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

(B) Given $x + y + z = 0$,we have $x = -(y + z)$.
Squaring both sides,we get $|x|^2 = |y + z|^2$.
$|x|^2 = |y|^2 + |z|^2 + 2(y \cdot z)$.
Since $|x| = |y| = |z| = 2$,we have $4 = 4 + 4 + 2(2)(2) \cos \theta$.
$4 = 8 + 8 \cos \theta$.
$8 \cos \theta = -4$,which gives $\cos \theta = -1/2$.
Thus,$\theta = 120^{\circ}$.
Now,$\csc^2(120^{\circ}) + \cot^2(120^{\circ}) = \left(\frac{2}{\sqrt{3}}\right)^2 + \left(-\frac{1}{\sqrt{3}}\right)^2$.
$= \frac{4}{3} + \frac{1}{3} = \frac{5}{3}$.

(C) The projection of vector $\vec{a} = 2i + j - 3k$ on vector $\vec{b} = i - 2j + k$ is given by the formula:
$\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
First,calculate the dot product $\vec{a} \cdot \vec{b}$:
$\vec{a} \cdot \vec{b} = (2)(1) + (1)(-2) + (-3)(1) = 2 - 2 - 3 = -3$
Next,calculate the magnitude of vector $\vec{b}$:
$|\vec{b}| = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$
Now,substitute these values into the formula:
$\text{Projection} = \frac{-3}{\sqrt{6}} = -\frac{3}{\sqrt{2 \times 3}} = -\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{2} \times \sqrt{3}} = -\frac{\sqrt{3}}{\sqrt{2}} = -\sqrt{\frac{3}{2}}$
Thus,the correct option is $C$.

(B) Given $|a| = 1$,$|b| = 1$,and $|a + b| = \sqrt{3}$.
Squaring the equation $|a + b| = \sqrt{3}$,we get $|a + b|^2 = 3$.
Using the property $|a + b|^2 = |a|^2 + |b|^2 + 2(a \cdot b)$,we have $1^2 + 1^2 + 2(a \cdot b) = 3$.
$2 + 2(a \cdot b) = 3 \implies 2(a \cdot b) = 1 \implies a \cdot b = \frac{1}{2}$.
Now,expand the expression $(3a - 4b) \cdot (2a + 5b)$:
$(3a - 4b) \cdot (2a + 5b) = 6(a \cdot a) + 15(a \cdot b) - 8(b \cdot a) - 20(b \cdot b)$.
Since $a \cdot a = |a|^2 = 1$,$b \cdot b = |b|^2 = 1$,and $a \cdot b = b \cdot a$,we get:
$6(1) + 7(a \cdot b) - 20(1) = 6 - 20 + 7 \left(\frac{1}{2}\right)$.
$= -14 + \frac{7}{2} = \frac{-28 + 7}{2} = -\frac{21}{2}$.

(D) Given that $a \perp (b + c)$,$b \perp (c + a)$,and $c \perp (a + b)$.
This implies:
$a \cdot (b + c) = 0 \implies a \cdot b + a \cdot c = 0$ .....$(i)$
$b \cdot (c + a) = 0 \implies b \cdot c + b \cdot a = 0$ .....$(ii)$
$c \cdot (a + b) = 0 \implies c \cdot a + c \cdot b = 0$ .....$(iii)$
Adding $(i), (ii),$ and $(iii)$,we get $2(a \cdot b + b \cdot c + c \cdot a) = 0$.
Now,consider the sum of the squares of the magnitudes:
$|a + b|^2 + |b + c|^2 + |c + a|^2 = 6^2 + 8^2 + 10^2 = 36 + 64 + 100 = 200$.
Expanding the left side:
$(|a|^2 + |b|^2 + 2a \cdot b) + (|b|^2 + |c|^2 + 2b \cdot c) + (|c|^2 + |a|^2 + 2c \cdot a) = 200$
$2(|a|^2 + |b|^2 + |c|^2) + 2(a \cdot b + b \cdot c + c \cdot a) = 200$.
Since $2(a \cdot b + b \cdot c + c \cdot a) = 0$,we have $2(|a|^2 + |b|^2 + |c|^2) = 200$,so $|a|^2 + |b|^2 + |c|^2 = 100$.
We know that $|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$.
Substituting the values: $|a + b + c|^2 = 100 + 0 = 100$.
Therefore,$|a + b + c| = \sqrt{100} = 10$.

(C) Given that $|\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b}$.
By the definitions of cross product and dot product,we have:
$|\vec{a}| |\vec{b}| \sin \theta = |\vec{a}| |\vec{b}| \cos \theta$
Assuming $\vec{a}$ and $\vec{b}$ are non-zero vectors,we can divide both sides by $|\vec{a}| |\vec{b}| \cos \theta$ (where $\cos \theta \neq 0$):
$\frac{\sin \theta}{\cos \theta} = 1$
$\tan \theta = 1$
Since $\theta$ is the angle between two vectors,$0 \leq \theta \leq \pi$. The value of $\theta$ for which $\tan \theta = 1$ is $\theta = \frac{\pi}{4}$.

(B) The component of vector $a$ along vector $b$ is the projection of $a$ on $b$,which is given by $a \cos \theta = \frac{a \cdot b}{|b|}$.
The component of vector $a$ perpendicular to vector $b$ is given by $a \sin \theta$. Since $|a \times b| = |a||b| \sin \theta$,we have $a \sin \theta = \frac{|a \times b|}{|b|}$.
Thus,the components are $\frac{a \cdot b}{|b|}$ and $\frac{|a \times b|}{|b|}$.

(B) We know that the magnitude of the cross product of two vectors $a$ and $b$ is given by $|a \times b| = |a| |b| \sin \theta$,where $\theta$ is the angle between them.
Therefore,$(a \times b)^2 = |a|^2 |b|^2 \sin^2 \theta$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get $(a \times b)^2 = |a|^2 |b|^2 (1 - \cos^2 \theta) = |a|^2 |b|^2 - |a|^2 |b|^2 \cos^2 \theta$.
Since $a \cdot b = |a| |b| \cos \theta$,we have $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 \theta$.
Thus,$(a \times b)^2 = |a|^2 |b|^2 - (a \cdot b)^2$.
Note that $|a|^2 = a \cdot a$ and $|b|^2 = b \cdot b$.
So,$(a \times b)^2 = (a \cdot a)(b \cdot b) - (a \cdot b)(b \cdot a)$.
This expression is equivalent to the determinant of the Gram matrix:
$\left| \begin{matrix} a \cdot a & a \cdot b \\ b \cdot a & b \cdot b \end{matrix} \right| = (a \cdot a)(b \cdot b) - (a \cdot b)(b \cdot a)$.
Hence,the correct option is $B$.

(C) Given that $a \cdot b = a \cdot c$ and $a \times b = a \times c$.
From the first equation,$a \cdot b - a \cdot c = 0$,which implies $a \cdot (b - c) = 0$.
This means either $a = 0$ or $(b - c) = 0$ or $a \perp (b - c)$.
From the second equation,$a \times b - a \times c = 0$,which implies $a \times (b - c) = 0$.
This means either $a = 0$ or $(b - c) = 0$ or $a \parallel (b - c)$.
Since $a \neq 0$,we have $a \perp (b - c)$ and $a \parallel (b - c)$ simultaneously.
$A$ non-zero vector cannot be both parallel and perpendicular to another vector unless the other vector is the zero vector.
Therefore,$b - c = 0$,which implies $b = c$.

(D) We know that $|a \times b| = |a| |b| \sin \theta$.
Given $|a| = 2$,$|b| = 5$,and $|a \times b| = 8$.
Substituting the values: $8 = 2 \times 5 \times \sin \theta$.
$8 = 10 \sin \theta \Rightarrow \sin \theta = \frac{8}{10} = \frac{4}{5}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\cos^2 \theta = 1 - (\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac{9}{25}$.
Thus,$\cos \theta = \pm \frac{3}{5}$.
Now,$a \cdot b = |a| |b| \cos \theta = 2 \times 5 \times (\pm \frac{3}{5}) = \pm 6$.
Given the options provided,the magnitude is $6$.

(B) Given that $|a \cdot b| = |a||b| \cos \theta = 3$ $(i)$
And $|a \times b| = |a||b| \sin \theta = 4$ $(ii)$
Dividing equation $(ii)$ by equation $(i)$,we get:
$\frac{|a||b| \sin \theta}{|a||b| \cos \theta} = \frac{4}{3}$
$\tan \theta = \frac{4}{3}$
Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3}$,the hypotenuse is $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Therefore,$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{5}$.
Thus,$\theta = \cos^{-1} \frac{3}{5}$.

(A) Given the equation $la + mb = c$.
To find $l$,take the cross product of both sides with $b$:
$(la + mb) \times b = c \times b$
$l(a \times b) + m(b \times b) = c \times b$
Since $b \times b = 0$,we have $l(a \times b) = c \times b$.
Taking the dot product of both sides with $(a \times b)$:
$l(a \times b) \cdot (a \times b) = (c \times b) \cdot (a \times b)$
$l|a \times b|^2 = (c \times b) \cdot (a \times b)$
$l = \frac{(c \times b) \cdot (a \times b)}{|a \times b|^2}$.
Similarly,to find $m$,take the cross product of the original equation with $a$:
$(la + mb) \times a = c \times a$
$l(a \times a) + m(b \times a) = c \times a$
Since $a \times a = 0$,we have $m(b \times a) = c \times a$.
Taking the dot product of both sides with $(b \times a)$:
$m|b \times a|^2 = (c \times a) \cdot (b \times a)$
$m = \frac{(c \times a) \cdot (b \times a)}{|b \times a|^2}$.

(D) We know the identity relating the cross product and dot product: $|a \times r|^2 + |a \cdot r|^2 = |a|^2 |r|^2$.
Given $a \times r = j$,we have $|a \times r| = |j| = 1$.
Substituting this into the identity: $1^2 + (a \cdot r)^2 = |a|^2 |r|^2$.
Thus,$(a \cdot r)^2 = |a|^2 |r|^2 - 1$.
Since the magnitude of $r$ is not fixed,the value of $a \cdot r$ depends on the choice of $r$.
Therefore,$a \cdot r$ is an arbitrary scalar.

(A) Given vectors are $\overrightarrow{A} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\overrightarrow{B} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.
First,calculate the dot product $\overrightarrow{A} \cdot \overrightarrow{B} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12$.
Next,calculate the magnitudes $|\overrightarrow{A}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$ and $|\overrightarrow{B}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
Using the formula $\cos \theta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}$,we get $\cos \theta = \frac{12}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{6}{\sqrt{84}} = \frac{6}{2\sqrt{21}} = \frac{3}{\sqrt{21}} = \sqrt{\frac{9}{21}} = \sqrt{\frac{3}{7}}$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we have $\sin^2 \theta = 1 - \frac{3}{7} = \frac{4}{7}$.
Therefore,$\sin \theta = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$.

(B) We know that the magnitude of the cross product of two vectors $a$ and $b$ is given by $|a \times b| = |a| |b| \sin \theta$,where $\theta$ is the angle between them.
Squaring both sides,we get $|a \times b|^2 = |a|^2 |b|^2 \sin^2 \theta$.
We also know that the dot product is given by $a \cdot b = |a| |b| \cos \theta$.
Squaring both sides,we get $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 \theta$.
Adding these two expressions:
$|a \times b|^2 + (a \cdot b)^2 = |a|^2 |b|^2 \sin^2 \theta + |a|^2 |b|^2 \cos^2 \theta$
$= |a|^2 |b|^2 (\sin^2 \theta + \cos^2 \theta)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$|a \times b|^2 + (a \cdot b)^2 = |a|^2 |b|^2$.
Since $|a|^2 = a \cdot a$ and $|b|^2 = b \cdot b$,the result is $(a \cdot a)(b \cdot b)$.

(A) Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$.
We know the definitions of the cross product and dot product:
$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$
$|\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta|$
Therefore,the ratio is:
$\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|} = \frac{|\vec{a}| |\vec{b}| \sin \theta}{|\vec{a}| |\vec{b}| |\cos \theta|} = \frac{\sin \theta}{|\cos \theta|} = |\tan \theta|$
Assuming $\theta$ is the angle in the range $[0, \pi]$,the expression simplifies to $\tan \theta$ for $0 \le \theta < \pi/2$.

(B) Let the required unit vector be $\hat{a}$. Since $\hat{a}$ lies in the plane of $\vec{u} = 2i + j + k$ and $\vec{v} = i - j + k$,it can be expressed as $\hat{a} = \alpha(2i + j + k) + \beta(i - j + k)$.
Simplifying,we get $\hat{a} = (2\alpha + \beta)i + (\alpha - \beta)j + (\alpha + \beta)k$.
Since $\hat{a}$ is a unit vector,its magnitude is $1$,so $(2\alpha + \beta)^2 + (\alpha - \beta)^2 + (\alpha + \beta)^2 = 1$.
Expanding this,we get $6\alpha^2 + 4\alpha\beta + 3\beta^2 = 1$ ... $(i)$.
Since $\hat{a}$ is orthogonal to $\vec{w} = 5i + 2j + 6k$,their dot product is zero: $5(2\alpha + \beta) + 2(\alpha - \beta) + 6(\alpha + \beta) = 0$.
This simplifies to $18\alpha + 9\beta = 0$,which gives $\beta = -2\alpha$.
Substituting $\beta = -2\alpha$ into equation $(i)$,we get $6\alpha^2 + 4\alpha(-2\alpha) + 3(-2\alpha)^2 = 1$.
$6\alpha^2 - 8\alpha^2 + 12\alpha^2 = 1$,which implies $10\alpha^2 = 1$,so $\alpha = \pm \frac{1}{\sqrt{10}}$.
Then $\beta = -2(\pm \frac{1}{\sqrt{10}}) = \mp \frac{2}{\sqrt{10}}$.
Substituting these values into the expression for $\hat{a}$,we get $\hat{a} = \pm \frac{1}{\sqrt{10}} [ (2 - 2)i + (1 + 2)j + (1 - 2)k ] = \pm \frac{3j - k}{\sqrt{10}}$.

(B) Given $a \times b = 2a \times c$,we can write $a \times (b - 2c) = 0$.
This implies that the vector $(b - 2c)$ is parallel to $a$.
Given $b - 2c = \lambda a$,we have $|b - 2c|^2 = \lambda^2 |a|^2$.
Expanding the left side: $|b|^2 + 4|c|^2 - 4(b \cdot c) = \lambda^2 (1)^2$.
Given $|b| = 4$ and $|c| = 1$,we have $16 + 4(1) - 4(b \cdot c) = \lambda^2$,so $20 - 4(b \cdot c) = \lambda^2$.
We are given $|b \times c| = \sqrt{15}$.
Since $|b \times c| = |b||c| \sin \alpha = \sqrt{15}$,where $\alpha$ is the angle between $b$ and $c$,we have $(4)(1) \sin \alpha = \sqrt{15}$,so $\sin \alpha = \frac{\sqrt{15}}{4}$.
Then $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{15}{16} = \frac{1}{16}$,so $\cos \alpha = \pm \frac{1}{4}$.
Since $b \cdot c = |b||c| \cos \alpha = (4)(1) \cos \alpha = 4 \cos \alpha$,we have $b \cdot c = 4(\pm \frac{1}{4}) = \pm 1$.
Substituting this into the equation for $\lambda^2$: $\lambda^2 = 20 - 4(\pm 1) = 20 \mp 4$.
If $b \cdot c = 1$,$\lambda^2 = 16$,so $\lambda = \pm 4$.
If $b \cdot c = -1$,$\lambda^2 = 24$,which is not among the options.
Thus,$\lambda = \pm 4$.

(C) The vector area of a triangle with vertices having position vectors $a, b, c$ is given by the formula:
$\text{Vector Area} = \frac{1}{2} \vec{AB} \times \vec{AC}$
Since $\vec{AB} = b - a$ and $\vec{AC} = c - a$,we have:
$\text{Vector Area} = \frac{1}{2} (b - a) \times (c - a)$
Expanding the cross product:
$\text{Vector Area} = \frac{1}{2} (b \times c - b \times a - a \times c + a \times a)$
Since $a \times a = 0$ and $-b \times a = a \times b$,and $-a \times c = c \times a$:
$\text{Vector Area} = \frac{1}{2} (a \times b + b \times c + c \times a)$
Thus,the correct option is $C$.

(D) The moment of a force $\overrightarrow{F}$ acting at a point $P$ about a point $C$ is defined as the cross product of the position vector of $P$ relative to $C$ and the force vector $\overrightarrow{F}$.
Mathematically,the moment $\overrightarrow{\tau}$ is given by $\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}$,where $\overrightarrow{r} = \overrightarrow{CP}$.
Therefore,the moment is $\overrightarrow{CP} \times \overrightarrow{F}$.

(B) Let the forces be $\vec{F_1} = i + 2j - 3k$,$\vec{F_2} = 2i + 3j + 4k$,and $\vec{F_3} = i - j + k$.
The resultant force $\vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} = (1+2+1)i + (2+3-1)j + (-3+4+1)k = 4i + 4j + 2k$.
The particle is at point $A(0, 1, 2)$ and we need the moment about point $B(1, -2, 0)$.
The position vector $\vec{r}$ of the point of application relative to the point about which the moment is calculated is $\vec{r} = \vec{OA} - \vec{OB} = (0-1)i + (1-(-2))j + (2-0)k = -i + 3j + 2k$.
The moment $\vec{M} = \vec{r} \times \vec{F} = \begin{vmatrix} i & j & k \\ -1 & 3 & 2 \\ 4 & 4 & 2 \end{vmatrix}$.
Calculating the determinant: $\vec{M} = i(6 - 8) - j(-2 - 8) + k(-4 - 12) = -2i + 10j - 16k$.
The magnitude of the moment is $|\vec{M}| = \sqrt{(-2)^2 + 10^2 + (-16)^2} = \sqrt{4 + 100 + 256} = \sqrt{360} = \sqrt{36 \times 10} = 6\sqrt{10}$.

(B) Given points are $A(-2, 2, 4)$,$B(2, 6, 3)$,and $P(1, 2, 1)$.
The force vector $\overrightarrow{F} = \overrightarrow{AB} = (2 - (-2))\hat{i} + (6 - 2)\hat{j} + (3 - 4)\hat{k} = 4\hat{i} + 4\hat{j} - \hat{k}$.
The position vector of $A$ with respect to $P$ is $\overrightarrow{r} = \overrightarrow{PA} = (-2 - 1)\hat{i} + (2 - 2)\hat{j} + (4 - 1)\hat{k} = -3\hat{i} + 0\hat{j} + 3\hat{k}$.
The moment of the force about $P$ is $\vec{M} = \overrightarrow{PA} \times \overrightarrow{F}$.
$\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & 3 \\ 4 & 4 & -1 \end{vmatrix} = \hat{i}(0 - 12) - \hat{j}(3 - 12) + \hat{k}(-12 - 0) = -12\hat{i} + 9\hat{j} - 12\hat{k}$.
The magnitude of the moment is $|\vec{M}| = \sqrt{(-12)^2 + 9^2 + (-12)^2} = \sqrt{144 + 81 + 144} = \sqrt{369} = \sqrt{9 \times 41} = 3\sqrt{41}$.

(A) The position vector of point $A$ relative to $O$ is $\overrightarrow{OA} = (4-1)i + (-1-(-3))j + (-7-2)k = 3i + 2j - 9k$.
The unit vector along the direction $(9, 6, -2)$ is $\hat{u} = \frac{9i + 6j - 2k}{\sqrt{9^2 + 6^2 + (-2)^2}} = \frac{9i + 6j - 2k}{\sqrt{81 + 36 + 4}} = \frac{9i + 6j - 2k}{11}$.
The force vector is $\vec{F} = 6 \hat{u} = \frac{6}{11}(9i + 6j - 2k)$.
The moment of the force about point $O$ is given by $\vec{M} = \overrightarrow{OA} \times \vec{F}$.
$\vec{M} = (3i + 2j - 9k) \times \frac{6}{11}(9i + 6j - 2k) = \frac{6}{11} \begin{vmatrix} i & j & k \\ 3 & 2 & -9 \\ 9 & 6 & -2 \end{vmatrix}$.
Calculating the determinant: $i(2(-2) - (-9)(6)) - j(3(-2) - (-9)(9)) + k(3(6) - 2(9)) = i(-4 + 54) - j(-6 + 81) + k(18 - 18) = 50i - 75j + 0k$.
Thus,$\vec{M} = \frac{6}{11}(50i - 75j) = \frac{6 \times 25}{11}(2i - 3j) = \frac{150}{11}(2i - 3j)$.

(D) Given that $a \cdot i = 4$.
We need to evaluate $(a \times j) \cdot (2j - 3k)$.
Using the property of scalar triple product $(A \times B) \cdot C = A \cdot (B \times C)$,we have:
$(a \times j) \cdot (2j - 3k) = a \cdot \{ j \times (2j - 3k) \}$.
Expanding the cross product:
$j \times (2j - 3k) = 2(j \times j) - 3(j \times k)$.
Since $j \times j = 0$ and $j \times k = i$,we get:
$j \times (2j - 3k) = 2(0) - 3(i) = -3i$.
Substituting this back into the expression:
$a \cdot (-3i) = -3(a \cdot i)$.
Given $a \cdot i = 4$,the result is:
$-3(4) = -12$.

(D) Given vectors are $a = 3i - j + 2k$ and $b = 2i + j - k$.
First,calculate the dot product $a \cdot b$:
$a \cdot b = (3)(2) + (-1)(1) + (2)(-1) = 6 - 1 - 2 = 3$.
Now,the expression is $a \times (a \cdot b) = a \times (3) = 3(a \times 1)$.
However,the cross product is defined between two vectors. Since $(a \cdot b)$ is a scalar $(3)$,the expression $a \times (3)$ is mathematically undefined because the cross product operation requires two vectors.
Therefore,the expression $a \times (a \cdot b)$ is not meaningful.

(B) Let the required vector be $\vec{v} = ai + bj + ck$.
Since $\vec{v}$,$i + j$,and $j + k$ are coplanar,their scalar triple product must be zero:
$\begin{vmatrix} a & b & c \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = 0$
$a(1 - 0) - b(1 - 0) + c(1 - 0) = 0 \Rightarrow a - b + c = 0$.
Also,since $\vec{v}$ is parallel to $2i - 2j - 4k$,we have $\vec{v} = \lambda(2i - 2j - 4k) = 2\lambda i - 2\lambda j - 4\lambda k$.
Comparing coefficients,$a = 2\lambda$,$b = -2\lambda$,$c = -4\lambda$.
Substitute these into the coplanarity condition $a - b + c = 0$:
$2\lambda - (-2\lambda) + (-4\lambda) = 0$
$2\lambda + 2\lambda - 4\lambda = 0 \Rightarrow 0 = 0$.
This confirms the vector is of the form $\lambda(2i - 2j - 4k)$.
Checking the options,for $\lambda = 0.5$,we get $i - j - 2k$,which matches option $(b)$.

(B) The given vectors are constructed using the Gram-Schmidt orthogonalization process.
For the set $\{a, b_1, c_2\}$:
$1$. $a \cdot b_1 = a \cdot (b - \frac{b \cdot a}{|a|^2} a) = a \cdot b - \frac{b \cdot a}{|a|^2} (a \cdot a) = a \cdot b - b \cdot a = 0$. Thus,$a \perp b_1$.
$2$. $a \cdot c_2 = a \cdot (c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1) = a \cdot c - \frac{c \cdot a}{|a|^2} (a \cdot a) - \frac{c \cdot b_1}{|b_1|^2} (a \cdot b_1) = a \cdot c - c \cdot a - 0 = 0$. Thus,$a \perp c_2$.
$3$. $b_1 \cdot c_2 = b_1 \cdot (c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1) = b_1 \cdot c - \frac{c \cdot a}{|a|^2} (b_1 \cdot a) - \frac{c \cdot b_1}{|b_1|^2} (b_1 \cdot b_1) = b_1 \cdot c - 0 - c \cdot b_1 = 0$. Thus,$b_1 \perp c_2$.
Since all pairs are orthogonal,the set $\{a, b_1, c_2\}$ is a set of mutually orthogonal vectors.
Therefore,option $(b)$ is the correct answer.

(D) Given equations are:
$(i) x + y = a$
$(ii) x \times y = b$
$(iii) x \cdot a = 1$
Taking the dot product of $(i)$ with $a$:
$a \cdot (x + y) = a \cdot a$
$a \cdot x + a \cdot y = |a|^2$
$1 + a \cdot y = |a|^2 \implies a \cdot y = |a|^2 - 1$ $(iv)$
Taking the cross product of $a$ with $(ii)$:
$a \times (x \times y) = a \times b$
$(a \cdot y)x - (a \cdot x)y = a \times b$
Substituting $(iii)$ and $(iv)$:
$(|a|^2 - 1)x - y = a \times b$ $(v)$
From $(i)$,$y = a - x$. Substitute this into $(v)$:
$(|a|^2 - 1)x - (a - x) = a \times b$
$(|a|^2 - 1 + 1)x = a + (a \times b)$
$|a|^2 x = a + (a \times b)$
$x = \frac{a + (a \times b)}{|a|^2}$
Then $y = a - x = a - \frac{a + (a \times b)}{|a|^2} = \frac{(|a|^2 - 1)a - (a \times b)}{|a|^2}$
Since the calculated values do not match any of the given options,the correct choice is $(d)$.

(A) Let $P(r)$ be a point equidistant from points $A(a)$ and $B(b)$.
Then,the distance $PA = PB$,which implies $PA^2 = PB^2$.
In vector form,this is $|r - a|^2 = |r - b|^2$.
Expanding this,we get $(r - a) \cdot (r - a) = (r - b) \cdot (r - b)$.
$|r|^2 - 2(r \cdot a) + |a|^2 = |r|^2 - 2(r \cdot b) + |b|^2$.
$-2(r \cdot a) + 2(r \cdot b) = |b|^2 - |a|^2$.
$2r \cdot (b - a) = |b|^2 - |a|^2$.
Alternatively,the locus is the perpendicular bisector of the segment $AB$.
Let $M$ be the midpoint of $AB$. The position vector of $M$ is $\frac{1}{2}(a + b)$.
Since $PM$ is perpendicular to $AB$,the vector $PM$ is perpendicular to the vector $AB$.
Thus,$(r - \frac{1}{2}(a + b)) \cdot (b - a) = 0$.
Multiplying by $-1$,we get $[r - \frac{1}{2}(a + b)] \cdot (a - b) = 0$.

(A) Since $a, b$ and $a \times b$ are non-coplanar,we can express $r$ as a linear combination: $r = xa + yb + z(a \times b)$ for some scalars $x, y, z$.
Given the equation $r \times a = b$,we substitute the expression for $r$:
$b = (xa + yb + z(a \times b)) \times a$
Using the distributive property of the cross product:
$b = x(a \times a) + y(b \times a) + z((a \times b) \times a)$
Since $a \times a = 0$ and $b \times a = -(a \times b)$:
$b = -y(a \times b) + z((a \cdot a)b - (a \cdot b)a)$
Given that $a$ and $b$ are perpendicular,$a \cdot b = 0$:
$b = -y(a \times b) + z(a \cdot a)b$
Comparing the coefficients of $b$ and $(a \times b)$ on both sides:
For the coefficient of $(a \times b)$,we have $-y = 0$,so $y = 0$.
For the coefficient of $b$,we have $z(a \cdot a) = 1$,so $z = \frac{1}{a \cdot a}$.
Substituting these values back into the expression for $r$:
$r = xa + 0b + \frac{1}{a \cdot a}(a \times b) = xa + \frac{1}{a \cdot a}(a \times b)$.

(C) Given vectors are $a = i + j$ and $b = 2i - k$. Let $r = xi + yj + zk$.
For the first line $r \times a = b \times a$, we have $(r - b) \times a = 0$. This implies $(r - b)$ is parallel to $a$.
$r - b = (x - 2)i + yj + (z + 1)k$.
Since $(r - b) \times a = 0$, the cross product is:
$\left| \begin{matrix} i & j & k \\ x-2 & y & z+1 \\ 1 & 1 & 0 \end{matrix} \right| = 0$
$-(z+1)i + (z+1)j + (x - 2 - y)k = 0$.
Comparing coefficients, we get $z + 1 = 0 \Rightarrow z = -1$ and $x - y - 2 = 0 \Rightarrow x - y = 2$.
For the second line $r \times b = a \times b$, we have $(r - a) \times b = 0$. This implies $(r - a)$ is parallel to $b$.
$r - a = (x - 1)i + (y - 1)j + zk$.
Since $(r - a) \times b = 0$, the cross product is:
$\left| \begin{matrix} i & j & k \\ x-1 & y-1 & z \\ 2 & 0 & -1 \end{matrix} \right| = 0$
$-(y-1)i + (x - 1 + 2z)j - 2(y-1)k = 0$.
Comparing coefficients, we get $y - 1 = 0 \Rightarrow y = 1$ and $x + 2z - 1 = 0$.
Substituting $y = 1$ into $x - y = 2$, we get $x - 1 = 2 \Rightarrow x = 3$.
Substituting $z = -1$ into $x + 2z = 1$, we get $x + 2(-1) = 1 \Rightarrow x = 3$.
Thus, the point of intersection is $r = 3i + j - k$.

(A) The angle between two faces of a tetrahedron is the angle between the normals to those faces.
First,find the normal vector $\vec{n_1}$ to the face $OAB$ using the cross product of vectors $\vec{OA}$ and $\vec{OB}$:
$\vec{OA} = \hat{i} + 2\hat{j} + \hat{k}$
$\vec{OB} = 2\hat{i} + \hat{j} + 3\hat{k}$
$\vec{n_1} = \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.
Next,find the normal vector $\vec{n_2}$ to the face $ABC$ using the cross product of vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}$
$\vec{AC} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$
$\vec{n_2} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.
The angle $\theta$ between the faces is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$:
$\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$.
$|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$.
$|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.
$\cos \theta = \frac{19}{\sqrt{35} \cdot \sqrt{35}} = \frac{19}{35}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{35}\right)$.

(B) Given that $a$,$b$,and $c$ are unit vectors,so $|a| = |b| = |c| = 1$.
We know that $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) = 1 + 1 - 2(a \cdot b) = 2 - 2(a \cdot b)$.
Similarly,$|b - c|^2 = 2 - 2(b \cdot c)$ and $|c - a|^2 = 2 - 2(c \cdot a)$.
Adding these three expressions:
$|a - b|^2 + |b - c|^2 + |c - a|^2 = 6 - 2(a \cdot b + b \cdot c + c \cdot a)$.
We also know that $|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 3 + 2(a \cdot b + b \cdot c + c \cdot a)$.
Therefore,$2(a \cdot b + b \cdot c + c \cdot a) = |a + b + c|^2 - 3$.
Substituting this into the sum:
$|a - b|^2 + |b - c|^2 + |c - a|^2 = 6 - (|a + b + c|^2 - 3) = 9 - |a + b + c|^2$.
Since $|a + b + c|^2 \ge 0$,the maximum value of the expression is $9 - 0 = 9$.
Thus,the expression does not exceed $9$.

(B) Given that $a$ and $b$ are unit vectors and are perpendicular,so $|a| = 1, |b| = 1$ and $a \cdot b = 0$.
Since $c$ is a unit vector,$|c| = 1$.
The vector $c$ is inclined at an angle $\theta$ to both $a$ and $b$,so $c \cdot a = |c||a| \cos \theta = \cos \theta$ and $c \cdot b = |c||b| \cos \theta = \cos \theta$.
Given $c = \alpha a + \beta b + \gamma (a \times b)$.
Taking the dot product with $a$: $c \cdot a = \alpha (a \cdot a) + \beta (b \cdot a) + \gamma ((a \times b) \cdot a) = \alpha (1) + \beta (0) + 0 = \alpha$. Thus,$\alpha = \cos \theta$.
Taking the dot product with $b$: $c \cdot b = \alpha (a \cdot b) + \beta (b \cdot b) + \gamma ((a \times b) \cdot b) = \alpha (0) + \beta (1) + 0 = \beta$. Thus,$\beta = \cos \theta$.
Since $c$ is a unit vector,$c \cdot c = 1$.
$c \cdot c = (\alpha a + \beta b + \gamma (a \times b)) \cdot (\alpha a + \beta b + \gamma (a \times b)) = \alpha^2 |a|^2 + \beta^2 |b|^2 + \gamma^2 |a \times b|^2 = 1$.
Since $|a \times b| = |a||b| \sin 90^\circ = 1$,we have $\alpha^2 + \beta^2 + \gamma^2 = 1$.
Substituting $\alpha = \beta = \cos \theta$,we get $2 \cos^2 \theta + \gamma^2 = 1$.
Therefore,$\gamma^2 = 1 - 2 \cos^2 \theta = - (2 \cos^2 \theta - 1) = - \cos 2\theta$.

(B) Let $\theta_1$ be the angle between $(a + b)$ and $a$,and $\theta_2$ be the angle between $(a + b)$ and $b$.
Since $(a + b)$ bisects the angle between $a$ and $b$,we have $\cos \theta_1 = \cos \theta_2$.
Using the definition of the dot product,$\cos \theta_1 = \frac{(a + b) \cdot a}{|a + b||a|}$ and $\cos \theta_2 = \frac{(a + b) \cdot b}{|a + b||b|}$.
Equating these,we get $\frac{|a|^2 + a \cdot b}{|a + b||a|} = \frac{a \cdot b + |b|^2}{|a + b||b|}$.
This simplifies to $\frac{|a|^2 + a \cdot b}{|a|} = \frac{a \cdot b + |b|^2}{|b|}$.
Multiplying both sides by $|a||b|$,we get $|b|(|a|^2 + a \cdot b) = |a|(a \cdot b + |b|^2)$.
$|a|^2|b| + |b|(a \cdot b) = |a|(a \cdot b) + |a||b|^2$.
$|a|^2|b| - |a||b|^2 + |b|(a \cdot b) - |a|(a \cdot b) = 0$.
$|a||b|(|a| - |b|) - (a \cdot b)(|a| - |b|) = 0$.
$(|a| - |b|)(|a||b| - a \cdot b) = 0$.
Since $a \cdot b = |a||b| \cos \theta$,where $\theta$ is the angle between $a$ and $b$,we have $(|a| - |b|)(|a||b| - |a||b| \cos \theta) = 0$.
$|a||b|(|a| - |b|)(1 - \cos \theta) = 0$.
This implies $|a| = |b|$ or $\cos \theta = 1$,which means $\theta = 0$.

(D) We have $\overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} = (a - 2b) - b = a - 3b$.
We have $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (2a + 3b) - a = a + 3b$.
Let $\theta$ be the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$.
Then $\cos \theta = \frac{\overrightarrow{BD} \cdot \overrightarrow{AC}}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = \frac{(a - 3b) \cdot (a + 3b)}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = \frac{|a|^2 - 9|b|^2}{|\overrightarrow{BD}| |\overrightarrow{AC}|}$.
Given $|a| = 3|b|$,we have $|a|^2 = 9|b|^2$.
Substituting this into the expression,we get $\cos \theta = \frac{9|b|^2 - 9|b|^2}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = 0$.
Since $\cos \theta = 0$,the angle $\theta = \frac{\pi}{2}$.
Therefore,the correct option is $(d)$.

(C) Let $\overrightarrow{V} = \overrightarrow{A} + t\overrightarrow{B}$.
$\overrightarrow{V} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-\hat{i} + 2\hat{j} + \hat{k})$
$\overrightarrow{V} = (1 - t)\hat{i} + (2 + 2t)\hat{j} + (3 + t)\hat{k}$.
Given that $\overrightarrow{V}$ is perpendicular to $\overrightarrow{D} = 3\hat{i} + 4\hat{j}$,their dot product must be zero.
$\overrightarrow{V} \cdot \overrightarrow{D} = 0$
$3(1 - t) + 4(2 + 2t) + 0(3 + t) = 0$
$3 - 3t + 8 + 8t = 0$
$11 + 5t = 0$
$5t = -11$
$t = -2.2$
Note: Re-evaluating the dot product with the provided vector $3\hat{i} + \hat{j}$ from the original text:
$3(1 - t) + 1(2 + 2t) = 0$
$3 - 3t + 2 + 2t = 0$
$5 - t = 0 \Rightarrow t = 5$.

(D) Let $r = \lambda b + \mu c$. Since $b = 4i + 3j$,$|b| = \sqrt{4^2 + 3^2} = 5$.
Since $c$ is perpendicular to $b$ in the $xy$-plane,$c$ must be parallel to $3i - 4j$. Let $c = k(3i - 4j)$. For projection of $r$ on $c$ to be $2$,we normalize $c$ such that $|c| = 1$,so $c = \pm \frac{1}{5}(3i - 4j)$.
Projection of $r$ on $b = \frac{r \cdot b}{|b|} = \lambda |b| = 5\lambda = 1 \Rightarrow \lambda = \frac{1}{5}$.
Projection of $r$ on $c = \frac{r \cdot c}{|c|} = \mu |c| = \mu = 2$.
Thus,$r = \frac{1}{5}(4i + 3j) \pm 2 \cdot \frac{1}{5}(3i - 4j)$.
Case $1$: $r = \frac{1}{5}(4i + 3j + 6i - 8j) = \frac{10i - 5j}{5} = 2i - j$.
Case $2$: $r = \frac{1}{5}(4i + 3j - 6i + 8j) = \frac{-2i + 11j}{5} = -\frac{2}{5}i + \frac{11}{5}j$.

(A) Any vector $r$ in the plane of $b$ and $c$ can be written as $r = \lambda b + \mu c$. For simplicity,let $r = b + tc$.
Substituting the vectors,$r = (i + 2j - k) + t(i + j - 2k) = (1 + t)i + (2 + t)j - (1 + 2t)k$.
The projection of $r$ on $a$ is given by $\frac{|r \cdot a|}{|a|} = \sqrt{2/3}$.
First,calculate $r \cdot a = (1 + t)(2) + (2 + t)(-1) + (-1 - 2t)(1) = 2 + 2t - 2 - t - 1 - 2t = -t - 1$.
Also,$|a| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.
Thus,$\frac{|-t - 1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}}$.
So,$|-t - 1| = 2$,which implies $-t - 1 = 2$ or $-t - 1 = -2$.
If $-t - 1 = 2$,then $t = -3$. Substituting $t = -3$ into $r$,we get $r = (1 - 3)i + (2 - 3)j - (1 - 6)k = -2i - j + 5k$.
If $-t - 1 = -2$,then $t = 1$. Substituting $t = 1$ into $r$,we get $r = (1 + 1)i + (2 + 1)j - (1 + 2)k = 2i + 3j - 3k$.
Therefore,the vectors are $2i + 3j - 3k$ and $-2i - j + 5k$.

(B) Given $a \times r = b + \lambda a$ and $a \cdot r = 3.$
Taking the dot product with $a$ on both sides of $a \times r = b + \lambda a$:
$(a \times r) \cdot a = b \cdot a + \lambda (a \cdot a)$
Since $(a \times r) \cdot a = 0$,we have $0 = b \cdot a + \lambda |a|^2$.
$a = 2i + j - k \implies |a|^2 = 2^2 + 1^2 + (-1)^2 = 6$.
$b \cdot a = (-i - 2j + k) \cdot (2i + j - k) = -2 - 2 - 1 = -5$.
Thus,$0 = -5 + \lambda(6) \implies \lambda = \frac{5}{6}$.
Now,$a \times r = b + \frac{5}{6}a$.
Taking the cross product with $a$ on both sides:
$a \times (a \times r) = a \times (b + \frac{5}{6}a) = a \times b + 0$.
Using the vector triple product formula $a \times (a \times r) = (a \cdot r)a - (a \cdot a)r$:
$(3)a - 6r = a \times b$.
$a \times b = \begin{vmatrix} i & j & k \\ 2 & 1 & -1 \\ -1 & -2 & 1 \end{vmatrix} = i(1-2) - j(2-1) + k(-4+1) = -i - j - 3k$.
$3(2i + j - k) - 6r = -i - j - 3k$.
$6i + 3j - 3k - 6r = -i - j - 3k$.
$6r = 7i + 4j \implies r = \frac{7}{6}i + \frac{2}{3}j$.

(A) Let $b = b_1 i + b_2 j + b_3 k$.
Given $a \times b = j - k$,we have:
$j - k = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix}$
$j - k = i(b_3 - b_2) - j(b_3 - b_1) + k(b_2 - b_1)$
Comparing the components:
$b_3 - b_2 = 0 \Rightarrow b_3 = b_2$
$b_1 - b_3 = 1 \Rightarrow b_1 = b_3 + 1 = b_2 + 1$
$b_2 - b_1 = -1 \Rightarrow b_2 - (b_2 + 1) = -1$ (This is consistent).
Given $a \cdot b = 1$,we have:
$(i + j + k) \cdot (b_1 i + b_2 j + b_3 k) = 1$
$b_1 + b_2 + b_3 = 1$
Substituting $b_1 = b_2 + 1$ and $b_3 = b_2$:
$(b_2 + 1) + b_2 + b_2 = 1$
$3b_2 + 1 = 1$
$3b_2 = 0 \Rightarrow b_2 = 0$
Thus,$b_3 = 0$ and $b_1 = 0 + 1 = 1$.
Therefore,$b = 1i + 0j + 0k = i$.

(A) The force vector $\overrightarrow{F}$ is given by $\overrightarrow{AB} = (3 - 1)i + (-4 - 2)j + (2 - (-3))k = 2i - 6j + 5k$.
The moment of force $\overrightarrow{F}$ about point $M$ is given by $\overrightarrow{\tau} = \overrightarrow{MA} \times \overrightarrow{F}$.
First,calculate the position vector $\overrightarrow{MA} = \overrightarrow{A} - \overrightarrow{M} = (1 - (-2))i + (2 - 4)j + (-3 - (-6))k = 3i - 2j + 3k$.
Now,calculate the cross product $\overrightarrow{MA} \times \overrightarrow{F} = \begin{vmatrix} i & j & k \\ 3 & -2 & 3 \\ 2 & -6 & 5 \end{vmatrix}$.
$= i((-2)(5) - (3)(-6)) - j((3)(5) - (3)(2)) + k((3)(-6) - (-2)(2))$.
$= i(-10 + 18) - j(15 - 6) + k(-18 + 4)$.
$= 8i - 9j - 14k$.

(A) Let $b$ and $c$ be two non-collinear unit vectors. Since they are non-collinear,$b, c,$ and $k = \frac{b \times c}{|b \times c|}$ form an orthonormal basis for the space $\mathbb{R}^3$.
Any vector $a$ can be expressed as a linear combination of these basis vectors:
$a = (a \cdot b)b + (a \cdot c)c + (a \cdot k)k$
Substituting $k = \frac{b \times c}{|b \times c|}$ into the expression,we get:
$a = (a \cdot b)b + (a \cdot c)c + \left(a \cdot \frac{b \times c}{|b \times c|}\right) \frac{b \times c}{|b \times c|}$
However,the term in the question is $\frac{a \cdot (b \times c)}{|b \times c|} (b \times c)$.
Note that $\frac{a \cdot (b \times c)}{|b \times c|} (b \times c) = (a \cdot k) (|b \times c| k) = (a \cdot k) (\sin \alpha) k$,where $\alpha$ is the angle between $b$ and $c$.
Since $|b \times c| = \sin \alpha$,the expression simplifies to:
$(a \cdot b)b + (a \cdot c)c + (a \cdot k)k = a$.

(A) The unit vectors along $a$ and $b$ are given by:
$\hat{a} = \frac{a}{|a|} = \frac{7i - 4j - 4k}{\sqrt{7^2 + (-4)^2 + (-4)^2}} = \frac{7i - 4j - 4k}{\sqrt{49 + 16 + 16}} = \frac{7i - 4j - 4k}{9}$
$\hat{b} = \frac{b}{|b|} = \frac{-2i - j + 2k}{\sqrt{(-2)^2 + (-1)^2 + 2^2}} = \frac{-2i - j + 2k}{\sqrt{4 + 1 + 4}} = \frac{-2i - j + 2k}{3} = \frac{-6i - 3j + 6k}{9}$
The internal bisector vector is proportional to $\hat{a} + \hat{b} = \frac{1}{9}(7i - 4j - 4k - 6i - 3j + 6k) = \frac{1}{9}(i - 7j + 2k)$.
Let $c = \lambda(\hat{a} + \hat{b}) = \frac{\lambda}{9}(i - 7j + 2k)$.
Given $|c| = 5\sqrt{6}$,we have $|c|^2 = 25 \times 6 = 150$.
$|c|^2 = \frac{\lambda^2}{81}(1^2 + (-7)^2 + 2^2) = \frac{\lambda^2}{81}(1 + 49 + 4) = \frac{54\lambda^2}{81} = \frac{2\lambda^2}{3}$.
Equating the two: $\frac{2\lambda^2}{3} = 150 \Rightarrow \lambda^2 = 225 \Rightarrow \lambda = 15$ (for internal bisector).
Thus,$c = \frac{15}{9}(i - 7j + 2k) = \frac{5}{3}(i - 7j + 2k)$.

(A) Since $c$ is coplanar with $a$ and $b$,we can write $c = xa + yb$ for some scalars $x$ and $y$.
Substituting the given vectors:
$c = x(2i + j + k) + y(i + 2j - k) = (2x + y)i + (x + 2y)j + (x - y)k$.
Given that $c$ is perpendicular to $a$,their dot product must be zero:
$a \cdot c = 0 \implies 2(2x + y) + 1(x + 2y) + 1(x - y) = 0$.
$4x + 2y + x + 2y + x - y = 0 \implies 6x + 3y = 0 \implies y = -2x$.
Substituting $y = -2x$ into the expression for $c$:
$c = (2x - 2x)i + (x - 4x)j + (x + 2x)k = -3xj + 3xk = 3x(-j + k)$.
Since $c$ is a unit vector,$|c| = 1$:
$|3x(-j + k)| = 1 \implies |3x| \sqrt{(-1)^2 + 1^2} = 1 \implies |3x| \sqrt{2} = 1 \implies x = \pm \frac{1}{3\sqrt{2}}$.
Thus,$c = 3(\pm \frac{1}{3\sqrt{2}})(-j + k) = \pm \frac{1}{\sqrt{2}}(-j + k)$.
Comparing with the given options,the correct answer is $\frac{1}{\sqrt{2}}(-j + k)$.

(A) Given equations are $r \times a = b \times a$ and $r \times b = a \times b$.
Adding these two equations,we get:
$r \times a + r \times b = b \times a + a \times b$
$r \times (a + b) = b \times a - b \times a = 0$
This implies that $r$ is parallel to $(a + b)$.
Given $a = i + j$ and $b = 2i - k$,we have $a + b = (i + j) + (2i - k) = 3i + j - k$.
Thus,$r = \lambda(3i + j - k)$.
For the intersection point,we check if $r = a + b$ satisfies the original equations.
If $r = 3i + j - k$,then $r \times a = (3i + j - k) \times (i + j) = 3(i \times i) + 3(i \times j) + (j \times i) + (j \times j) - (k \times i) - (k \times j) = 0 + 3k - k + 0 - j + i = i - j + 2k$.
Also $b \times a = (2i - k) \times (i + j) = 2(i \times i) + 2(i \times j) - (k \times i) - (k \times j) = 0 + 2k - j + i = i - j + 2k$.
Since $r \times a = b \times a$ and $r \times b = a \times b$ are satisfied for $r = 3i + j - k$,the point of intersection is $3i + j - k$.

(B) The coordinates of $O$ are $(0, 0, 0)$,$P$ are $(1, -2, 1)$,and $Q$ are $(2, 3, 4)$.
The direction ratios of line $OP$ are $(1-0, -2-0, 1-0) = (1, -2, 1)$.
The direction ratios of line $OQ$ are $(2-0, 3-0, 4-0) = (2, 3, 4)$.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Calculating the dot product: $(1 \times 2) + (-2 \times 3) + (1 \times 4) = 2 - 6 + 4 = 0$.
Since the dot product is $0$,the lines $OP$ and $OQ$ are perpendicular,i.e.,$OP \perp OQ$.

(B) First,calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$AB = \sqrt{(7-5)^2 + (-4 - (-1))^2 + (7-1)^2} = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$BC = \sqrt{(1-7)^2 + (-6 - (-4))^2 + (10-7)^2} = \sqrt{(-6)^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
$CD = \sqrt{(-1-1)^2 + (-3 - (-6))^2 + (4-10)^2} = \sqrt{(-2)^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$DA = \sqrt{(5 - (-1))^2 + (-1 - (-3))^2 + (1-4)^2} = \sqrt{6^2 + 2^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Since all sides are equal $(AB = BC = CD = DA = 7)$,the quadrilateral is either a square or a rhombus.
Next,check the dot product of adjacent vectors to see if the angle is $90^\circ$.
$\overrightarrow{AB} = (7-5, -4-(-1), 7-1) = (2, -3, 6)$.
$\overrightarrow{BC} = (1-7, -6-(-4), 10-7) = (-6, -2, 3)$.
$\overrightarrow{AB} \cdot \overrightarrow{BC} = (2)(-6) + (-3)(-2) + (6)(3) = -12 + 6 + 18 = 12 \neq 0$.
Since the dot product is not zero,the angle is not $90^\circ$. Therefore,$ABCD$ is a rhombus.

(B) The vector $\vec{PQ} = (4-3)\hat{i} + (6-4)\hat{j} + (3-5)\hat{k} = 1\hat{i} + 2\hat{j} - 2\hat{k}$.
The magnitude of $\vec{PQ}$ is $|\vec{PQ}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The vector $\vec{RS} = (1 - (-1))\hat{i} + (0-2)\hat{j} + (5-4)\hat{k} = 2\hat{i} - 2\hat{j} + 1\hat{k}$.
The projection of $\vec{RS}$ on $\vec{PQ}$ is given by the formula $\frac{\vec{RS} \cdot \vec{PQ}}{|\vec{PQ}|}$.
Calculate the dot product: $\vec{RS} \cdot \vec{PQ} = (2)(1) + (-2)(2) + (1)(-2) = 2 - 4 - 2 = -4$.
Therefore,the projection is $\frac{-4}{3}$.

(C) Let the direction ratios of the two lines be $\vec{a} = (1, 1, 2)$ and $\vec{b} = (\sqrt{3} - 1, -\sqrt{3} - 1, 4)$.
The cosine of the angle $\theta$ between the lines is given by the formula:
$\cos \theta = \frac{|a_1b_1 + a_2b_2 + a_3b_3|}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}$
Calculating the dot product:
$a_1b_1 + a_2b_2 + a_3b_3 = 1(\sqrt{3} - 1) + 1(-\sqrt{3} - 1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$.
Calculating the magnitudes:
$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
$|\vec{b}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2} = \sqrt{(3 + 1 - 2\sqrt{3}) + (3 + 1 + 2\sqrt{3}) + 16} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
Substituting the values:
$\cos \theta = \frac{6}{\sqrt{6} \times 2\sqrt{6}} = \frac{6}{2 \times 6} = \frac{6}{12} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^o$.

(A) The direction ratios of the line segment $AB$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (3-2, 5-3, -3-(-1)) = (1, 2, -2)$.
The direction ratios of the line segment $CD$ are given by $(x_4-x_3, y_4-y_3, z_4-z_3) = (3-1, 5-2, 7-3) = (2, 3, 4)$.
Let the direction ratios of $AB$ be $(a_1, b_1, c_1) = (1, 2, -2)$ and the direction ratios of $CD$ be $(a_2, b_2, c_2) = (2, 3, 4)$.
Calculate the dot product of the direction vectors: $a_1 a_2 + b_1 b_2 + c_1 c_2 = (1)(2) + (2)(3) + (-2)(4) = 2 + 6 - 8 = 0$.
Since the dot product is $0$,the lines $AB$ and $CD$ are perpendicular to each other.
Therefore,the angle $\theta$ between them is $\frac{\pi}{2}$.