3$.
The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{1^2 + 0^2 + 4^2} = \sqrt{1 + 0 + 16} = \sqrt{17}$.
The magnitude of $\overrightarrow{CD}$ is $|\overrightarrow{CD}| = \sqrt{7^2 + (-1)^2 + (-1)^2} = \sqrt{49 + 1 + 1} = \sqrt{51}$.
Since $\sqrt{51} = \sqrt{17 \times 3} = \sqrt{17} \sqrt{3}$,we have $|\overrightarrow{CD}| = \sqrt{17} \sqrt{3}$.
The cosine of the angle $\theta$ is given by $\cos \theta = \frac{|\overrightarrow{AB} \cdot \overrightarrow{CD}|}{|\overrightarrow{AB}| |\overrightarrow{CD}|} = \frac{3}{\sqrt{17} \cdot \sqrt{17} \sqrt{3}} = \frac{3}{17 \sqrt{3}}$.

(A) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at point $M$.
$M$ is the midpoint of $AC$,so $M = \left(\frac{3+7}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) = (5, 3, 4)$.
Since $M$ is also the midpoint of $BD$,let $D = (x, y, z)$. Then $\left(\frac{x+1}{2}, \frac{y+4}{2}, \frac{z+6}{2}\right) = (5, 3, 4)$.
Solving for $x, y, z$: $x+1=10 \Rightarrow x=9$; $y+4=6 \Rightarrow y=2$; $z+6=8 \Rightarrow z=2$. Thus,$D = (9, 2, 2)$.
The vector $\vec{DC} = (7-9, 4-2, 5-2) = (-2, 2, 3)$. The direction ratios are $a_1 = -2, b_1 = 2, c_1 = 3$.
The vector $\vec{DB} = (1-9, 4-2, 6-2) = (-8, 2, 4)$. The direction ratios are $a_2 = -8, b_2 = 2, c_2 = 4$.
The angle $\theta$ between $\vec{DC}$ and $\vec{DB}$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
$\cos \theta = \frac{|(-2)(-8) + (2)(2) + (3)(4)|}{\sqrt{(-2)^2 + 2^2 + 3^2} \sqrt{(-8)^2 + 2^2 + 4^2}} = \frac{|16 + 4 + 12|}{\sqrt{4+4+9} \sqrt{64+4+16}} = \frac{32}{\sqrt{17} \sqrt{84}} = \frac{32}{\sqrt{17} \cdot 2\sqrt{21}} = \frac{16}{\sqrt{357}}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{16}{\sqrt{357}}\right)$.

(C) Given position vectors are:
$A = \hat{i}+2\hat{j}+\hat{k}$
$B = 2\hat{i}-\hat{j}+2\hat{k}$
$C = \hat{i}+\hat{j}+2\hat{k}$
First,find the vectors $\overrightarrow{AC}$ and $\overrightarrow{AB}$:
$\overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} = (\hat{i}+\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}+\hat{k}) = 0\hat{i} - \hat{j} + \hat{k}$
$\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} = (2\hat{i}-\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}+\hat{k}) = \hat{i} - 3\hat{j} + \hat{k}$
Now,calculate the cross product $\overrightarrow{AC} \times \overrightarrow{AB}$:
$\overrightarrow{AC} \times \overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -1 & 1 \\ 1 & -3 & 1 \end{vmatrix} = \hat{i}(-1+3) - \hat{j}(0-1) + \hat{k}(0+1) = 2\hat{i} + \hat{j} + \hat{k}$
The magnitude of the cross product is $|\overrightarrow{AC} \times \overrightarrow{AB}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$.
The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{1^2 + (-3)^2 + 1^2} = \sqrt{1+9+1} = \sqrt{11}$.
The perpendicular distance from point $C$ to line $AB$ is given by the formula $d = \frac{|\overrightarrow{AC} \times \overrightarrow{AB}|}{|\overrightarrow{AB}|}$.
$d = \frac{\sqrt{6}}{\sqrt{11}} = \sqrt{\frac{6}{11}}$.

(A) Let $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. Since $a$ is a unit vector,$|a|^2 = a_1^2 + a_2^2 + a_3^2 = 1$.
Now,$a \times \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{i} = -a_2 \hat{k} + a_3 \hat{j}$.
So,$|a \times \hat{i}|^2 = a_2^2 + a_3^2$.
Similarly,$|a \times \hat{j}|^2 = a_1^2 + a_3^2$ and $|a \times \hat{k}|^2 = a_1^2 + a_2^2$.
Adding these,we get $|a \times \hat{i}|^2 + |a \times \hat{j}|^2 + |a \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2)$.
Since $a_1^2 + a_2^2 + a_3^2 = 1$,the sum is $2(1) = 2$.

(D) Since the vector $r$ lies in the plane of $a$ and $b$,it can be expressed as a linear combination: $r = a + tb$ (assuming the vector is not parallel to $b$).
$r = (i + 2j + k) + t(i - j + k) = (1 + t)i + (2 - t)j + (1 + t)k$.
Given that the projection of $r$ on $c$ is $\frac{1}{\sqrt{3}}$,we use the formula: $\frac{r \cdot c}{|c|} = \frac{1}{\sqrt{3}}$.
First,calculate $|c| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$.
Now,calculate the dot product $r \cdot c = ((1 + t)i + (2 - t)j + (1 + t)k) \cdot (i + j - k) = (1 + t) + (2 - t) - (1 + t) = 2 - t$.
Substituting these into the projection formula: $\frac{2 - t}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
This implies $2 - t = 1$,so $t = 1$.
Substituting $t = 1$ back into the expression for $r$: $r = (1 + 1)i + (2 - 1)j + (1 + 1)k = 2i + j + 2k$.

(B) Let the unit vector along the line be $\hat{u} = \cos \alpha \hat{i} + \cos \alpha \hat{j} + \cos \alpha \hat{k}$.
Since $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$,we have $3 \cos^2 \alpha = 1$,which implies $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Thus,the unit vector along the line is $\hat{u} = \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$.
The projection of vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line is given by the magnitude of the dot product $\vec{a} \cdot \hat{u}$.
$|\vec{a} \cdot \hat{u}| = |(4\hat{i} - 3\hat{j} + 2\hat{k}) \cdot \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})|$
$= |\pm \frac{1}{\sqrt{3}}(4 - 3 + 2)|$
$= |\pm \frac{3}{\sqrt{3}}| = \sqrt{3}$.

(B) First,we calculate the vectors $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$.
$2 \overrightarrow{a}-\overrightarrow{c} = 2(-\hat{i}+\hat{j}+2 \hat{k}) - (-2 \hat{i}+\hat{j}+3 \hat{k}) = (-2\hat{i}+2\hat{j}+4\hat{k}) + (2\hat{i}-\hat{j}-3\hat{k}) = \hat{j}+\hat{k}$.
$\overrightarrow{a}+\overrightarrow{b} = (-\hat{i}+\hat{j}+2 \hat{k}) + (2 \hat{i}-\hat{j}-\hat{k}) = \hat{i}+\hat{k}$.
Let $\theta$ be the angle between these two vectors. The cosine of the angle is given by $\cos \theta = \frac{(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k})}{|\hat{j}+\hat{k}| |\hat{i}+\hat{k}|}$.
Calculating the dot product: $(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k}) = (0)(1) + (1)(0) + (1)(1) = 1$.
Calculating the magnitudes: $|\hat{j}+\hat{k}| = \sqrt{0^2+1^2+1^2} = \sqrt{2}$ and $|\hat{i}+\hat{k}| = \sqrt{1^2+0^2+1^2} = \sqrt{2}$.
Thus,$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(B) For Assertion $(A)$: The lines are $\bar{r}=\bar{a}+t\bar{b}$ and $\bar{r}=\bar{b}+s\bar{a}$. These lines pass through points with position vectors $\bar{a}$ and $\bar{b}$ respectively,and are parallel to vectors $\bar{b}$ and $\bar{a}$. Since both lines pass through the point with position vector $\bar{a}+\bar{b}$ (by setting $t=1$ in the first and $s=1$ in the second),they intersect. Thus,$(A)$ is true.
For Reason $(R)$: The shortest distance between two skew lines $\bar{r}=\bar{p}+t\bar{q}$ and $\bar{r}=\bar{c}+s\bar{d}$ is given by the formula $d = \frac{|(\bar{p}-\bar{c}) \cdot (\bar{q} \times \bar{d})|}{|\bar{q} \times \bar{d}|}$. This is indeed the length of the projection of the vector $(\bar{p}-\bar{c})$ onto the vector $(\bar{q} \times \bar{d})$. Thus,$(R)$ is true.
However,the intersection of the lines in $(A)$ is a specific property of these lines,while $(R)$ provides a general formula for the shortest distance between skew lines. Therefore,$(R)$ is not the correct explanation for $(A)$.
The correct option is $(B)$.

(D) First,we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$:
$\overrightarrow{AB} = B - A = (3-1, 4-(-1), -2-2) = (2, 5, -4)$
$\overrightarrow{CD} = D - C = (3-0, 5-3, 6-2) = (3, 2, 4)$
Let $\theta$ be the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$. The cosine of the angle is given by the formula:
$\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|}$
Calculate the dot product:
$\overrightarrow{AB} \cdot \overrightarrow{CD} = (2)(3) + (5)(2) + (-4)(4) = 6 + 10 - 16 = 0$
Since the dot product is $0$,the vectors are perpendicular.
Therefore,$\cos \theta = 0$,which implies $\theta = 90^{\circ}$.

(D) Given the two skew lines are $r = b + ta$ and $r = d + sc$.
The formula for the shortest distance between two skew lines $r = a_1 + \lambda b_1$ and $r = a_2 + \mu b_2$ is given by $d = \left| \frac{(a_2 - a_1) \cdot (b_1 \times b_2)}{|b_1 \times b_2|} \right|$.
Here,$a_1 = b$,$b_1 = a$,$a_2 = d$,and $b_2 = c$.
Substituting these values,the shortest distance is $\left| \frac{(d - b) \cdot (a \times c)}{|a \times c|} \right|$.
Since $|x| = |-x|$,this is equivalent to $\left| \frac{(b - d) \cdot (a \times c)}{|a \times c|} \right|$.
This expression represents the magnitude of the orthogonal projection of the vector $(b - d)$ onto the vector $(a \times c)$.

(A) The position vectors of points $P$ and $Q$ with respect to the origin $O(0,0,0)$ are given by $\vec{OP} = 0\hat{i} + 1\hat{j} + 2\hat{k}$ and $\vec{OQ} = 4\hat{i} - 2\hat{j} + 1\hat{k}$.
To find the angle $\theta = \angle POQ$,we use the dot product formula: $\cos \theta = \frac{\vec{OP} \cdot \vec{OQ}}{|\vec{OP}| |\vec{OQ}|}$.
First,calculate the dot product:
$\vec{OP} \cdot \vec{OQ} = (0)(4) + (1)(-2) + (2)(1) = 0 - 2 + 2 = 0$.
Since the dot product is $0$,the vectors $\vec{OP}$ and $\vec{OQ}$ are perpendicular to each other.
Therefore,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(A) Let the points be $A(2, 1, 1)$,$B(1, -1, 1)$,and $C(-1, 1, -1)$.
The normal vector $\vec{n}$ to the plane is given by $\vec{AB} \times \vec{AC}$.
$\vec{AB} = (1-2)\hat{i} + (-1-1)\hat{j} + (1-1)\hat{k} = -\hat{i} - 2\hat{j}$.
$\vec{AC} = (-1-2)\hat{i} + (1-1)\hat{j} + (-1-1)\hat{k} = -3\hat{i} - 2\hat{k}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & 0 \\ -3 & 0 & -2 \end{vmatrix} = \hat{i}(4-0) - \hat{j}(2-0) + \hat{k}(0-6) = 4\hat{i} - 2\hat{j} - 6\hat{k}$.
Dividing by $2$,we get the normal vector $\vec{n}' = 2\hat{i} - \hat{j} - 3\hat{k}$.
The unit vector $e$ perpendicular to the plane is $e = \pm \frac{2\hat{i} - \hat{j} - 3\hat{k}}{\sqrt{2^2 + (-1)^2 + (-3)^2}} = \pm \frac{1}{\sqrt{14}}(2\hat{i} - \hat{j} - 3\hat{k})$.
Given $a = 2\hat{i} - 3\hat{j} + 6\hat{k}$,the projection vector of $a$ on $e$ is $(a \cdot e)e$.
$a \cdot e = \pm \frac{1}{\sqrt{14}}(2(2) + (-3)(-1) + 6(-3)) = \pm \frac{1}{\sqrt{14}}(4 + 3 - 18) = \mp \frac{11}{\sqrt{14}}$.
Projection vector $= (a \cdot e)e = \left(\mp \frac{11}{\sqrt{14}}\right) \left(\pm \frac{1}{\sqrt{14}}(2\hat{i} - \hat{j} - 3\hat{k})\right) = -\frac{11}{14}(2\hat{i} - \hat{j} - 3\hat{k}) = \frac{11}{14}(-2\hat{i} + \hat{j} + 3\hat{k})$.

(A) The vector equation of a plane passing through three non-collinear points with position vectors $a, b, c$ is given by $r = (1 - x - y)a + xb + yc$.
If four points $a, b, c, d$ are coplanar,then $d$ must be expressible as a linear combination of $a, b, c$ such that the sum of the coefficients is $1$.
Given the equation $2a - 3b + 7c - 6d = 0$,we can rewrite it as $6d = 2a - 3b + 7c$,which implies $d = \frac{2}{6}a - \frac{3}{6}b + \frac{7}{6}c = \frac{1}{3}a - \frac{1}{2}b + \frac{7}{6}c$.
The sum of the coefficients is $\frac{1}{3} - \frac{1}{2} + \frac{7}{6} = \frac{2 - 3 + 7}{6} = \frac{6}{6} = 1$.
Since the sum of the coefficients is $1$,the point $d$ lies in the plane formed by $a, b, c$.
Thus,$(A)$ is true,$(R)$ is true,and $(R)$ is the correct explanation of $(A)$.

(A) The equation $|\vec{r}-\vec{a}| - |\vec{r}-\vec{b}| = c$ represents the definition of a hyperbola where $A$ and $B$ are the foci.
The distance between the foci is $2ae = |\vec{a}-\vec{b}|$.
The distance between the vertices (transverse axis length) is $2a = c$.
Therefore,the eccentricity $e$ is given by:
$e = \frac{\text{distance between foci}}{\text{distance between vertices}} = \frac{|\vec{a}-\vec{b}|}{c}$.

(B) Given that $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$,there exists a scalar $k_{1}$ such that $\vec{\alpha}+3 \vec{\beta}=k_{1} \vec{\gamma}$.
This implies $\vec{\beta}=\frac{k_{1}}{3} \vec{\gamma}-\frac{1}{3} \vec{\alpha}$.
Also,$\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$,so there exists a scalar $k_{2}$ such that $\vec{\beta}+2 \vec{\gamma}=k_{2} \vec{\alpha}$.
This implies $\vec{\beta}=k_{2} \vec{\alpha}-2 \vec{\gamma}$.
Equating the two expressions for $\vec{\beta}$,we get $\frac{k_{1}}{3} \vec{\gamma}-\frac{1}{3} \vec{\alpha}=k_{2} \vec{\alpha}-2 \vec{\gamma}$.
Rearranging the terms,we have $\vec{\alpha}(k_{2}+\frac{1}{3})=\vec{\gamma}(\frac{k_{1}}{3}+2)$.
Since $\vec{\alpha}$ and $\vec{\gamma}$ are non-collinear,the coefficients must be zero: $k_{2}+\frac{1}{3}=0 \Rightarrow k_{2}=-\frac{1}{3}$ and $\frac{k_{1}}{3}+2=0 \Rightarrow k_{1}=-6$.
Substituting $k_{1}=-6$ into the first equation,we get $\vec{\alpha}+3 \vec{\beta}=-6 \vec{\gamma}$.
Therefore,$\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}=\overrightarrow{0}$.

(C) Given the equation $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$.
Let $\vec{u} = \vec{\alpha} \times \vec{\beta}$,$\vec{v} = \vec{\beta} \times \vec{\gamma}$,and $\vec{w} = \vec{\gamma} \times \vec{\alpha}$.
The given equation is $a\vec{u} + b\vec{v} + c\vec{w} = \overrightarrow{0}$.
Since $a, b, c$ are non-zero scalars,the vectors $\vec{u}, \vec{v}, \vec{w}$ are linearly dependent,which implies they are coplanar.
For any three vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$,the cross products $\vec{\alpha} \times \vec{\beta}$,$\vec{\beta} \times \vec{\gamma}$,and $\vec{\gamma} \times \vec{\alpha}$ are coplanar if and only if $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are coplanar.
Therefore,the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are coplanar.

(B) Let the unit vector in the $ZOX$ plane be $\vec{r} = x \hat{i} + z \hat{k}$.
Since it is a unit vector,$|\vec{r}|^2 = x^2 + z^2 = 1$.
Given $\vec{\alpha} = 2 \hat{i} + 2 \hat{j} - \hat{k}$,we have $|\vec{\alpha}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4+4+1} = 3$.
The angle between $\vec{r}$ and $\vec{\alpha}$ is $45^{\circ}$,so $\vec{r} \cdot \vec{\alpha} = |\vec{r}| |\vec{\alpha}| \cos 45^{\circ}$.
$(x \hat{i} + z \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - \hat{k}) = 1 \cdot 3 \cdot \frac{1}{\sqrt{2}} \Rightarrow 2x - z = \frac{3}{\sqrt{2}}$.
Given $\vec{\beta} = \hat{j} - \hat{k}$,we have $|\vec{\beta}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.
The angle between $\vec{r}$ and $\vec{\beta}$ is $60^{\circ}$,so $\vec{r} \cdot \vec{\beta} = |\vec{r}| |\vec{\beta}| \cos 60^{\circ}$.
$(x \hat{i} + z \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} - 1 \hat{k}) = 1 \cdot \sqrt{2} \cdot \frac{1}{2} \Rightarrow -z = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow z = -\frac{1}{\sqrt{2}}$.
Substituting $z$ into $2x - z = \frac{3}{\sqrt{2}}$,we get $2x - (-\frac{1}{\sqrt{2}}) = \frac{3}{\sqrt{2}} \Rightarrow 2x = \frac{2}{\sqrt{2}} = \sqrt{2} \Rightarrow x = \frac{1}{\sqrt{2}}$.
Thus,$\vec{r} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{k}$.

(D) Given,$|a+b| < |a-b|$.
Squaring both sides,we get $|a+b|^2 < |a-b|^2$.
Using the property $|x|^2 = x \cdot x$,we have $(a+b) \cdot (a+b) < (a-b) \cdot (a-b)$.
Expanding the dot product,$|a|^2 + |b|^2 + 2(a \cdot b) < |a|^2 + |b|^2 - 2(a \cdot b)$.
Subtracting $|a|^2 + |b|^2$ from both sides,$2(a \cdot b) < -2(a \cdot b)$.
This simplifies to $4(a \cdot b) < 0$,which implies $a \cdot b < 0$.
Since $a \cdot b = |a||b| \cos \theta < 0$ and $|a|, |b| > 0$,we must have $\cos \theta < 0$.
Therefore,the angle $\theta$ between $a$ and $b$ is an obtuse angle.

(D) Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$.
We know that $\vec{a} \cdot \vec{b} = k^2 \cos \alpha$,$\vec{b} \cdot \vec{c} = k^2 \cos \beta$,and $\vec{c} \cdot \vec{a} = k^2 \cos \gamma$.
Consider the magnitude of the sum of the vectors: $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$.
Expanding this,we get: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \geq 0$.
$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$.
Substituting the values: $3k^2 + 2k^2(\cos \alpha + \cos \beta + \cos \gamma) \geq 0$.
Dividing by $2k^2$ (since $k > 0$): $\frac{3}{2} + (\cos \alpha + \cos \beta + \cos \gamma) \geq 0$.
Therefore,$\cos \alpha + \cos \beta + \cos \gamma \geq -\frac{3}{2}$.