Vector or Cross product of two vectors and its applications Questions in English

(A) Given $\vec{u} \cdot \hat{n} = 0$ and $\vec{v} \cdot \hat{n} = 0$,it implies that $\hat{n}$ is perpendicular to both $\vec{u}$ and $\vec{v}$.
Therefore,$\hat{n}$ must be parallel to the cross product $\vec{u} \times \vec{v}$.
$\vec{u} \times \vec{v} = (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = -\hat{i} \times \hat{j} + \hat{j} \times \hat{i} = -\hat{k} - \hat{k} = -2\hat{k}$.
The unit vector $\hat{n}$ is given by $\hat{n} = \pm \frac{\vec{u} \times \vec{v}}{|\vec{u} \times \vec{v}|} = \pm \frac{-2\hat{k}}{2} = \pm \hat{k}$.
Now,we calculate $|\vec{w} \cdot \hat{n}| = |(\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (\pm \hat{k})| = |\pm 3| = 3$.

(A) Given vectors are $a = i + j + 2k$ and $b = 3i + j + k$.
The cross product $a \times b$ is calculated using the determinant:
$a \times b = \begin{vmatrix} i & j & k \\ 1 & 1 & 2 \\ 3 & 1 & 1 \end{vmatrix}$
Expanding the determinant along the first row:
$a \times b = i(1 \times 1 - 2 \times 1) - j(1 \times 1 - 2 \times 3) + k(1 \times 1 - 1 \times 3)$
$a \times b = i(1 - 2) - j(1 - 6) + k(1 - 3)$
$a \times b = -i - j(-5) + k(-2)$
$a \times b = -i + 5j - 2k$

(B) In $\Delta ABC$,by the triangle law of vector addition,we have $\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}$.
Given $\vec{BC} = \vec{a}$,$\vec{CA} = \vec{b}$,and $\vec{AB} = \vec{c}$,we have $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
Taking the cross product with $\vec{a}$ on both sides: $\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times \vec{0}$.
Since $\vec{a} \times \vec{a} = \vec{0}$,we get $\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}$,which implies $\vec{a} \times \vec{b} = \vec{c} \times \vec{a}$.
Similarly,taking the cross product with $\vec{b}$ on both sides: $\vec{b} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{b} \times \vec{0}$.
This gives $\vec{b} \times \vec{a} + \vec{b} \times \vec{c} = \vec{0}$,which implies $\vec{b} \times \vec{c} = \vec{a} \times \vec{b}$.
Therefore,$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$.

(D) Given that $\hat{u}$ and $\hat{v}$ are unit vectors,so $|\hat{u}| = 1$ and $|\hat{v}| = 1$.
Let $\vec{w} = 2\hat{u} \times 3\hat{v} = 6(\hat{u} \times \hat{v})$.
We are given that $\vec{w}$ is a unit vector,so $|\vec{w}| = 1$.
Therefore,$|6(\hat{u} \times \hat{v})| = 1$.
$6|\hat{u}||\hat{v}|\sin\theta = 1$,where $\theta$ is the angle between $\hat{u}$ and $\hat{v}$.
Since $|\hat{u}| = 1$ and $|\hat{v}| = 1$,we have $6(1)(1)\sin\theta = 1$.
$6\sin\theta = 1$,which implies $\sin\theta = \frac{1}{6}$.
Since $\theta$ is an acute angle,$\sin\theta$ is positive,and there is exactly one value of $\theta$ in the interval $(0, \frac{\pi}{2})$ such that $\sin\theta = \frac{1}{6}$,which is $\theta = \arcsin(\frac{1}{6})$.
Thus,there is exactly one value of $\theta$.

(C) Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$.
$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix}$
$= \hat{i}(0 - 1) - \hat{j}(0 - 1) + \hat{k}(1 - 1)$
$= -\hat{i} + \hat{j} + 0\hat{k} = -(\hat{i} - \hat{j})$.
Any vector perpendicular to both is a scalar multiple of this vector,which can be written as $c(\hat{i} - \hat{j})$ for some scalar $c$.

(C) Let $\vec{a} = 3i + 2j - k$ and $\vec{b} = 12i + 5j - 5k$.
$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 12 & 5 & -5 \end{vmatrix}$
$= i(2(-5) - (-1)(5)) - j(3(-5) - (-1)(12)) + k(3(5) - 2(12))$
$= i(-10 + 5) - j(-15 + 12) + k(15 - 24)$
$= -5i + 3j - 9k$.
The magnitude of this vector is $|\vec{a} \times \vec{b}| = \sqrt{(-5)^2 + 3^2 + (-9)^2} = \sqrt{25 + 9 + 81} = \sqrt{115}$.
The unit vector perpendicular to both is $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{-5i + 3j - 9k}{\sqrt{115}}$.
Comparing with the given options,the correct option is $\frac{-5i + 3j - 9k}{\sqrt{115}}$.

(B) In a triangle $ABC$,the sides are represented by vectors $\vec{BC} = a$,$\vec{CA} = b$,and $\vec{AB} = c$.
By the triangle law of vector addition,the sum of vectors representing the sides of a triangle taken in order is zero.
Thus,$\vec{BC} + \vec{CA} + \vec{AB} = 0$,which implies $a + b + c = 0$.
Taking the cross product of this equation with $a$:
$a \times (a + b + c) = a \times 0$
$a \times a + a \times b + a \times c = 0$
Since $a \times a = 0$,we get $a \times b = c \times a$.
Similarly,taking the cross product with $b$:
$b \times (a + b + c) = b \times 0$
$b \times a + b \times b + b \times c = 0$
$b \times a + b \times c = 0 \implies a \times b = b \times c$.
Therefore,$a \times b = b \times c = c \times a$.

(A) Given $\vec{r} \times \vec{a} = \vec{b}$.
Since $\vec{a}$ and $\vec{b}$ are perpendicular,$\vec{a} \cdot \vec{b} = 0$.
We can express $\vec{r}$ as a linear combination of $\vec{a}$,$\vec{b}$,and $\vec{a} \times \vec{b}$:
$\vec{r} = x\vec{a} + y\vec{b} + z(\vec{a} \times \vec{b})$ for some scalars $x, y, z$.
Taking the cross product with $\vec{a}$ on both sides:
$(\vec{r} \times \vec{a}) = (x\vec{a} + y\vec{b} + z(\vec{a} \times \vec{b})) \times \vec{a}$
$\vec{b} = x(\vec{a} \times \vec{a}) + y(\vec{b} \times \vec{a}) + z((\vec{a} \times \vec{b}) \times \vec{a})$
Since $\vec{a} \times \vec{a} = 0$ and $(\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{b} \cdot \vec{a})\vec{a}$:
$\vec{b} = y(\vec{b} \times \vec{a}) + z((\vec{a} \cdot \vec{a})\vec{b} - 0)$
$\vec{b} = -y(\vec{a} \times \vec{b}) + z(\vec{a} \cdot \vec{a})\vec{b}$
Comparing coefficients of $\vec{b}$ and $(\vec{a} \times \vec{b})$:
$-y = 0 \implies y = 0$
$z(\vec{a} \cdot \vec{a}) = 1 \implies z = \frac{1}{\vec{a} \cdot \vec{a}}$
Substituting these into the expression for $\vec{r}$:
$\vec{r} = x\vec{a} + \frac{1}{\vec{a} \cdot \vec{a}} (\vec{a} \times \vec{b})$.

(C) The unit vector $\hat{n}$ perpendicular to the plane containing vectors $\vec{a}$ and $\vec{b}$ is given by $\hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -3 \\ 4 & 3 & -1 \end{vmatrix}$
$= \hat{i}((-6)(-1) - (-3)(3)) - \hat{j}((2)(-1) - (-3)(4)) + \hat{k}((2)(3) - (-6)(4))$
$= \hat{i}(6 + 9) - \hat{j}(-2 + 12) + \hat{k}(6 + 24)$
$= 15\hat{i} - 10\hat{j} + 30\hat{k}$.
Now,find the magnitude $|\vec{a} \times \vec{b}| = \sqrt{15^2 + (-10)^2 + 30^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35$.
The unit vector is $\pm \frac{15\hat{i} - 10\hat{j} + 30\hat{k}}{35} = \pm \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7}$.
Comparing this with the given options,option $C$ is $\frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7}$.

(C) The formula for the sine of the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by $\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(-2 - 3) - \hat{j}(1 - 6) + \hat{k}(1 - (-4)) = -5\hat{i} + 5\hat{j} + 5\hat{k}$.
Next,find the magnitude of the cross product:
$|\vec{a} \times \vec{b}| = \sqrt{(-5)^2 + 5^2 + 5^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}$.
Now,find the magnitudes of vectors $\vec{a}$ and $\vec{b}$:
$|\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
$|\vec{b}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Finally,substitute these values into the formula:
$\sin \theta = \frac{5\sqrt{3}}{\sqrt{14} \cdot \sqrt{6}} = \frac{5\sqrt{3}}{\sqrt{84}} = \frac{5\sqrt{3}}{2\sqrt{21}} = \frac{5\sqrt{3}}{2\sqrt{7}\sqrt{3}} = \frac{5}{2\sqrt{7}}$.

(C) Let the position vectors of the points be $\vec{r}_1 = \hat{i} - \hat{j} + \hat{k}$ and $\vec{r}_2 = 2\hat{i} - 3\hat{j} - \hat{k}$.
The force $\vec{F} = 3\hat{i} + 2\hat{j} - \hat{k}$ acts at $\vec{r}_1$ and $-\vec{F}$ acts at $\vec{r}_2$.
The torque $\vec{\tau}$ of the couple is given by $\vec{\tau} = \vec{r} \times \vec{F}$,where $\vec{r} = \vec{r}_1 - \vec{r}_2$.
$\vec{r} = (\hat{i} - \hat{j} + \hat{k}) - (2\hat{i} - 3\hat{j} - \hat{k}) = -\hat{i} + 2\hat{j} + 2\hat{k}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 3 & 2 & -1 \end{vmatrix} = \hat{i}(-2 - 4) - \hat{j}(1 - 6) + \hat{k}(-2 - 6) = -6\hat{i} + 5\hat{j} - 8\hat{k}$.
The magnitude of the torque is $|\vec{\tau}| = \sqrt{(-6)^2 + 5^2 + (-8)^2} = \sqrt{36 + 25 + 64} = \sqrt{125} = 5\sqrt{5}$.

(A) Let the points be $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$.
We find two vectors in the plane:
$\vec{AB} = (2-1)i + (0-(-1))j + (-1-2)k = i + j - 3k$
$\vec{AC} = (0-1)i + (2-(-1))j + (1-2)k = -i + 3j - k$
The vector perpendicular to the plane is given by the cross product $\vec{n} = \vec{AB} \times \vec{AC}$.
$\vec{n} = \begin{vmatrix} i & j & k \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix} = i(-1 - (-9)) - j(-1 - 3) + k(3 - (-1)) = 8i + 4j + 4k$.
The unit vector perpendicular to the plane is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|}$.
$|\vec{n}| = \sqrt{8^2 + 4^2 + 4^2} = \sqrt{64 + 16 + 16} = \sqrt{96} = 4\sqrt{6}$.
$\hat{n} = \pm \frac{8i + 4j + 4k}{4\sqrt{6}} = \pm \frac{2i + j + k}{\sqrt{6}}$.

(B) The area of the quadrilateral $OABC$ can be calculated as the sum of the areas of $\triangle OAB$ and $\triangle OBC$.
Area of $\triangle OAB = \frac{1}{2} |\overline{OA} \times \overline{OB}| = \frac{1}{2} |\vec{a} \times (10\vec{a} + 2\vec{b})| = \frac{1}{2} |10(\vec{a} \times \vec{a}) + 2(\vec{a} \times \vec{b})| = \frac{1}{2} |0 + 2(\vec{a} \times \vec{b})| = |\vec{a} \times \vec{b}|$.
Area of $\triangle OBC = \frac{1}{2} |\overline{OB} \times \overline{OC}| = \frac{1}{2} |(10\vec{a} + 2\vec{b}) \times \vec{b}| = \frac{1}{2} |10(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{b})| = \frac{1}{2} |10(\vec{a} \times \vec{b}) + 0| = 5|\vec{a} \times \vec{b}|$.
Thus,$p = |\vec{a} \times \vec{b}| + 5|\vec{a} \times \vec{b}| = 6|\vec{a} \times \vec{b}|$.
The area $q$ of the parallelogram with adjacent sides $OA$ and $OC$ is given by $q = |\vec{a} \times \vec{b}|$.
Therefore,$p/q = \frac{6|\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|} = 6$.

(B) The cross product $\vec{a} \times \vec{b}$ is given by the determinant of the matrix formed by the unit vectors and the components of the vectors:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 6 & -3 & 2 \end{vmatrix}$
Expanding the determinant along the first row:
$= \hat{i} ((2)(2) - (-1)(-3)) - \hat{j} ((2)(2) - (-1)(6)) + \hat{k} ((2)(-3) - (2)(6))$
$= \hat{i} (4 - 3) - \hat{j} (4 + 6) + \hat{k} (-6 - 12)$
$= \hat{i} (1) - \hat{j} (10) + \hat{k} (-18)$
$= \hat{i} - 10\hat{j} - 18\hat{k}$

(B) To find the unit vector perpendicular to the plane containing points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$,we first find two vectors lying in the plane:
$\vec{PQ} = (2-1)\hat{i} + (0-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$
$\vec{PR} = (0-1)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$
Now,find the cross product $\vec{n} = \vec{PQ} \times \vec{PR}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1)) = 8\hat{i} + 4\hat{j} + 4\hat{k}$
Simplifying the vector by dividing by $4$,we get a vector parallel to the normal: $2\hat{i} + \hat{j} + \hat{k}$.
The magnitude is $\sqrt{2^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$.
Thus,the unit vector is $\frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$.

(B) First,calculate the cross product $a \times b$:
$a \times b = \begin{vmatrix} i & j & k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = i(0 - (-2)) - j(0 - (-2)) + k(2 - 1) = 2i - 2j + k$.
The magnitude is $|a \times b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$.
Given $|c - a| = 2\sqrt{2}$,squaring both sides gives $|c - a|^2 = 8$.
$|c|^2 - 2(a \cdot c) + |a|^2 = 8$.
Since $a \cdot c = |c|$ and $|a|^2 = 2^2 + 1^2 + (-2)^2 = 9$,we have:
$|c|^2 - 2|c| + 9 = 8 \implies |c|^2 - 2|c| + 1 = 0 \implies (|c| - 1)^2 = 0 \implies |c| = 1$.
Finally,the magnitude of the cross product is:
$|(a \times b) \times c| = |a \times b| |c| \sin(30^{\circ}) = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(C) The vector product (cross product) of two vectors $\vec{a}$ and $\vec{b}$,denoted by $\vec{a} \times \vec{b}$,results in a vector that is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$.
To find a unit vector perpendicular to both,we divide this cross product by its magnitude.
The formula for the unit vector $\hat{n}$ perpendicular to both $\vec{a}$ and $\vec{b}$ is given by:
$\hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$.
Comparing this with the given options,the correct expression is $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$.

(B) 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 the vectors.
Given $|a| = 4$,$|b| = 2$,and $\theta = \pi/6$.
Substituting these values,we get $|a \times b| = 4 \times 2 \times \sin(\pi/6)$.
Since $\sin(\pi/6) = 1/2$,we have $|a \times b| = 4 \times 2 \times (1/2) = 4$.
Therefore,$|a \times b|^{2} = (4)^{2} = 16$.

(B) Let the position vectors be $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - 4\hat{k}$,and $\vec{c} = 7\hat{i} + 4\hat{j} + 9\hat{k}$.
Two vectors in the plane are $\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (3-1)\hat{j} + (-4-1)\hat{k} = \hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{AC} = \vec{c} - \vec{a} = (7-1)\hat{i} + (4-1)\hat{j} + (9-1)\hat{k} = 6\hat{i} + 3\hat{j} + 8\hat{k}$.
$A$ vector perpendicular to the plane is $\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -5 \\ 6 & 3 & 8 \end{vmatrix}$.
$\vec{n} = \hat{i}(16 - (-15)) - \hat{j}(8 - (-30)) + \hat{k}(3 - 12) = 31\hat{i} - 38\hat{j} - 9\hat{k}$.
The magnitude is $|\vec{n}| = \sqrt{31^2 + (-38)^2 + (-9)^2} = \sqrt{961 + 1444 + 81} = \sqrt{2486}$.
The unit vector is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{31\hat{i} - 38\hat{j} - 9\hat{k}}{\sqrt{2486}}$.
Comparing with the options,option $B$ is correct.

(A) The first plane is defined by vectors $\vec{n}_1 = \vec{i}$ and $\vec{n}_2 = \vec{i} + \vec{j}$. The normal to this plane is $\vec{N}_1 = \vec{n}_1 \times \vec{n}_2 = \vec{i} \times (\vec{i} + \vec{j}) = \vec{k}$.
The second plane is defined by vectors $\vec{n}_3 = \vec{i} - \vec{j}$ and $\vec{n}_4 = \vec{i} + \vec{k}$. The normal to this plane is $\vec{N}_2 = \vec{n}_3 \times \vec{n}_4 = (\vec{i} - \vec{j}) \times (\vec{i} + \vec{k}) = \vec{i} \times \vec{i} + \vec{i} \times \vec{k} - \vec{j} \times \vec{i} - \vec{j} \times \vec{k} = 0 - \vec{j} + \vec{k} - \vec{i} = -\vec{i} - \vec{j} + \vec{k}$.
The vector $\vec{a}$ is parallel to the line of intersection,so $\vec{a}$ is parallel to $\vec{N}_1 \times \vec{N}_2 = \vec{k} \times (-\vec{i} - \vec{j} + \vec{k}) = -(\vec{k} \times \vec{i}) - (\vec{k} \times \vec{j}) + (\vec{k} \times \vec{k}) = -\vec{j} + \vec{i} + 0 = \vec{i} - \vec{j}$.
Let $\vec{b} = \vec{i} - 2\vec{j} + 2\vec{k}$. The angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (1)(1) + (-1)(-2) + (0)(2) = 1 + 2 = 3$.
$|\vec{a}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$.
$\cos \theta = \frac{3}{\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}}$.
Thus,$\theta = \frac{\pi}{4}$. Since the vector $\vec{a}$ could be in the opposite direction,the angle could also be $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

(B) The direction vectors of lines $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$ respectively.
$A$ vector perpendicular to both $L_1$ and $L_2$ is given by the cross product $\vec{n} = \vec{v_1} \times \vec{v_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.
The magnitude of this vector is $|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.
The unit vector perpendicular to both lines is $\pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$.

(A) Given that $2\vec{u} \times 3\vec{v}$ is a unit vector,its magnitude must be $1$.
Thus,$|2\vec{u} \times 3\vec{v}| = 1$.
Since $\vec{u}$ and $\vec{v}$ are unit vectors,we have $|\vec{u}| = 1$ and $|\vec{v}| = 1$.
Using the property of the cross product,$|2\vec{u} \times 3\vec{v}| = 6|\vec{u}||\vec{v}|\sin\theta = 6(1)(1)\sin\theta = 6\sin\theta$.
According to the problem,$6\sin\theta = 1$,which implies $\sin\theta = \frac{1}{6}$.
Since $\theta$ is an acute angle,there exists exactly one value of $\theta$ such that $\theta = \arcsin(\frac{1}{6})$.

(D) Let $\vec{b} = x\vec{i} + y\vec{j} + z\vec{k}$.
Given $\vec{a} \cdot \vec{b} = 3$,where $\vec{a} = 0\vec{i} + 1\vec{j} - 1\vec{k}$.
So,$(0)(x) + (1)(y) + (-1)(z) = 3 \Rightarrow y - z = 3 \quad \dots(1)$.
Also,$\vec{a} \times \vec{b} = -\vec{c}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 1 & -1 \\ x & y & z \end{vmatrix} = \vec{i}(z + y) - \vec{j}(z) + \vec{k}(-x) = (y + z)\vec{i} - z\vec{j} - x\vec{k}$.
Given $\vec{a} \times \vec{b} = -\vec{c} = -(\vec{i} - \vec{j} - \vec{k}) = -\vec{i} + \vec{j} + \vec{k}$.
Comparing components:
$y + z = -1 \quad \dots(2)$
$-z = 1 \Rightarrow z = -1$
$-x = 1 \Rightarrow x = -1$
From $(1)$,$y - (-1) = 3 \Rightarrow y + 1 = 3 \Rightarrow y = 2$.
Wait,checking the cross product again: $\vec{a} \times \vec{b} = (z + y)\vec{i} - (z)\vec{j} - (x)\vec{k}$.
Given $\vec{a} \times \vec{b} = -\vec{c} = -\vec{i} + \vec{j} + \vec{k}$.
$y + z = -1$,$-z = 1 \Rightarrow z = -1$,$-x = 1 \Rightarrow x = -1$.
Then $y - 1 = -1 \Rightarrow y = 0$. Check $\vec{a} \cdot \vec{b} = y - z = 0 - (-1) = 1 \neq 3$.
Re-evaluating: $\vec{a} \times \vec{b} + \vec{c} = 0 \Rightarrow \vec{a} \times \vec{b} = -\vec{c} = -\vec{i} + \vec{j} + \vec{k}$.
Correct calculation: $\vec{a} \times \vec{b} = (z+y)\vec{i} - (z)\vec{j} - (x)\vec{k}$.
Equating: $z+y = -1$,$-z = 1 \Rightarrow z = -1$,$-x = 1 \Rightarrow x = -1$.
$y - 1 = -1 \Rightarrow y = 0$. This contradicts $\vec{a} \cdot \vec{b} = 3$.
Let's re-read the cross product: $\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 1 & -1 \\ x & y & z \end{vmatrix} = (z+y)\vec{i} - (z)\vec{j} - (x)\vec{k}$.
If $\vec{a} \times \vec{b} = -\vec{c} = -\vec{i} + \vec{j} + \vec{k}$,then $z+y = -1, -z = 1, -x = 1$. Result $\vec{b} = -\vec{i} + 0\vec{j} - \vec{k}$.
Given the options,let's check $\vec{b} = -\vec{i} + \vec{j} - 2\vec{k}$ (Option $D$).
$\vec{a} \cdot \vec{b} = (1)(1) + (-1)(-2) = 1 + 2 = 3$. (Matches)
$\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 1 & -1 \\ -1 & 1 & -2 \end{vmatrix} = \vec{i}(-2 + 1) - \vec{j}(0 - 1) + \vec{k}(0 - (-1)) = -\vec{i} + \vec{j} + \vec{k}$.
$\vec{a} \times \vec{b} + \vec{c} = (-\vec{i} + \vec{j} + \vec{k}) + (\vec{i} - \vec{j} - \vec{k}) = 0$. (Matches)
Thus,$\vec{b} = -\vec{i} + \vec{j} - 2\vec{k}$.

(C) Given $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$.
First,calculate the magnitude of $\vec a$: $|\vec a| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$.
Next,calculate the cross product $\vec a \times \vec b$:
$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat i(0 - (-2)) - \hat j(0 - (-2)) + \hat k(2 - 1) = 2\hat i - 2\hat j + \hat k$.
The magnitude is $|\vec a \times \vec b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$.
Given $|(\vec a \times \vec b) \times \vec c| = 3$ and the angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$,we use the formula $|\vec u \times \vec v| = |\vec u||\vec v| \sin \theta$:
$|\vec a \times \vec b||\vec c| \sin 30^\circ = 3 \implies 3 \cdot |\vec c| \cdot \frac{1}{2} = 3 \implies |\vec c| = 2$.
Now,use the condition $|\vec c - \vec a| = 3$. Squaring both sides:
$|\vec c - \vec a|^2 = 3^2 \implies |\vec c|^2 + |\vec a|^2 - 2(\vec a \cdot \vec c) = 9$.
Substitute the known values $|\vec c| = 2$ and $|\vec a| = 3$:
$2^2 + 3^2 - 2(\vec a \cdot \vec c) = 9 \implies 4 + 9 - 2(\vec a \cdot \vec c) = 9 \implies 13 - 2(\vec a \cdot \vec c) = 9$.
$2(\vec a \cdot \vec c) = 4 \implies \vec a \cdot \vec c = 2$.

(D) Let $b = b_1 i + b_2 j + b_3 k$.
Given $a \cdot b = 3$,we have $b_1 + b_2 + b_3 = 3$ ......$(i)$.
Given $a \times b = c$,where $a = (1, 1, 1)$ and $c = (0, 1, -1)$,we calculate the cross product:
$a \times b = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} = (b_3 - b_2)i + (b_1 - b_3)j + (b_2 - b_1)k$.
Comparing this with $c = (0, 1, -1)$,we get:
$b_3 - b_2 = 0$ ......$(ii)$
$b_1 - b_3 = 1$ ......$(iii)$
$b_2 - b_1 = -1$ ......$(iv)$
From $(ii)$,$b_2 = b_3$. Substituting into $(i)$,$b_1 + 2b_2 = 3$.
From $(iii)$,$b_1 = 1 + b_3 = 1 + b_2$.
Substituting $b_1$ into $b_1 + 2b_2 = 3$,we get $(1 + b_2) + 2b_2 = 3$,so $3b_2 = 2$,which means $b_2 = \frac{2}{3}$.
Then $b_3 = \frac{2}{3}$ and $b_1 = 1 + \frac{2}{3} = \frac{5}{3}$.
Thus,$b = \left( \frac{5}{3}, \frac{2}{3}, \frac{2}{3} \right)$.

(B) Given vectors are $a = 3i - 5j$ and $b = 6i + 3j$.
First,we calculate the cross product $c = a \times b$:
$c = \begin{vmatrix} i & j & k \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{vmatrix} = i(0 - 0) - j(0 - 0) + k(9 - (-30)) = 39k$.
Now,we find the magnitudes of the vectors:
$|a| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$.
$|b| = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45}$.
$|c| = \sqrt{0^2 + 0^2 + 39^2} = 39$.
Therefore,the ratio $|a|:|b|:|c|$ is $\sqrt{34} : \sqrt{45} : 39$.

(A) The plane containing vectors $i$ and $i + j$ has a normal vector $n_1 = i \times (i + j) = i \times i + i \times j = 0 + k = k.$ The equation of this plane is $r \cdot k = 0,$ which implies $z = 0.$
The plane containing vectors $i - j$ and $i + k$ has a normal vector $n_2 = (i - j) \times (i + k) = i \times i + i \times k - j \times i - j \times k = 0 - j + k - i = -i - j + k.$ The equation of this plane is $r \cdot (-i - j + k) = 0,$ which implies $-x - y + z = 0$ or $x + y - z = 0.$
The vector $a$ is parallel to the line of intersection of these two planes,so $a$ must be parallel to the cross product of the normals $n_1$ and $n_2$: $v = n_1 \times n_2 = k \times (-i - j + k) = -(k \times i) - (k \times j) + (k \times k) = -j + i + 0 = i - j.$
Thus,$a$ is parallel to $i - j.$
Let $\theta$ be the angle between $a = i - j$ and $b = i - 2j + 2k.$
$\cos \theta = \frac{a \cdot b}{|a| |b|} = \frac{(1)(1) + (-1)(-2) + (0)(2)}{\sqrt{1^2 + (-1)^2} \sqrt{1^2 + (-2)^2 + 2^2}} = \frac{1 + 2}{\sqrt{2} \sqrt{1 + 4 + 4}} = \frac{3}{\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}}.$
Since $a$ can be in the direction of $i - j$ or $-(i - j)$,the angle $\theta$ satisfies $\cos \theta = \pm \frac{1}{\sqrt{2}}.$
Therefore,$\theta = \frac{\pi}{4}$ or $\theta = \frac{3\pi}{4}$.

(C) Given $2\hat{a} = \hat{b} \times \hat{c} + 2\hat{b}$.
Taking the dot product with $\hat{b}$ on both sides:
$2(\hat{a} \cdot \hat{b}) = (\hat{b} \times \hat{c}) \cdot \hat{b} + 2(\hat{b} \cdot \hat{b})$
Since $(\hat{b} \times \hat{c}) \cdot \hat{b} = 0$ and $\hat{b} \cdot \hat{b} = 1$,we get $2(\hat{a} \cdot \hat{b}) = 2$,so $\hat{a} \cdot \hat{b} = 1$.
Since $\hat{a}$ and $\hat{b}$ are unit vectors,$\hat{a} \cdot \hat{b} = 1$ implies $\hat{a} = \hat{b}$.
Substituting $\hat{a} = \hat{b}$ into the original equation: $2\hat{b} = \hat{b} \times \hat{c} + 2\hat{b}$,which implies $\hat{b} \times \hat{c} = 0$.
This means $\hat{c} = \hat{b}$ or $\hat{c} = -\hat{b}$.
Case $1$: If $\hat{c} = \hat{b}$,then $|2\hat{a} + \hat{b} + \hat{c}| = |2\hat{b} + \hat{b} + \hat{b}| = |4\hat{b}| = 4$.
Case $2$: If $\hat{c} = -\hat{b}$,then $|2\hat{a} + \hat{b} + \hat{c}| = |2\hat{b} + \hat{b} - \hat{b}| = |2\hat{b}| = 2$.
The sum of possible values is $4 + 2 = 6$.

(B) Given $|\vec{a}| = |\vec{b}| = \sqrt{2}$ and $|\vec{a} + \vec{b}| = \sqrt{5}$.
Squaring both sides,$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 5$.
$2 + 2 + 2(\vec{a} \cdot \vec{b}) = 5 \Rightarrow 2(\vec{a} \cdot \vec{b}) = 1 \Rightarrow \vec{a} \cdot \vec{b} = \frac{1}{2}$.
Now,$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = (2)(2) - (\frac{1}{2})^2 = 4 - \frac{1}{4} = \frac{15}{4}$.
Given $\vec{c} = \vec{a} + 2\vec{b} + 2(\vec{a} \times \vec{b})$.
Since $\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,we have $\vec{c} \cdot (\vec{a} \times \vec{b}) = 2|\vec{a} \times \vec{b}|^2 = 2(\frac{15}{4}) = \frac{15}{2}$.
$|\vec{c}|^2 = |\vec{a} + 2\vec{b}|^2 + |2(\vec{a} \times \vec{b})|^2 = |\vec{a}|^2 + 4|\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) + 4|\vec{a} \times \vec{b}|^2$.
$|\vec{c}|^2 = 2 + 4(2) + 4(\frac{1}{2}) + 4(\frac{15}{4}) = 2 + 8 + 2 + 15 = 27$.
$|\vec{c}| = \sqrt{27} = 3\sqrt{3}$.

(D) The vector $\vec{B}$ is defined as $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y, A_z$ are the areas of the projections of $\Delta PQR$ on the $yz, zx, xy$ planes respectively.
By the property of vector areas,$\vec{B} = \frac{1}{2} (\vec{PQ} \times \vec{PR})$.
Given $P(1, 2, -3)$,$Q(-2, 1, -4)$,and $R(3, 4, -2)$,we have:
$\vec{PQ} = (-2-1)\hat{i} + (1-2)\hat{j} + (-4 - (-3))\hat{k} = -3\hat{i} - \hat{j} - \hat{k}$.
$\vec{PR} = (3-1)\hat{i} + (4-2)\hat{j} + (-2 - (-3))\hat{k} = 2\hat{i} + 2\hat{j} + \hat{k}$.
Now,$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -1 & -1 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(-1 - (-2)) - \hat{j}(-3 - (-2)) + \hat{k}(-6 - (-2)) = \hat{i} + \hat{j} - 4\hat{k}$.
Thus,$\vec{B} = \frac{1}{2} (\hat{i} + \hat{j} - 4\hat{k})$.
Then,$|\vec{B}|^2 = \left( \frac{1}{2} \right)^2 (1^2 + 1^2 + (-4)^2) = \frac{1}{4} (1 + 1 + 16) = \frac{18}{4} = \frac{9}{2}$.

(C) Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ and the two vectors defining the plane be $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by $\vec{a} \times \vec{b}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) = 5\hat{i} - 5\hat{j} - 5\hat{k}$.
We can take the normal vector as $\vec{n} = \hat{i} - \hat{j} - \hat{k}$.
Let $\theta$ be the angle between the vector $\vec{v}$ and the plane. Then the angle between $\vec{v}$ and the normal $\vec{n}$ is $90^{\circ} - \theta$.
Given $\cot \theta = \sqrt{2}$,we have $\tan \theta = \frac{1}{\sqrt{2}}$,so $\sin \theta = \frac{1}{\sqrt{3}}$ and $\cos \theta = \sqrt{\frac{2}{3}}$.
Using the dot product formula: $\cos(90^{\circ} - \theta) = \sin \theta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.
$\frac{1}{\sqrt{3}} = \frac{|x - y - z|}{\sqrt{x^2 + y^2 + z^2} \sqrt{1^2 + (-1)^2 + (-1)^2}}$.
$\frac{1}{\sqrt{3}} = \frac{|x - y - z|}{\sqrt{x^2 + y^2 + z^2} \sqrt{3}}$.
Squaring both sides: $\frac{1}{3} = \frac{(x - y - z)^2}{3(x^2 + y^2 + z^2)}$.
$x^2 + y^2 + z^2 = x^2 + y^2 + z^2 - 2xy + 2yz - 2zx$.
$0 = -2xy + 2yz - 2zx$.
$xy + zx = yz$,which is $x(y + z) = yz$.

(A) Given $|\vec c - \vec a| = \sqrt{14}$. Squaring both sides,we get $|\vec c|^2 + |\vec a|^2 - 2(\vec c \cdot \vec a) = 14$.
Since $\vec a = 2\hat i + \hat j - 2\hat k$,$|\vec a| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3$.
Thus,$|\vec c|^2 + 9 - 2(\vec c \cdot \vec a) = 14$,which implies $|\vec c|^2 - 2(\vec c \cdot \vec a) = 5$ ........$(1)$.
Given $\vec a \cdot \vec c + 2|\vec c| = 0$,so $\vec a \cdot \vec c = -2|\vec c|$.
Substituting this into $(1)$,we get $|\vec c|^2 - 2(-2|\vec c|) = 5$,which simplifies to $|\vec c|^2 + 4|\vec c| - 5 = 0$.
Factoring gives $(|\vec c| + 5)(|\vec c| - 1) = 0$. Since $|\vec c| > 0$,we have $|\vec c| = 1$.
Now,$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat i(0 - (-2)) - \hat j(0 - (-2)) + \hat k(2 - 1) = 2\hat i - 2\hat j + \hat k$.
The magnitude $|\vec a \times \vec b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$.
The magnitude of the cross product is $|(\vec a \times \vec b) \times \vec c| = |\vec a \times \vec b| |\vec c| \sin(30^o) = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(A) Let $\vec{r}_1 = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{r}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Given $a^2 + b^2 + c^2 = 1$,we have $|\vec{r}_1| = 1$.
The cross product is given by $\vec{r}_1 \times \vec{r}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & c \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(4b - 3c) - \hat{j}(4a - 2c) + \hat{k}(3a - 2b)$.
The square of the magnitude is $|\vec{r}_1 \times \vec{r}_2|^2 = (4b - 3c)^2 + (4a - 2c)^2 + (3a - 2b)^2$.
Using the property $|\vec{r}_1 \times \vec{r}_2| = |\vec{r}_1| |\vec{r}_2| \sin \theta$,where $\theta$ is the angle between the vectors.
Since $|\vec{r}_1| = 1$ and $|\vec{r}_2| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$,we have $|\vec{r}_1 \times \vec{r}_2| = \sqrt{29} \sin \theta$.
Therefore,$|\vec{r}_1 \times \vec{r}_2|^2 = 29 \sin^2 \theta$.
The maximum value of $\sin^2 \theta$ is $1$,so the maximum value of the expression is $29 \times 1 = 29$.

(B) Let $\vec{x} = a\hat{i} + b\hat{j} + c\hat{k}$.
Since $\vec{x}$ is a unit vector,$a^2 + b^2 + c^2 = 1$.
The cross product is given by:
$\vec{x} \times (\hat{i} - 2\hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & c \\ 1 & -2 & 1 \end{vmatrix} = (b+2c)\hat{i} - (a-c)\hat{j} + (-2a-b)\hat{k}$.
Equating this to $-\hat{i} + \hat{k}$,we get:
$b + 2c = -1$,$a - c = 0$,and $-2a - b = 1$.
From $a = c$,substituting into the other equations: $b + 2a = -1$ and $-2a - b = 1$,which are the same equation.
Using $a^2 + b^2 + c^2 = 1$,we get $a^2 + (-1-2a)^2 + a^2 = 1$.
$a^2 + 1 + 4a + 4a^2 + a^2 = 1 \Rightarrow 6a^2 + 4a = 0$.
So,$2a(3a + 2) = 0$,which gives $a = 0$ or $a = -\frac{2}{3}$.
If $a = 0$,then $c = 0$ and $b = -1$,so $\vec{x} = -\hat{j}$.
If $a = -\frac{2}{3}$,then $c = -\frac{2}{3}$ and $b = -1 - 2(-\frac{2}{3}) = -1 + \frac{4}{3} = \frac{1}{3}$.
Thus,$\vec{x} = -\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} = -\frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$.

(B) Given,$2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$.
From the given equation,we have $\vec{c} = -(2\vec{a} + 3\vec{b})$.
Now,consider the expression $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$.
Substitute $\vec{c} = -(2\vec{a} + 3\vec{b})$ into the expression:
$= \vec{a} \times \vec{b} + \vec{b} \times (-(2\vec{a} + 3\vec{b})) + (-(2\vec{a} + 3\vec{b})) \times \vec{a}$
Using the distributive property of the cross product:
$= \vec{a} \times \vec{b} - 2(\vec{b} \times \vec{a}) - 3(\vec{b} \times \vec{b}) - 2(\vec{a} \times \vec{a}) - 3(\vec{b} \times \vec{a})$
Since $\vec{x} \times \vec{x} = \vec{0}$ and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:
$= \vec{a} \times \vec{b} - 2(-(\vec{a} \times \vec{b})) - 3(\vec{0}) - 2(\vec{0}) - 3(-(\vec{a} \times \vec{b}))$
$= \vec{a} \times \vec{b} + 2(\vec{a} \times \vec{b}) + 3(\vec{a} \times \vec{b})$
$= 6(\vec{a} \times \vec{b})$.
Alternatively,from $2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$,taking cross product with $\vec{b}$:
$2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{b}) + (\vec{c} \times \vec{b}) = \vec{0} \Rightarrow 2(\vec{a} \times \vec{b}) = \vec{b} \times \vec{c} \Rightarrow \vec{a} \times \vec{b} = \frac{1}{2}(\vec{b} \times \vec{c})$.
Substituting this into $6(\vec{a} \times \vec{b})$ gives $6 \times \frac{1}{2}(\vec{b} \times \vec{c}) = 3(\vec{b} \times \vec{c})$.

(C) Let $\vec{u} = \sqrt{3}(\vec{a} \times \vec{b})$ and $\vec{v} = \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.
We know that $\vec{v} = \vec{a} \times (\vec{b} \times \vec{a})$.
Since $\vec{a} \times \vec{b}$ is perpendicular to $\vec{a}$ and $\vec{b}$,and $\vec{a} \times (\vec{b} \times \vec{a})$ is also perpendicular to $\vec{a}$,we observe that $\vec{u} \perp \vec{v}$.
Thus,the triangle is a right-angled triangle with one angle equal to $\frac{\pi}{2}$.
Now,$|\vec{u}| = \sqrt{3}|\vec{a} \times \vec{b}| = \sqrt{3}|\vec{a}||\vec{b}|\sin \theta = \sqrt{3}|\vec{b}|\sin \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Also,$|\vec{v}| = |\vec{a} \times (\vec{b} \times \vec{a})| = |\vec{a}| |\vec{b} \times \vec{a}| \sin(90^{\circ}) = |\vec{b}|\sin \theta$.
In the right-angled triangle,let $\alpha$ be the angle between the sides $\vec{u}$ and $\vec{v}$.
Then $\tan \alpha = \frac{|\vec{u}|}{|\vec{v}|} = \frac{\sqrt{3}|\vec{b}|\sin \theta}{|\vec{b}|\sin \theta} = \sqrt{3}$.
Therefore,$\alpha = \frac{\pi}{3}$.
The third angle is $\pi - \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.
Thus,the angles are $\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}$.

(B) Let the diagonals be $\vec{d_1} = 8\hat{i} - 6\hat{j} + 0\hat{k}$ and $\vec{d_2} = 3\hat{i} + 4\hat{j} - 12\hat{k}$.
The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula $\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
First,calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & -6 & 0 \\ 3 & 4 & -12 \end{vmatrix}$
$= \hat{i}((-6)(-12) - (0)(4)) - \hat{j}((8)(-12) - (0)(3)) + \hat{k}((8)(4) - (-6)(3))$
$= \hat{i}(72 - 0) - \hat{j}(-96 - 0) + \hat{k}(32 + 18)$
$= 72\hat{i} + 96\hat{j} + 50\hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{d_1} \times \vec{d_2}| = \sqrt{72^2 + 96^2 + 50^2}$
$= \sqrt{5184 + 9216 + 2500}$
$= \sqrt{16900} = 130$.
Therefore,the area of the parallelogram is $\frac{1}{2} \times 130 = 65$ sq. units.

(B) Given $\vec{b} = 3\hat{j} + 4\hat{k}$ and $\vec{a} = \hat{i} + \hat{j}$.
$\vec{b_1}$ is parallel to $\vec{a}$,so $\vec{b_1} = \text{proj}_{\vec{a}} \vec{b} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a}$.
$\vec{b} \cdot \vec{a} = (3\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j}) = 3$.
$|\vec{a}|^2 = |\hat{i} + \hat{j}|^2 = 1^2 + 1^2 = 2$.
Thus,$\vec{b_1} = \frac{3}{2}(\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}$.
Since $\vec{b} = \vec{b_1} + \vec{b_2}$,we have $\vec{b_2} = \vec{b} - \vec{b_1} = (3\hat{j} + 4\hat{k}) - (\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}) = -\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} + 4\hat{k}$.
Now,calculate the cross product $\vec{b_1} \times \vec{b_2}$:
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3/2 & 3/2 & 0 \\ -3/2 & 3/2 & 4 \end{vmatrix}$.
$= \hat{i} \left( (3/2)(4) - (0)(3/2) \right) - \hat{j} \left( (3/2)(4) - (0)(-3/2) \right) + \hat{k} \left( (3/2)(3/2) - (3/2)(-3/2) \right)$.
$= \hat{i}(6) - \hat{j}(6) + \hat{k}(9/4 + 9/4) = 6\hat{i} - 6\hat{j} + \frac{9}{2}\hat{k}$.

(D) Given $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$.
First,calculate the magnitude of $\vec a$: $|\vec a| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = 3$.
Next,calculate the cross product $\vec a \times \vec b$:
$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat i(0 - (-2)) - \hat j(0 - (-2)) + \hat k(2 - 1) = 2\hat i - 2\hat j + \hat k$.
The magnitude is $|\vec a \times \vec b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$.
Given $|\vec c - \vec a| = 2\sqrt 2$,squaring both sides gives $|\vec c - \vec a|^2 = 8$.
Expanding this,we get $|\vec c|^2 + |\vec a|^2 - 2(\vec c \cdot \vec a) = 8$.
Substituting $|\vec a| = 3$ and $\vec a \cdot \vec c = |\vec c|$,we have $|\vec c|^2 + 9 - 2|\vec c| = 8$.
This simplifies to $|\vec c|^2 - 2|\vec c| + 1 = 0$,which is $(|\vec c| - 1)^2 = 0$,so $|\vec c| = 1$.
Finally,the magnitude of the cross product is $|(\vec a \times \vec b) \times \vec c| = |\vec a \times \vec b| |\vec c| \sin 30^o$.
Substituting the values: $3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(D) Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ be the required unit vector.
Since $\vec{v}$ is perpendicular to $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$,we have $\vec{v} \cdot \vec{a} = 0$.
$2x - y + 2z = 0$ ...... $(i)$
Since $\vec{v}$ is coplanar with $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + 2\hat{j} - \hat{k}$,$\vec{v}$ can be written as a linear combination of $\vec{b}$ and $\vec{c}$.
$\vec{v} = p(\hat{i} + \hat{j} - \hat{k}) + q(2\hat{i} + 2\hat{j} - \hat{k}) = (p + 2q)\hat{i} + (p + 2q)\hat{j} - (p + q)\hat{k}$.
Comparing components,$x = p + 2q$,$y = p + 2q$,$z = -(p + q)$.
Substitute into $(i)$: $2(p + 2q) - (p + 2q) + 2(-p - q) = 0$.
$2p + 4q - p - 2q - 2p - 2q = 0 \Rightarrow -p = 0 \Rightarrow p = 0$.
Thus,$x = 2q, y = 2q, z = -q$.
Since $\vec{v}$ is a unit vector,$x^2 + y^2 + z^2 = 1$.
$(2q)^2 + (2q)^2 + (-q)^2 = 1 \Rightarrow 4q^2 + 4q^2 + q^2 = 1 \Rightarrow 9q^2 = 1 \Rightarrow q = \pm \frac{1}{3}$.
For $q = \frac{1}{3}$,$\vec{v} = \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$.
This matches option $D$.

(B) Let the vertices be $O(0, 0, 0)$,$P(1, 2, 1)$,$Q(2, 1, 3)$,and $R(-1, 1, 2)$.
To find the angle between faces $OPQ$ and $PQR$,we find the normal vectors to these faces.
For face $OPQ$,the normal vector $\vec{n_1} = \vec{OP} \times \vec{OQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.
For face $PQR$,the normal vector $\vec{n_2} = \vec{PQ} \times \vec{PR}$.
$\vec{PQ} = (2-1)\hat{i} + (1-2)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}$.
$\vec{PR} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$.
$\vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.
The angle $\theta$ between the faces is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25+1+9} = \sqrt{35}$.
$|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1+25+9} = \sqrt{35}$.
$\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$.
$\cos \theta = \frac{19}{\sqrt{35} \cdot \sqrt{35}} = \frac{19}{35}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{35}\right)$.

(A) Given that $\vec{a}, \vec{b},$ and $\vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$
Using the vector triple product formula,$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}.$
Given $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b},$ we have $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{2} \vec{b}.$
Since $\vec{b}$ and $\vec{c}$ are non-parallel,they are linearly independent. Comparing the coefficients of $\vec{b}$ and $\vec{c},$ we get $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = 0.$
We know $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \alpha = 1 \cdot 1 \cdot \cos \alpha = 0 \implies \alpha = 90^{\circ}.$
We know $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \beta = 1 \cdot 1 \cdot \cos \beta = \frac{1}{2} \implies \beta = 60^{\circ}.$
Thus,$|\alpha - \beta| = |90^{\circ} - 60^{\circ}| = 30^{\circ}.$

(D) Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.
The magnitude of the projection of vector $\vec{v} = 2\hat{i} + 3\hat{j} + \hat{k}$ on vector $\vec{n}$ is given by $\frac{|\vec{v} \cdot \vec{n}|}{|\vec{n}|}$.
$\vec{v} \cdot \vec{n} = (2)(1) + (3)(-2) + (1)(1) = 2 - 6 + 1 = -3$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
Therefore,the magnitude of the projection is $\frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{9}{6}} = \sqrt{\frac{3}{2}}$.

(A) First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(2 + x) - \hat{j}(3 - x) + \hat{k}(-3 - 2) = (x + 2)\hat{i} + (x - 3)\hat{j} - 5\hat{k}$.
Next,find the magnitude $r = |\vec{a} \times \vec{b}|$:
$r^2 = (x + 2)^2 + (x - 3)^2 + (-5)^2$
$r^2 = (x^2 + 4x + 4) + (x^2 - 6x + 9) + 25$
$r^2 = 2x^2 - 2x + 38 = 2(x^2 - x + 19)$.
Complete the square for the expression inside the parenthesis:
$r^2 = 2\left((x - \frac{1}{2})^2 + 19 - \frac{1}{4}\right) = 2\left((x - \frac{1}{2})^2 + \frac{75}{4}\right) = 2(x - \frac{1}{2})^2 + \frac{75}{2}$.
Since $(x - \frac{1}{2})^2 \geq 0$,the minimum value of $r^2$ is $\frac{75}{2}$.
Therefore,$r^2 \geq \frac{75}{2} \implies r \geq \sqrt{\frac{75}{2}} = \sqrt{\frac{25 \times 3}{2}} = 5\sqrt{\frac{3}{2}}$.
Thus,the condition is $r \geq 5\sqrt{\frac{3}{2}}$.

(A) Given $\vec{\alpha} = 3\hat{i} + \hat{j}$ and $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}.$
Since $\vec{\beta}_1$ is parallel to $\vec{\alpha},$ we have $\vec{\beta}_1 = \lambda \vec{\alpha} = \lambda(3\hat{i} + \hat{j}).$
Given $\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2,$ we have $\vec{\beta}_2 = \vec{\beta}_1 - \vec{\beta} = \lambda(3\hat{i} + \hat{j}) - (2\hat{i} - \hat{j} + 3\hat{k}) = (3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k}.$
Since $\vec{\beta}_2$ is perpendicular to $\vec{\alpha},$ $\vec{\beta}_2 \cdot \vec{\alpha} = 0.$
$((3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k}) \cdot (3\hat{i} + \hat{j}) = 0.$
$3(3\lambda - 2) + 1(\lambda + 1) = 0 \implies 9\lambda - 6 + \lambda + 1 = 0 \implies 10\lambda = 5 \implies \lambda = \frac{1}{2}.$
Thus,$\vec{\beta}_1 = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}$ and $\vec{\beta}_2 = (\frac{3}{2} - 2)\hat{i} + (\frac{1}{2} + 1)\hat{j} - 3\hat{k} = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}.$
Now,$\vec{\beta}_1 \times \vec{\beta}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix}.$
$= \hat{i}(-\frac{3}{2} - 0) - \hat{j}(-\frac{9}{2} - 0) + \hat{k}(\frac{9}{4} - (-\frac{1}{4})) = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{10}{4}\hat{k} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k}.$
$= \frac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k}).$

(A) First,calculate $\vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j}$.
Next,calculate $\vec{a} - \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 4\hat{k}$.
$A$ vector perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ is given by their cross product:
$\vec{v} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16 - 0) - \hat{j}(16 - 0) + \hat{k}(0 - 8) = 16\hat{i} - 16\hat{j} - 8\hat{k}$.
Simplify the vector: $8(2\hat{i} - 2\hat{j} - \hat{k})$.
The magnitude of this vector is $8 \sqrt{2^2 + (-2)^2 + (-1)^2} = 8 \sqrt{4+4+1} = 8 \times 3 = 24$.
We need a vector with magnitude $12$,which is half of the magnitude of $\vec{v}$.
Thus,the required vector is $\pm \frac{12}{24} \times 8(2\hat{i} - 2\hat{j} - \hat{k}) = \pm 4(2\hat{i} - 2\hat{j} - \hat{k})$.
Comparing with the options,$4(2\hat{i} - 2\hat{j} - \hat{k})$ is the correct choice.

(C) Given $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$,we can write $\vec{b} \times (\vec{c} - \vec{a}) = \vec{0}$.
This implies that $\vec{c} - \vec{a}$ is parallel to $\vec{b}$,so $\vec{c} - \vec{a} = k\vec{b}$ for some scalar $k$.
Thus,$\vec{c} = \vec{a} + k\vec{b}$.
Given $\vec{c} \cdot \vec{a} = 0$,we substitute $\vec{c}$:
$(\vec{a} + k\vec{b}) \cdot \vec{a} = 0 \Rightarrow |\vec{a}|^2 + k(\vec{b} \cdot \vec{a}) = 0$.
Calculating magnitudes and dot products:
$|\vec{a}|^2 = 1^2 + (-2)^2 + 1^2 = 6$.
$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(-2) + (1)(1) = 1 + 2 + 1 = 4$.
Substituting these values: $6 + k(4) = 0 \Rightarrow 4k = -6 \Rightarrow k = -\frac{3}{2}$.
Now,$\vec{c} \cdot \vec{b} = (\vec{a} + k\vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} + k|\vec{b}|^2$.
$|\vec{b}|^2 = 1^2 + (-1)^2 + 1^2 = 3$.
$\vec{c} \cdot \vec{b} = 4 + (-\frac{3}{2})(3) = 4 - \frac{9}{2} = \frac{8-9}{2} = -\frac{1}{2}$.