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

(A) First,calculate the cross product $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{vmatrix} = \hat{i}(-12 - (-2)) - \hat{j}(-8 - 1) + \hat{k}(4 - (-3)) = -10\hat{i} + 9\hat{j} + 7\hat{k}$
Next,calculate the cross product $\bar{a} \times \bar{c}$:
$\bar{a} \times \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(3 - (-1)) - \hat{j}(2 - (-1)) + \hat{k}(2 - 3) = 4\hat{i} - 3\hat{j} - \hat{k}$
Now,compute the dot product of the two resulting vectors:
$(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c}) = (-10\hat{i} + 9\hat{j} + 7\hat{k}) \cdot (4\hat{i} - 3\hat{j} - \hat{k})$
$= (-10)(4) + (9)(-3) + (7)(-1)$
$= -40 - 27 - 7 = -74$

(A) Let $\vec{a}$ and $\vec{b}$ be the vectors along the lines whose direction ratios are $2, 3, 1$ and $1, 2, 1$ respectively.
$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.
$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(2-1) + \hat{k}(4-3) = \hat{i} - \hat{j} + \hat{k}$.
Thus,the direction ratios are $1, -1, 1$.
Comparing this with the given options,the correct choice is $A$.

(A) Given $u = a - b$ and $v = a + b$.
We need to find $|u \times v|$.
$|u \times v| = |(a - b) \times (a + b)|$
$= |a \times a + a \times b - b \times a - b \times b|$
Since $a \times a = 0$ and $b \times b = 0$,and $b \times a = -(a \times b)$,we have:
$|u \times v| = |0 + a \times b + a \times b - 0| = |2(a \times b)| = 2|a \times b|$.
We know that $|a \times b|^{2} + (a \cdot b)^{2} = |a|^{2} |b|^{2}$.
Given $|a| = |b| = 2$,so $|a|^{2} = 4$ and $|b|^{2} = 4$.
$|a \times b|^{2} + (a \cdot b)^{2} = 4 \times 4 = 16$.
$|a \times b| = \sqrt{16 - (a \cdot b)^{2}}$.
Therefore,$|u \times v| = 2 \sqrt{16 - (a \cdot b)^{2}}$.

(C) Let the vertices of the triangle be $O(0), A(\overrightarrow{a}+\overrightarrow{b}),$ and $B(\overrightarrow{a}-\overrightarrow{b})$.
The area of the triangle is given by $\frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}|$.
Substituting the vectors,we get $\text{Area} = \frac{1}{2} |(\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b})|$.
Expanding the cross product: $(\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b}) = \overrightarrow{a} \times \overrightarrow{a} - \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{a} - \overrightarrow{b} \times \overrightarrow{b}$.
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$ and $\overrightarrow{b} \times \overrightarrow{b} = 0$,and $\overrightarrow{b} \times \overrightarrow{a} = -(\overrightarrow{a} \times \overrightarrow{b})$,we have:
$= 0 - (\overrightarrow{a} \times \overrightarrow{b}) - (\overrightarrow{a} \times \overrightarrow{b}) - 0 = -2(\overrightarrow{a} \times \overrightarrow{b})$.
Thus,$\text{Area} = \frac{1}{2} |-2(\overrightarrow{a} \times \overrightarrow{b})| = |\overrightarrow{a} \times \overrightarrow{b}|$.
Since $\overrightarrow{a} \perp \overrightarrow{b}$,$|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin(90^{\circ}) = 2 \times 3 \times 1 = 6$.

(B) Let the diagonals of the parallelogram be $\overrightarrow{d_1} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\overrightarrow{d_2} = -\hat{i} + 2\hat{j}$.
The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $\text{Area} = \frac{1}{2} |\overrightarrow{d_1} \times \overrightarrow{d_2}|$.
First,calculate the cross product $\overrightarrow{d_1} \times \overrightarrow{d_2}$:
$\overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix}$
$= \hat{i}((-3)(0) - (2)(2)) - \hat{j}((1)(0) - (2)(-1)) + \hat{k}((1)(2) - (-3)(-1))$
$= \hat{i}(0 - 4) - \hat{j}(0 + 2) + \hat{k}(2 - 3)$
$= -4\hat{i} - 2\hat{j} - \hat{k}$.
Now,calculate the magnitude of the cross product:
$|\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-4)^2 + (-2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$.
Therefore,the area of the parallelogram is $\frac{1}{2} \times \sqrt{21} = \frac{\sqrt{21}}{2}$.

(C) Let $\bar{u} = \bar{a} \times \bar{b}$ and $\bar{v} = \bar{c} \times \bar{d}$.
Given that $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are unit vectors,$|\bar{a}| = |\bar{b}| = |\bar{c}| = |\bar{d}| = 1$.
Since $\bar{a} \cdot \bar{b} = \frac{1}{2}$,the angle between $\bar{a}$ and $\bar{b}$ is $\frac{\pi}{3}$,so $|\bar{a} \times \bar{b}| = |\bar{a}||\bar{b}| \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.
Similarly,$|\bar{c} \times \bar{d}| = \frac{\sqrt{3}}{2}$.
Using the vector triple product identity,$[\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d} = (\bar{a} \times \bar{b}) \times (\bar{c} \times \bar{d}) = \bar{u} \times \bar{v}$.
The magnitude is $|\bar{u} \times \bar{v}| = |\bar{u}| |\bar{v}| \sin(\theta)$,where $\theta = \frac{\pi}{6}$.
$|\bar{u} \times \bar{v}| = (\frac{\sqrt{3}}{2}) (\frac{\sqrt{3}}{2}) \sin(\frac{\pi}{6}) = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$.

(B) Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$. Squaring both sides,we get $|\overline{c}|^2+|\overline{a}|^2-2(\overline{a} \cdot \overline{c})=8$.
Since $|\overline{a}|^2 = 2^2+1^2+(-2)^2 = 9$ and $\overline{a} \cdot \overline{c}=|\overline{c}|$,the equation becomes $|\overline{c}|^2+9-2|\overline{c}|=8$.
This simplifies to $(|\overline{c}|-1)^2=0$,so $|\overline{c}|=1$.
Next,calculate $\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i}-2\hat{j}+\hat{k}$.
The magnitude is $|\overline{a} \times \overline{b}| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{4+4+1} = 3$.
Finally,$|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin(30^{\circ}) = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(C) We are given $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \frac{\pi}{3} = 10$.
Substituting $|\vec{b}| = 5$ and $\cos \frac{\pi}{3} = \frac{1}{2}$,we get $5 \times |\vec{c}| \times \frac{1}{2} = 10$,which implies $|\vec{c}| = 4$.
Now,the magnitude of the cross product is $|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin \frac{\pi}{3} = 5 \times 4 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$.
Since $\vec{a}$ is perpendicular to $\vec{b} \times \vec{c}$,the angle between $\vec{a}$ and $(\vec{b} \times \vec{c})$ is $\frac{\pi}{2}$.
Therefore,$|\vec{a} \times(\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b} \times \vec{c}| \sin \frac{\pi}{2} = \sqrt{3} \times (10\sqrt{3}) \times 1 = 30$.

(B) First,calculate $|\bar{a}|^2 = 4^2 + 3^2 + 1^2 = 16 + 9 + 1 = 26$.
Also,$\bar{a} \cdot \bar{b} = (4)(1) + (3)(-2) + (1)(2) = 4 - 6 + 2 = 0$.
Using the vector triple product formula $\bar{a} \times (\bar{a} \times \bar{b}) = (\bar{a} \cdot \bar{b})\bar{a} - (\bar{a} \cdot \bar{a})\bar{b}$,since $\bar{a} \cdot \bar{b} = 0$,we have $\bar{a} \times (\bar{a} \times \bar{b}) = -|\bar{a}|^2 \bar{b} = -26\bar{b}$.
Now,let $\bar{v} = \bar{a} \times (\bar{a} \times \bar{b}) = -26\bar{b}$.
Then $\bar{a} \times (\bar{a} \times \bar{v}) = \bar{a} \times (\bar{a} \times (-26\bar{b})) = -26(\bar{a} \times (\bar{a} \times \bar{b}))$.
Substituting the previous result: $-26(-26\bar{b}) = 676\bar{b}$.
Thus,the correct option is $B$.

(A) Given that $\bar{b} \cdot \bar{c} = |\bar{b}| |\bar{c}| \cos(45^{\circ}) = 8$.
Substituting the values,we get $2 \times |\bar{c}| \times \frac{1}{\sqrt{2}} = 8$,which implies $|\bar{c}| = 4 \sqrt{2}$.
Since $\bar{a}, \bar{b}, \bar{c}$ are coplanar,the vector $\bar{b} \times \bar{c}$ is perpendicular to the plane containing $\bar{a}, \bar{b}, \bar{c}$.
Therefore,$\bar{a}$ is perpendicular to $\bar{b} \times \bar{c}$.
Using the property $|\bar{a} \times (\bar{b} \times \bar{c})| = |\bar{a}| |\bar{b} \times \bar{c}| \sin(90^{\circ}) = |\bar{a}| |\bar{b} \times \bar{c}|$.
We know $|\bar{b} \times \bar{c}| = |\bar{b}| |\bar{c}| \sin(45^{\circ}) = 2 \times 4 \sqrt{2} \times \frac{1}{\sqrt{2}} = 8$.
Thus,$|\bar{a} \times (\bar{b} \times \bar{c})| = 1 \times 8 = 8$.

(A) Given that $\bar{b} \cdot \bar{c} = |\bar{b}| |\bar{c}| \cos(45^{\circ}) = 8$.
Substituting the values,we get $2 \cdot |\bar{c}| \cdot \frac{1}{\sqrt{2}} = 8$,which implies $|\bar{c}| = 4\sqrt{2}$.
Since $\bar{a}, \bar{b}, \bar{c}$ are coplanar,the vector $\bar{v} = (\bar{b} \times \bar{c})$ is perpendicular to the plane containing $\bar{a}, \bar{b}, \bar{c}$.
Thus,$\bar{a}$ is perpendicular to $(\bar{b} \times \bar{c})$.
Therefore,$|\bar{a} \times (\bar{b} \times \bar{c})| = |\bar{a}| |\bar{b} \times \bar{c}| \sin(90^{\circ}) = |\bar{a}| |\bar{b} \times \bar{c}|$.
We know $|\bar{b} \times \bar{c}| = |\bar{b}| |\bar{c}| \sin(45^{\circ}) = 2 \cdot 4\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 8$.
Since $|\bar{a}| = 1$,the value is $1 \cdot 8 = 8$.

(A) Using the vector triple product formula,$\bar{a} \times(\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}$.
Given $\bar{a} \times(\bar{b} \times \bar{c}) = \frac{\bar{b}+\bar{c}}{\sqrt{2}}$,we have:
$(\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} = \frac{1}{\sqrt{2}} \bar{b} + \frac{1}{\sqrt{2}} \bar{c}$.
Rearranging the terms,we get:
$(\bar{a} \cdot \bar{c} - \frac{1}{\sqrt{2}}) \bar{b} - (\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}}) \bar{c} = 0$.
Since $\bar{b}$ and $\bar{c}$ are non-coplanar (and thus linearly independent),the coefficients must be zero:
$\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}} = 0 \Rightarrow \bar{a} \cdot \bar{b} = -\frac{1}{\sqrt{2}}$.
Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.
Thus,$|\bar{a}| |\bar{b}| \cos \theta = -\frac{1}{\sqrt{2}}$,where $\theta$ is the angle between $\bar{a}$ and $\bar{b}$.
$\cos \theta = -\frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{3 \pi}{4}$.

(D) Given the vector triple product formula: $\bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}$.
Comparing this with the given equation $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{\sqrt{3}}{2} \bar{b} + \frac{\sqrt{3}}{2} \bar{c}$,we have:
$(\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} = \frac{\sqrt{3}}{2} \bar{b} + \frac{\sqrt{3}}{2} \bar{c}$.
Since $\bar{b}$ and $\bar{c}$ are not parallel,we can equate the coefficients of $\bar{b}$ and $\bar{c}$:
$\bar{a} \cdot \bar{c} = \frac{\sqrt{3}}{2}$ and $-(\bar{a} \cdot \bar{b}) = \frac{\sqrt{3}}{2}$,which implies $\bar{a} \cdot \bar{b} = -\frac{\sqrt{3}}{2}$.
Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$. Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.
$\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos \theta = 1 \times 1 \times \cos \theta = \cos \theta$.
Thus,$\cos \theta = -\frac{\sqrt{3}}{2}$.
Since $\cos \theta = -\frac{\sqrt{3}}{2}$,the angle $\theta = \frac{5 \pi}{6}$.

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

(A) Given $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,so $|\overline{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4+1+4 = 9$,which implies $|\overline{a}|=3$.
Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$,squaring both sides gives $|\overline{c}-\overline{a}|^2 = (2 \sqrt{2})^2 = 8$.
Expanding this,we get $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{a} \cdot \overline{c}) = 8$.
Substituting $\overline{a} \cdot \overline{c} = |\overline{c}|$ and $|\overline{a}|^2 = 9$,we have $|\overline{c}|^2 + 9 - 2|\overline{c}| = 8$,which simplifies to $|\overline{c}|^2 - 2|\overline{c}| + 1 = 0$.
This is $(|\overline{c}|-1)^2 = 0$,so $|\overline{c}| = 1$.
Next,calculate $\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i} - 2 \hat{j} + \hat{k}$.
Thus,$|\overline{a} \times \overline{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.
Finally,$|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin(\frac{\pi}{6}) = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(D) Given $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$.
First,calculate the magnitude of $\bar{a}$:
$|\bar{a}|=\sqrt{2^2+1^2+(-2)^2}=\sqrt{4+1+4}=3$.
Next,calculate the cross product $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i}-2 \hat{j}+\hat{k}$.
The magnitude is $|\bar{a} \times \bar{b}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3$.
Given $|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle $\theta$ between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $\frac{\pi}{6}$.
Using the formula $|\bar{u} \times \bar{v}| = |\bar{u}||\bar{v}| \sin \theta$:
$|(\bar{a} \times \bar{b}) \times \bar{c}| = |\bar{a} \times \bar{b}| |\bar{c}| \sin \frac{\pi}{6} = 3$.
$3 \cdot |\bar{c}| \cdot \frac{1}{2} = 3 \Rightarrow |\bar{c}|=2$.
Now,use the given $|\bar{c}-\bar{a}|=4$:
$|\bar{c}-\bar{a}|^2 = |\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{a} \cdot \bar{c}) = 4^2$.
$2^2 + 3^2 - 2(\bar{a} \cdot \bar{c}) = 16$.
$4 + 9 - 2(\bar{a} \cdot \bar{c}) = 16$.
$13 - 2(\bar{a} \cdot \bar{c}) = 16$.
$-2(\bar{a} \cdot \bar{c}) = 3$.
$\bar{a} \cdot \bar{c} = -\frac{3}{2}$.

(A) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,$|\vec{a} \times \vec{b}|$.
First,we calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}$
$= \hat{i}(3(-2) - 4(-1)) - \hat{j}(2(-2) - 4(0)) + \hat{k}(2(-1) - 3(0))$
$= \hat{i}(-6 + 4) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0)$
$= -2\hat{i} + 4\hat{j} - 2\hat{k}$.
Now,we find the magnitude of this vector:
$|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2}$
$= \sqrt{4 + 16 + 4} = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$.
Thus,the area is $2\sqrt{6}$ sq. units.

(A) We are given that $|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$,and $\vec{a} \times \vec{b}$ is a unit vector,which means $|\vec{a} \times \vec{b}|=1$.
We know the formula for the magnitude of the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Substituting the given values: $1 = 3 \times \frac{\sqrt{2}}{3} \times \sin \theta$.
This simplifies to $1 = \sqrt{2} \sin \theta$.
Therefore,$\sin \theta = \frac{1}{\sqrt{2}}$.
Since $\sin \theta = \frac{1}{\sqrt{2}}$,the angle $\theta = \frac{\pi}{4}$.

(B) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.
Then,$\vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} = -a_2\hat{k} + a_3\hat{j}$.
So,$|\vec{a} \times \hat{i}|^2 = a_2^2 + a_3^2$.
Similarly,$|\vec{a} \times \hat{j}|^2 = a_1^2 + a_3^2$ and $|\vec{a} \times \hat{k}|^2 = a_1^2 + a_2^2$.
Adding these,we get $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2)$.
Since $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$,the expression equals $2|\vec{a}|^2$.

(B) The area of a triangle with vertices $A$,$B$,and $C$ is given by the formula $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Given vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$.
First,find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k} = \hat{j} + 2\hat{k}$.
$\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k} = \hat{i} + 2\hat{j}$.
Now,calculate the cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - 1\hat{k}$.
The magnitude of the cross product is $|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$.
Therefore,the area of the triangle is $\frac{1}{2} \sqrt{21}$ sq. units.

(A) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,$|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$
$= \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2))$
$= \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2)$
$= 20\hat{i} + 5\hat{j} - 5\hat{k}$
Now,find the magnitude of this vector:
$|\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2}$
$= \sqrt{400 + 25 + 25} = \sqrt{450}$
$= \sqrt{225 \times 2} = 15\sqrt{2}$
Thus,the area of the parallelogram is $15\sqrt{2}$ square units.

(C) Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.
Given that $|\vec{a}| = 3$,we have $a_1^2 + a_2^2 + a_3^2 = 3^2 = 9$.
Now,$\vec{a} \times \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{i} = -a_2 \hat{k} + a_3 \hat{j}$.
So,$|\vec{a} \times \hat{i}|^2 = a_2^2 + a_3^2$.
Similarly,$|\vec{a} \times \hat{j}|^2 = a_1^2 + a_3^2$ and $|\vec{a} \times \hat{k}|^2 = a_1^2 + a_2^2$.
Adding these,we get $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2)$.
Substituting the value,$2(9) = 18$.

(D) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,$|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$
$= \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2))$
$= \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2)$
$= 20\hat{i} + 5\hat{j} - 5\hat{k}$
Now,calculate the magnitude $|\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2}$
$= \sqrt{400 + 25 + 25} = \sqrt{450}$
$= \sqrt{225 \times 2} = 15\sqrt{2}$ square units.
Therefore,the correct option is $D$.

(A) We know that the magnitude of the cross product of two vectors $\bar{a}$ and $\bar{b}$ is given by $|\bar{a} \times \bar{b}| = |\bar{a}| |\bar{b}| \sin(\theta)$,where $\theta$ is the angle between the vectors.
Given that $|\bar{a}| = \frac{2}{3}$,$|\bar{b}| = 3$,and $|\bar{a} \times \bar{b}| = 1$.
Substituting these values into the formula:
$1 = (\frac{2}{3}) \times 3 \times \sin(\theta)$
$1 = 2 \sin(\theta)$
$\sin(\theta) = \frac{1}{2}$
Since $\sin(\theta) = \frac{1}{2}$,the angle $\theta$ is $\frac{\pi}{6}$ (or $30^{\circ}$).
Therefore,the correct option is $A$.

(C) The area of a parallelogram with adjacent sides $\bar{a}$ and $\bar{b}$ is given by the magnitude of their cross product,$|\bar{a} \times \bar{b}|$.
Given $\bar{a} = 0\hat{i} + 1\hat{j} + 2\hat{k}$ and $\bar{b} = 1\hat{i} + 2\hat{j} + 0\hat{k}$.
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$
$= \hat{i}(1 \times 0 - 2 \times 2) - \hat{j}(0 \times 0 - 2 \times 1) + \hat{k}(0 \times 2 - 1 \times 1)$
$= \hat{i}(-4) - \hat{j}(-2) + \hat{k}(-1) = -4\hat{i} + 2\hat{j} - \hat{k}$.
The area is $|\bar{a} \times \bar{b}| = \sqrt{(-4)^2 + (2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$ square units.

(C) Given that $|\vec{a}| = 3$,$|\vec{b}| = \frac{\sqrt{2}}{3}$,and $|\vec{a} \times \vec{b}| = 1$ (since it is a unit vector).
We know the formula for the magnitude of the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta)$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Substituting the given values: $1 = 3 \times \frac{\sqrt{2}}{3} \times \sin(\theta)$.
This simplifies to: $1 = \sqrt{2} \sin(\theta)$.
Therefore,$\sin(\theta) = \frac{1}{\sqrt{2}}$.
Since $\sin(\theta) = \frac{1}{\sqrt{2}}$,the angle $\theta = \frac{\pi}{4}$.

(A) The area of a triangle with vertices $A$,$B$,and $C$ is given by the formula: $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Given vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$.
First,we find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k} = \langle 0, 1, 2 \rangle$.
$\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k} = \langle 1, 2, 0 \rangle$.
Now,calculate the cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - 1\hat{k}$.
The magnitude of the cross product is $|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$.
Therefore,the area of the triangle is $\frac{1}{2} \times \sqrt{21} = \frac{\sqrt{21}}{2}$.

(C) Given $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-3\hat{k}$.
First,calculate $\vec{p}=\vec{a}-\vec{b} = (1-1)\hat{i} + (-1-2)\hat{j} + (1-(-3))\hat{k} = 0\hat{i}-3\hat{j}+4\hat{k}$.
Next,calculate $\vec{q}=\vec{a}+\vec{b} = (1+1)\hat{i} + (-1+2)\hat{j} + (1-3)\hat{k} = 2\hat{i}+1\hat{j}-2\hat{k}$.
The vector perpendicular to both $\vec{p}$ and $\vec{q}$ is given by $\vec{n} = \vec{p} \times \vec{q}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & 4 \\ 2 & 1 & -2 \end{vmatrix} = \hat{i}(6-4) - \hat{j}(0-8) + \hat{k}(0-(-6)) = 2\hat{i}+8\hat{j}+6\hat{k}$.
The unit vector is $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{\sqrt{2^2+8^2+6^2}} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{\sqrt{4+64+36}} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{\sqrt{104}} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{2\sqrt{26}} = \frac{1}{\sqrt{26}}\hat{i}+\frac{4}{\sqrt{26}}\hat{j}+\frac{3}{\sqrt{26}}\hat{k}$.
Thus,the correct option is $C$.

(C) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,$|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}$
$= \hat{i}(1(1) - 4(-1)) - \hat{j}(3(1) - 4(1)) + \hat{k}(3(-1) - 1(1))$
$= \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1)$
$= 5\hat{i} + \hat{j} - 4\hat{k}$
Now,calculate the magnitude $|\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2}$
$= \sqrt{25 + 1 + 16} = \sqrt{42}$ sq. units.
Thus,the correct option is $C$.

(C) We know that for any two vectors $\vec{a}$ and $\vec{b}$,the relationship between their dot product and cross product is given by the identity:
$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$
Given values are $|\vec{a}|=10$,$|\vec{b}|=2$,and $\vec{a} \cdot \vec{b}=12$.
Substituting these values into the identity:
$|\vec{a} \times \vec{b}|^2 + (12)^2 = (10)^2 (2)^2$
$|\vec{a} \times \vec{b}|^2 + 144 = 100 \times 4$
$|\vec{a} \times \vec{b}|^2 + 144 = 400$
$|\vec{a} \times \vec{b}|^2 = 400 - 144$
$|\vec{a} \times \vec{b}|^2 = 256$
Taking the square root on both sides:
$|\vec{a} \times \vec{b}| = \sqrt{256} = 16$
Therefore,the correct option is $C$.

(C) Given the cross product of two vectors is zero,the vectors must be collinear.
So,$(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) = k (\hat{i} + \lambda \hat{j} + \mu \hat{k})$ for some scalar $k$.
Comparing the components:
$2 = k \implies k = 2$
$6 = k \lambda \implies 6 = 2 \lambda \implies \lambda = 3$
$27 = k \mu \implies 27 = 2 \mu \implies \mu = \frac{27}{2}$
Therefore,$\lambda + \mu = 3 + \frac{27}{2} = \frac{6 + 27}{2} = \frac{33}{2}$.

(D) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula:
Area $= \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
Given $\vec{d_1} = 0\hat{i} + 1\hat{j} + 1\hat{k}$ and $\vec{d_2} = 1\hat{i} + 1\hat{j} + 0\hat{k}$.
First,calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(0 - 1) + \hat{k}(0 - 1) = -\hat{i} + \hat{j} - \hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$.
Finally,the area is $\frac{1}{2} \times \sqrt{3} = \frac{\sqrt{3}}{2}$ sq. units.
Thus,the correct option is $D$.

(B) Given the adjacent sides of the parallelogram are $\vec{a} = \hat{i} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$.
The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product: $|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix}$
$= \hat{i}(0(1) - 1(1)) - \hat{j}(1(1) - 1(2)) + \hat{k}(1(1) - 0(2))$
$= \hat{i}(-1) - \hat{j}(-1) + \hat{k}(1)$
$= -\hat{i} + \hat{j} + \hat{k}$.
Now,calculate the magnitude of the resulting vector:
$|\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (1)^2 + (1)^2}$
$= \sqrt{1 + 1 + 1} = \sqrt{3}$.
Thus,the area of the parallelogram is $\sqrt{3}$ square units.

(D) We know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
Substituting the given values: $12 = 10 \times 2 \times \cos \theta$.
$12 = 20 \cos \theta \implies \cos \theta = \frac{12}{20} = \frac{3}{5}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Now,$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
$|\vec{a} \times \vec{b}| = 10 \times 2 \times \frac{4}{5} = 20 \times \frac{4}{5} = 16$.

(C) Given the equation $a+2b+3c=0$.
Taking the cross product of the equation with $c$:
$(a+2b+3c) \times c = 0 \times c = 0$
$a \times c + 2(b \times c) + 3(c \times c) = 0$
Since $c \times c = 0$,we have $a \times c + 2(b \times c) = 0$,which implies $c \times a = 2(b \times c)$.
Taking the cross product of the equation with $b$:
$(a+2b+3c) \times b = 0 \times b = 0$
$a \times b + 2(b \times b) + 3(c \times b) = 0$
Since $b \times b = 0$,we have $a \times b - 3(b \times c) = 0$,which implies $a \times b = 3(b \times c)$.
Now,substitute these into the given expression:
$(a \times b) + (b \times c) + (c \times a) = 3(b \times c) + (b \times c) + 2(b \times c) = 6(b \times c)$.
Comparing this with $\lambda(b \times c)$,we get $\lambda = 6$.

(C) The area of a parallelogram with adjacent sides $a$ and $b$ is given by $|a \times b| = 15$ sq units.
Now,the area of the parallelogram with adjacent sides $(3a+2b)$ and $(a+3b)$ is given by the magnitude of their cross product:
Area $= |(3a+2b) \times (a+3b)|$
$= |3a \times a + 9a \times b + 2b \times a + 6b \times b|$
Since $a \times a = 0$ and $b \times b = 0$,the expression simplifies to:
$= |9(a \times b) + 2(b \times a)|$
Using the property $b \times a = -(a \times b)$:
$= |9(a \times b) - 2(a \times b)|$
$= |7(a \times b)|$
$= 7 |a \times b|$
Substituting the given value $|a \times b| = 15$:
$= 7 \times 15 = 105$ sq units.

(D) Given,$\overrightarrow{a}+2 \overrightarrow{b}+3 \overrightarrow{c}=\overrightarrow{0} \quad \dots(i)$
Taking the cross product of equation $(i)$ with $\overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} + 2(\overrightarrow{b} \times \overrightarrow{b}) + 3(\overrightarrow{c} \times \overrightarrow{b}) = \overrightarrow{0} \times \overrightarrow{b}$
Since $\overrightarrow{b} \times \overrightarrow{b} = \overrightarrow{0}$ and $\overrightarrow{c} \times \overrightarrow{b} = -(\overrightarrow{b} \times \overrightarrow{c})$:
$\overrightarrow{a} \times \overrightarrow{b} - 3(\overrightarrow{b} \times \overrightarrow{c}) = \overrightarrow{0} \implies \overrightarrow{a} \times \overrightarrow{b} = 3(\overrightarrow{b} \times \overrightarrow{c})$
Taking the cross product of equation $(i)$ with $\overrightarrow{c}$:
$\overrightarrow{a} \times \overrightarrow{c} + 2(\overrightarrow{b} \times \overrightarrow{c}) + 3(\overrightarrow{c} \times \overrightarrow{c}) = \overrightarrow{0} \times \overrightarrow{c}$
Since $\overrightarrow{c} \times \overrightarrow{c} = \overrightarrow{0}$ and $\overrightarrow{a} \times \overrightarrow{c} = -(\overrightarrow{c} \times \overrightarrow{a})$:
$-(\overrightarrow{c} \times \overrightarrow{a}) + 2(\overrightarrow{b} \times \overrightarrow{c}) = \overrightarrow{0} \implies \overrightarrow{c} \times \overrightarrow{a} = 2(\overrightarrow{b} \times \overrightarrow{c})$
Now,substituting these into the expression $\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a}$:
$= 3(\overrightarrow{b} \times \overrightarrow{c}) + (\overrightarrow{b} \times \overrightarrow{c}) + 2(\overrightarrow{b} \times \overrightarrow{c})$
$= 6(\overrightarrow{b} \times \overrightarrow{c})$

(D) Let $\overrightarrow{a} = \hat{i} + \hat{j}$ and $\overrightarrow{b} = \hat{j} + \hat{k}$.
The vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by the cross product $\overrightarrow{a} \times \overrightarrow{b}$.
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k}$.
The magnitude of this vector is $|\overrightarrow{a} \times \overrightarrow{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The required unit vector is $\pm \frac{\overrightarrow{a} \times \overrightarrow{b}}{|\overrightarrow{a} \times \overrightarrow{b}|} = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.
Comparing this with the given options,the correct option is $\frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.

(B) To find a vector perpendicular to the plane containing points $A, B$,and $C$,we calculate the cross product of two vectors lying in the plane,such as $\vec{AB}$ and $\vec{AC}$.
Given points: $A(1, -1, 2)$,$B(2, 0, -1)$,$C(0, 2, 1)$.
$\vec{AB} = (2-1)\hat{i} + (0 - (-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$
$\vec{AC} = (0-1)\hat{i} + (2 - (-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$
The normal vector $\vec{n} = \vec{AB} \times \vec{AC} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{array}\right|$
$\vec{n} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1))$
$\vec{n} = \hat{i}(8) - \hat{j}(-4) + \hat{k}(4) = 8\hat{i} + 4\hat{j} + 4\hat{k}$
Thus,the vector perpendicular to the plane is $8\hat{i} + 4\hat{j} + 4\hat{k}$.

(B) Let $\vec{u} = i+j+k$ and $\vec{v} = 2i+j+3k$.
$A$ vector perpendicular to both $\vec{u}$ and $\vec{v}$ is given by the cross product $\vec{w} = \vec{u} \times \vec{v}$.
$\vec{w} = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ 2 & 1 & 3 \end{vmatrix} = i(3-1) - j(3-2) + k(1-2) = 2i - j - k$.
The magnitude of $\vec{w}$ is $|\vec{w}| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4+1+1} = \sqrt{6}$.
The unit vector perpendicular to both is $\pm \frac{\vec{w}}{|\vec{w}|} = \pm \frac{2i-j-k}{\sqrt{6}}$.
Comparing with the given options,the correct answer is $\frac{2i-j-k}{\sqrt{6}}$.

(D) vector that lies in the plane of $b$ and $c$ is given by a linear combination of $b$ and $c$. $A$ vector perpendicular to $a$ and lying in the plane of $b$ and $c$ is given by the vector product $a \times (b \times c)$.
First,we calculate $b \times c$:
$b \times c = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -5 \\ 3 & 5 & -1 \end{vmatrix} = \hat{i}(-2 + 25) - \hat{j}(-1 + 15) + \hat{k}(5 - 6) = 23 \hat{i} - 14 \hat{j} - \hat{k}$.
Now,we calculate $a \times (b \times c)$:
$a \times (b \times c) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 23 & -14 & -1 \end{vmatrix} = \hat{i}(-3 - 14) - \hat{j}(-2 + 23) + \hat{k}(-28 - 69) = -17 \hat{i} - 21 \hat{j} - 97 \hat{k}$.
Thus,the required vector is $-17 \hat{i} - 21 \hat{j} - 97 \hat{k}$.

(A) Given the vertices of the quadrilateral $ABCD$ are $A(0,4,1)$,$B(2,3,-1)$,$C(4,5,0)$,and $D(2,6,2)$.
We can divide the quadrilateral into two triangles,$\triangle ABD$ and $\triangle BCD$.
The area of the quadrilateral is the sum of the areas of these two triangles.
Area of $\triangle ABD = \frac{1}{2} |\vec{AB} \times \vec{AD}|$.
$\vec{AB} = (2-0)\hat{i} + (3-4)\hat{j} + (-1-1)\hat{k} = 2\hat{i} - \hat{j} - 2\hat{k}$.
$\vec{AD} = (2-0)\hat{i} + (6-4)\hat{j} + (2-1)\hat{k} = 2\hat{i} + 2\hat{j} + \hat{k}$.
$\vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(-1+4) - \hat{j}(2+4) + \hat{k}(4+2) = 3\hat{i} - 6\hat{j} + 6\hat{k}$.
$|\vec{AB} \times \vec{AD}| = \sqrt{3^2 + (-6)^2 + 6^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9$.
Area of $\triangle ABD = \frac{1}{2} \times 9 = 4.5 \text{ sq units}$.
Area of $\triangle BCD = \frac{1}{2} |\vec{BC} \times \vec{BD}|$.
$\vec{BC} = (4-2)\hat{i} + (5-3)\hat{j} + (0-(-1))\hat{k} = 2\hat{i} + 2\hat{j} + \hat{k}$.
$\vec{BD} = (2-2)\hat{i} + (6-3)\hat{j} + (2-(-1))\hat{k} = 0\hat{i} + 3\hat{j} + 3\hat{k}$.
$\vec{BC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 0 & 3 & 3 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(6-0) + \hat{k}(6-0) = 3\hat{i} - 6\hat{j} + 6\hat{k}$.
$|\vec{BC} \times \vec{BD}| = \sqrt{3^2 + (-6)^2 + 6^2} = 9$.
Area of $\triangle BCD = \frac{1}{2} \times 9 = 4.5 \text{ sq units}$.
Total Area = $4.5 + 4.5 = 9 \text{ sq units}$.

(D) Let $\vec{a} = \hat{i}+2\hat{j}+\hat{k}$ and $\vec{b} = -2\hat{i}+\hat{j}+3\hat{k}$.
The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3 - (-2)) + \hat{k}(1 - (-4)) = 5\hat{i} - 5\hat{j} + 5\hat{k}$.
The magnitude of this vector is $|\vec{a} \times \vec{b}| = \sqrt{5^2 + (-5)^2 + 5^2} = \sqrt{25+25+25} = \sqrt{75} = 5\sqrt{3}$.
The unit vector perpendicular to the plane is $\hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{5\hat{i}-5\hat{j}+5\hat{k}}{5\sqrt{3}} = \pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$.
Comparing this with the given options,option $D$ represents $\frac{5\hat{i}-5\hat{j}+5\hat{k}}{5\sqrt{3}}$,which simplifies to $\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$.

(A) Given,$|\vec{a}|=|\vec{b}|=1$.
Since $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$,let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$. Then $\vec{a} \cdot \vec{b} = \cos \theta$.
So,$\vec{u} = \vec{a} - \cos \theta \vec{b}$.
Calculating the magnitude squared: $|\vec{u}|^2 = |\vec{a}|^2 + \cos^2 \theta |\vec{b}|^2 - 2 \cos \theta (\vec{a} \cdot \vec{b}) = 1 + \cos^2 \theta - 2 \cos^2 \theta = 1 - \cos^2 \theta = \sin^2 \theta$.
Thus,$|\vec{u}| = \sin \theta$ (since $\sin \theta > 0$ for non-collinear vectors).
Also,$|\vec{v}| = |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = \sin \theta$.
Therefore,$|\vec{v}| = |\vec{u}|$.
Now,$\vec{u} \cdot \vec{v} = (\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) \cdot (\vec{a} \times \vec{b}) = \vec{a} \cdot (\vec{a} \times \vec{b}) - (\vec{a} \cdot \vec{b}) \vec{b} \cdot (\vec{a} \times \vec{b}) = 0 - 0 = 0$.
Hence,$|\vec{v}| = |\vec{u}| + |\vec{u} \cdot \vec{v}| = |\vec{u}| + 0 = |\vec{u}|$.

(D) First,calculate the cross product $\vec{a} \times \vec{c}$:
$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & -3 & -1 \end{vmatrix} = \hat{i}(1 - (-3)) - \hat{j}(-1 - 2) + \hat{k}(-3 - (-2)) = 4 \hat{i} + 3 \hat{j} - \hat{k}$.
Next,calculate the cross product $\vec{b} \times \vec{d}$:
$\vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(1 - (-2)) - \hat{j}(1 - (-4)) + \hat{k}(1 - 2) = 3 \hat{i} - 5 \hat{j} - \hat{k}$.
Finally,calculate the cross product of the two resulting vectors:
$(\vec{a} \times \vec{c}) \times(\vec{b} \times \vec{d}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & -1 \\ 3 & -5 & -1 \end{vmatrix} = \hat{i}(-3 - 5) - \hat{j}(-4 - (-3)) + \hat{k}(-20 - 9) = -8 \hat{i} + \hat{j} - 29 \hat{k}$.

(D) Given,$F=2 \hat{i}+2 \hat{j}+5 \hat{k}$,$A=(1,2,5)$,and $B=(-1,-2,-3)$.
First,we find the vector $BA = A - B$:
$BA = (1 - (-1)) \hat{i} + (2 - (-2)) \hat{j} + (5 - (-3)) \hat{k} = 2 \hat{i} + 4 \hat{j} + 8 \hat{k}$.
Now,we calculate the cross product $BA \times F$:
$BA \times F = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 8 \\ 2 & 2 & 5 \end{vmatrix}$
$= \hat{i}(4 \times 5 - 8 \times 2) - \hat{j}(2 \times 5 - 8 \times 2) + \hat{k}(2 \times 2 - 4 \times 2)$
$= \hat{i}(20 - 16) - \hat{j}(10 - 16) + \hat{k}(4 - 8)$
$= 4 \hat{i} + 6 \hat{j} - 4 \hat{k}$.
Comparing this with the given expression $4 \hat{i} + 6 \hat{j} + 2 \lambda \hat{k}$,we get:
$2 \lambda = -4$.
Therefore,$\lambda = -2$.

(A) Let the diagonals of the parallelogram be $\overrightarrow{d_1} = \overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{d_2} = \overrightarrow{b} + \overrightarrow{c}$.
$\overrightarrow{d_1} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 3\hat{j} + 5\hat{k}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$
$\overrightarrow{d_2} = (\hat{i} + 3\hat{j} + 5\hat{k}) + (7\hat{i} + 9\hat{j} + 11\hat{k}) = 8\hat{i} + 12\hat{j} + 16\hat{k}$
The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $\frac{1}{2} |\overrightarrow{d_1} \times \overrightarrow{d_2}|$.
$\overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 6 \\ 8 & 12 & 16 \end{vmatrix} = \hat{i}(64 - 72) - \hat{j}(32 - 48) + \hat{k}(24 - 32) = -8\hat{i} + 16\hat{j} - 8\hat{k}$
Magnitude $|\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-8)^2 + 16^2 + (-8)^2} = \sqrt{64 + 256 + 64} = \sqrt{384} = \sqrt{64 \times 6} = 8\sqrt{6}$
Area $= \frac{1}{2} \times 8\sqrt{6} = 4\sqrt{6}$ sq units.

(B) Let $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.
The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 3 & -2 & 1 \end{vmatrix}$
$= \hat{i}(1(1) - 2(-2)) - \hat{j}(1(1) - 2(3)) + \hat{k}(1(-2) - 1(3))$
$= \hat{i}(1 + 4) - \hat{j}(1 - 6) + \hat{k}(-2 - 3)$
$= 5\hat{i} + 5\hat{j} - 5\hat{k}$
Now,calculate the magnitude $|\vec{a} \times \vec{b}| = \sqrt{5^2 + 5^2 + (-5)^2}$
$= \sqrt{25 + 25 + 25} = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$.
Thus,the area is $5\sqrt{3}$ square units.

(C) Given that $p \times q = r$ and $q \times p = r$.
We know that the cross product is anticommutative,so $q \times p = -(p \times q)$.
Substituting the given values,we get $r = -r$,which implies $2r = 0$,so $r = 0$.
However,the problem states that $r \neq 0$.
Since the condition $p \times q = q \times p = r$ leads to a contradiction $(r = 0)$ given $r \neq 0$,the premises provided in the question are mathematically inconsistent.
Assuming the question intended to imply properties of cross products where $p \times q = r$,then $r$ is orthogonal to both $p$ and $q$.
However,based on the strict logical interpretation of the provided equations,both statements $(i)$ and (ii) cannot be satisfied simultaneously under the constraint $r \neq 0$.