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

(B) Let the required vector be $\vec{v} = ai + bj + ck$.
For $\vec{v}$ to be coplanar with $\vec{u_1} = i + j + 2k$ and $\vec{u_2} = i + 2j + k$,it must be a linear combination of $\vec{u_1}$ and $\vec{u_2}$.
So,$\vec{v} = p(i + j + 2k) + r(i + 2j + k) = (p+r)i + (p+2r)j + (2p+r)k$.
Comparing components,we get $a = p+r$,$b = p+2r$,and $c = 2p+r$.
Since $\vec{v}$ is perpendicular to $\vec{w} = i + j + k$,their dot product is zero:
$(ai + bj + ck) \cdot (i + j + k) = a + b + c = 0$.
Substituting the expressions for $a, b, c$:
$(p+r) + (p+2r) + (2p+r) = 4p + 4r = 0$,which implies $p = -r$.
Substituting $p = -r$ into the expressions for $a, b, c$:
$a = -r + r = 0$,
$b = -r + 2r = r$,
$c = 2(-r) + r = -r$.
Thus,$\vec{v} = r(j - k)$.
For $\vec{v}$ to be a unit vector,$|\vec{v}| = 1$:
$\sqrt{0^2 + r^2 + (-r)^2} = 1 \Rightarrow \sqrt{2r^2} = 1 \Rightarrow |r|\sqrt{2} = 1 \Rightarrow r = \pm \frac{1}{\sqrt{2}}$.
Therefore,the required unit vector is $\pm \frac{1}{\sqrt{2}}(j - k)$.

(A) Let the required vector be $\vec{v} = x\,i + y\,j + z\,k$.
Since $\vec{v}$ is perpendicular to $\vec{a} = 3\,i + j - 4\,k$ and $\vec{b} = 6\,i + 5\,j - 2\,k$,it must be parallel to $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{vmatrix} = i(-2 - (-20)) - j(-6 - (-24)) + k(15 - 6) = 18\,i - 18\,j + 9\,k$.
Simplifying,we can take the direction vector $\vec{u} = 2\,i - 2\,j + k$.
The magnitude of $\vec{u}$ is $|\vec{u}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Since the required vector has length $3$,the vector is $\pm(2\,i - 2\,j + k)$.
Comparing with the given options,$2\,i - 2\,j + k$ is the correct choice.

(D) Let the two vectors be $\vec{a} = i + 2j + k$ and $\vec{b} = i + j + 2k$.
Any vector in the plane of $\vec{a}$ and $\vec{b}$ is given by $\vec{v} = \vec{a} + \lambda \vec{b}$ or $\vec{v} = \vec{a} \times \vec{b}$.
However,a vector perpendicular to both $\vec{a}$ and $\vec{b}$ is $\vec{n} = \vec{a} \times \vec{b}$.
We need a vector $\vec{r}$ in the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{r} \perp (2i + j + k)$.
Let $\vec{r} = \vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$.
Alternatively,consider the vector $\vec{v} = \vec{a} - \vec{b} = (i + 2j + k) - (i + j + 2k) = j - k$.
Check if $\vec{v}$ is perpendicular to $2i + j + k$: $(j - k) \cdot (2i + j + k) = (0)(2) + (1)(1) + (-1)(1) = 1 - 1 = 0$.
Since the dot product is $0$,the vector $j - k$ lies in the plane and is perpendicular to $2i + j + k$.
The unit vector is $\frac{j - k}{|j - k|} = \frac{j - k}{\sqrt{0^2 + 1^2 + (-1)^2}} = \frac{j - k}{\sqrt{2}}$.

(C) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
We know that the magnitude of the cross product is given by $|a \times b| = |a||b| \sin \theta$,where $\theta$ is the angle between $a$ and $b$.
Since $a \times b$ is a unit vector,$|a \times b| = 1$.
Substituting the values,we get $1 = (1)(1) \sin \theta$.
This implies $\sin \theta = 1$.
Therefore,$\theta = \frac{\pi}{2}$.

(B) For three points $A(a), B(b), C(c)$ to be collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.
This implies that their cross product must be zero: $\vec{AB} \times \vec{BC} = 0$.
Substituting the position vectors,we get $(b - a) \times (c - b) = 0$.
Expanding the cross product: $b \times c - b \times b - a \times c + a \times b = 0$.
Since $b \times b = 0$,the expression simplifies to $b \times c - a \times c + a \times b = 0$.
Rearranging the terms,we get $a \times b + b \times c + c \times a = 0$.

(A) Using the distributive property of the cross product:
$(a - b) \times (a + b) = a \times a + a \times b - b \times a - b \times b$
Since the cross product of any vector with itself is zero ($a \times a = 0$ and $b \times b = 0$):
$= 0 + a \times b - b \times a - 0$
$= a \times b - b \times a$
Since the cross product is anticommutative $(b \times a = -(a \times b))$:
$= a \times b - (-(a \times b))$
$= a \times b + a \times b$
$= 2(a \times b)$

(C) Given that $a + b + c = 0.$
Taking the cross product with $a$ on both sides:
$a \times (a + b + c) = a \times 0$
$a \times a + a \times b + a \times c = 0$
Since $a \times a = 0,$ we have $a \times b + a \times c = 0,$ which implies $a \times b = - (a \times c) = c \times a$ .....$(i)$
Similarly,taking the cross product with $b$ on both sides:
$b \times (a + b + c) = b \times 0$
$b \times a + b \times b + b \times c = 0$
Since $b \times b = 0,$ we have $b \times a + b \times c = 0,$ which implies $-(a \times b) = b \times c,$ or $a \times b = b \times c$ .....$(ii)$
From $(i)$ and $(ii),$ we get $a \times b = b \times c = c \times a.$

(B) Using the distributive property of the cross product:
$(2a + 3b) \times (5a + 7b) = 2a \times (5a + 7b) + 3b \times (5a + 7b)$
$= 2a \times 5a + 2a \times 7b + 3b \times 5a + 3b \times 7b$
Since $a \times a = 0$ and $b \times b = 0$:
$= 0 + 14(a \times b) + 15(b \times a) + 0$
Using the property $a \times b = -(b \times a)$:
$= 14(a \times b) - 15(a \times b) = -1(a \times b) = b \times a$.

(C) The vector perpendicular to $a$ and $b$ is given by the cross product $a \times b$.
$a \times b = \begin{vmatrix} i & j & k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = i(1 - 0) - j(1 - 0) + k(1 - 0) = i - j + k$.
The magnitude of this vector is $|a \times b| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The unit vectors perpendicular to both $a$ and $b$ are given by $\pm \frac{a \times b}{|a \times b|}$.
Thus,the unit vectors are $\pm \frac{1}{\sqrt{3}}(i - j + k)$.
Therefore,there are exactly $2$ such vectors.

(D) Let $b = b_1 i + b_2 j + b_3 k$.
Given $a \cdot b = 1$,we have $(i - j + k) \cdot (b_1 i + b_2 j + b_3 k) = 1 \Rightarrow b_1 - b_2 + b_3 = 1$ ... $(i)$
Given $a \times b = c$,we calculate the cross product:
$a \times b = \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} = i(-b_3 - b_2) - j(b_3 - b_1) + k(b_2 + b_1) = (-b_2 - b_3)i + (b_1 - b_3)j + (b_1 + b_2)k$.
Equating this to $c = -i - j + 0k$,we get:
$-b_2 - b_3 = -1 \Rightarrow b_2 + b_3 = 1$ ... $(ii)$
$b_1 - b_3 = -1$ ... $(iii)$
$b_1 + b_2 = 0$ ... $(iv)$
From $(iv)$,$b_2 = -b_1$. Substituting into $(ii)$,$-b_1 + b_3 = 1 \Rightarrow b_3 - b_1 = 1$.
Adding this to $(iii)$,$(b_3 - b_1) + (b_1 - b_3) = 1 - 1 \Rightarrow 0 = 0$.
Using $(i)$,$b_1 - (-b_1) + b_3 = 1 \Rightarrow 2b_1 + b_3 = 1$.
From $(iii)$,$b_3 = b_1 - 1$.
Substituting into $2b_1 + b_3 = 1$: $2b_1 + b_1 - 1 = 1 \Rightarrow 3b_1 = 2 \Rightarrow b_1 = 2/3$.
This leads to $b_2 = -2/3$ and $b_3 = -1/3$.
Checking the options,none match $(2/3, -2/3, -1/3)$. Thus,the correct answer is None of these.

(A) Given that $a \times b = b \times c \ne 0.$
This can be rewritten as $a \times b - b \times c = 0.$
Using the property of cross product $b \times c = -(c \times b)$,we get $a \times b + c \times b = 0.$
Factoring out the cross product,we have $(a + c) \times b = 0.$
Since the cross product of two vectors is zero,the vectors must be parallel.
Therefore,$a + c$ is parallel to $b$,which implies $a + c = k\,b$ for some scalar $k$.

(C) Given the properties of the vector cross product,we know that the cross product is anticommutative,i.e.,$u \times v = -(v \times u)$.
Consider the expression $a \times (b - c)$.
Using the distributive property,we have $a \times (b - c) = a \times b - a \times c$.
Now consider the right-hand side: $(c - b) \times a = c \times a - b \times a$.
Using the anticommutative property,$c \times a = -(a \times c)$ and $b \times a = -(a \times b)$.
Substituting these into the expression,we get $(c - b) \times a = -(a \times c) - (-(a \times b)) = a \times b - a \times c$.
Since both sides are equal,the statement $a \times (b - c) = (c - b) \times a$ is true.

(B) Given that $a \times b = b \times c \ne 0$.
This can be rewritten as $a \times b - b \times c = 0$.
Using the property of cross product $b \times c = -c \times b$,we get $a \times b + c \times b = 0$.
Factoring out the cross product,we have $(a + c) \times b = 0$.
Since the cross product of two vectors is zero,the vectors must be parallel.
Therefore,$(a + c) \parallel b$.

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

(C) Given $a = 2i + 3j - 5k$ and $b = mi + nj + 12k$. Since $a \times b = 0$,the vectors are collinear.
The cross product is given by the determinant:
$a \times b = \begin{vmatrix} i & j & k \\ 2 & 3 & -5 \\ m & n & 12 \end{vmatrix} = 0$
Expanding the determinant:
$i(36 - (-5n)) - j(24 - (-5m)) + k(2n - 3m) = 0$
$(36 + 5n)i - (24 + 5m)j + (2n - 3m)k = 0$
For the vector to be zero,each component must be zero:
$36 + 5n = 0 \Rightarrow n = -\frac{36}{5}$
$24 + 5m = 0 \Rightarrow m = -\frac{24}{5}$
$2n - 3m = 2(-\frac{36}{5}) - 3(-\frac{24}{5}) = -\frac{72}{5} + \frac{72}{5} = 0$ (This confirms the values).
Thus,$(m, n) = \left( -\frac{24}{5}, -\frac{36}{5} \right)$.

(D) Let $\vec{a} = i + 2j - 2k$ and $\vec{b} = -i + 2j + 2k$.
To find a vector perpendicular to both $\vec{a}$ and $\vec{b}$,we calculate the cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 1 & 2 & -2 \\ -1 & 2 & 2 \end{vmatrix} = i(4 - (-4)) - j(2 - 2) + k(2 - (-2)) = 8i - 0j + 4k = 8i + 4k$.
The magnitude of the cross product is $|\vec{a} \times \vec{b}| = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$.
The unit vector perpendicular to both is $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{8i + 4k}{4\sqrt{5}} = \pm \frac{2i + k}{\sqrt{5}}$.
Comparing this with the given options,the correct option is $\frac{1}{\sqrt{5}}(2i + k)$.

(C) Let $\vec{a} = 6i + 2j + 3k$ and $\vec{b} = 3i - 6j - 2k$.
The vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 6 & 2 & 3 \\ 3 & -6 & -2 \end{vmatrix} = i(-4 - (-18)) - j(-12 - 9) + k(-36 - 6) = 14i + 21j - 42k$.
The magnitude of this vector is $|\vec{a} \times \vec{b}| = \sqrt{14^2 + 21^2 + (-42)^2} = \sqrt{196 + 441 + 1764} = \sqrt{2401} = 49$.
The unit vector perpendicular to both is $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{14i + 21j - 42k}{49} = \pm \frac{2i + 3j - 6k}{7}$.
Comparing with the given options,the correct unit vector is $\frac{2i + 3j - 6k}{7}$.

(C) We know that the magnitude of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,where $\theta$ is the angle between the vectors.
Squaring both sides,we get $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$.
Using the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$,we have:
$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 (1 - \cos^2 \theta)$
$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$
Since the dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$,it follows that $(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$.
Substituting this into the equation,we get:
$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$.

(C) Let $\vec{a} = 3i + 2j - k$ and $\vec{b} = 12i + 5j - 5k$.
The 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 $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-5i + 3j - 9k}{\sqrt{115}}$.

(A) Let $\vec{a} = 3i + 2j - k$ and $\vec{b} = 12i + 5j - 5k$.
The cross product $\vec{a} \times \vec{b}$ is given by:
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 12 & 5 & -5 \end{vmatrix} = i(-10 - (-5)) - j(-15 - (-12)) + k(15 - 24) = -5i + 3j - 9k$.
The magnitude of the cross product is $|\vec{a} \times \vec{b}| = \sqrt{(-5)^2 + 3^2 + (-9)^2} = \sqrt{25 + 9 + 81} = \sqrt{115}$.
The magnitudes of the vectors are $|\vec{a}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$ and $|\vec{b}| = \sqrt{12^2 + 5^2 + (-5)^2} = \sqrt{144 + 25 + 25} = \sqrt{194}$.
Since $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,we have $\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{\sqrt{115}}{\sqrt{14}\sqrt{194}}$.

(D) The cross product of two vectors $a$ and $b$ is defined as $a \times b = |a||b| \sin(\theta) \hat{n}$,where $\theta$ is the angle between the vectors.
If $a \times b = 0$,then $|a||b| \sin(\theta) = 0$.
This implies that either $|a| = 0$,$|b| = 0$,or $\sin(\theta) = 0$.
If $\sin(\theta) = 0$,then $\theta = 0$ or $\theta = \pi$,which means the vectors $a$ and $b$ are parallel or collinear.
Since the condition $a \times b = 0$ can hold even if $a \neq 0$ and $b \neq 0$ (when they are parallel),none of the options $A, B,$ or $C$ are necessarily true.
Therefore,the correct option is $D$.

(A) Using the distributive property of the cross product over vector addition,we have:
$a \times (b + c) = a \times b + a \times c$
$b \times (c + a) = b \times c + b \times a$
$c \times (a + b) = c \times a + c \times b$
Summing these expressions:
$(a \times b + a \times c) + (b \times c + b \times a) + (c \times a + c \times b)$
Using the anticommutative property $u \times v = -(v \times u)$:
$a \times b + a \times c + b \times c - a \times b - c \times a + c \times b$
Since $a \times c = -(c \times a)$,the terms cancel out:
$a \times b - a \times b + a \times c - a \times c + b \times c - b \times c = 0$
Thus,the result is $0$.

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

(B) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.
Then,$|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$.
Now,$\vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} = a_1(\hat{i} \times \hat{i}) + a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) = 0 - a_2\hat{k} + a_3\hat{j}$.
So,$|\vec{a} \times \hat{i}|^2 = |a_3\hat{j} - a_2\hat{k}|^2 = a_3^2 + a_2^2$.
Similarly,$|\vec{a} \times \hat{j}|^2 = |a_1\hat{k} - a_3\hat{i}|^2 = a_1^2 + a_3^2$.
And,$|\vec{a} \times \hat{k}|^2 = |a_2\hat{i} - a_1\hat{j}|^2 = a_2^2 + a_1^2$.
Adding these three,we get:
$|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_3^2 + a_2^2) + (a_1^2 + a_3^2) + (a_2^2 + a_1^2) = 2(a_1^2 + a_2^2 + a_3^2) = 2|\vec{a}|^2$.

(B) Let the points be $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$.
First,we find two vectors in the plane: $\overrightarrow{PQ} = (2-1)i + (0-(-1))j + (-1-2)k = i + j - 3k$ and $\overrightarrow{PR} = (0-1)i + (2-(-1))j + (1-2)k = -i + 3j - k$.
$A$ vector perpendicular to the plane is given by the cross product $\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$.
$\vec{n} = \begin{vmatrix} i & j & k \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix} = i(-1 - (-9)) - j(-1 - 3) + k(3 - (-1)) = 8i + 4j + 4k$.
To simplify,we can take a parallel vector $\vec{v} = 2i + j + k$.
The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
The unit vector perpendicular to the plane is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{2i + j + k}{\sqrt{6}}$.
Comparing this with the given options,the correct option is $\frac{2i + j + k}{\sqrt{6}}$.

(B) Let $\vec{a} = 4i - j + 3k$ and $\vec{b} = -2i + j - 2k$.
The unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by $\hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix}$
$= i((-1)(-2) - (3)(1)) - j((4)(-2) - (3)(-2)) + k((4)(1) - (-1)(-2))$
$= i(2 - 3) - j(-8 + 6) + k(4 - 2)$
$= -i + 2j + 2k$.
Next,calculate the magnitude $|\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Thus,the unit vector is $\frac{-i + 2j + 2k}{3} = \frac{1}{3}(-i + 2j + 2k)$.
Comparing this with the given options,the correct option is $(b)$.

(C) First,calculate the vectors $a + b$ and $b + c$:
$a + b = (i + j - k) + (-i + 2j + k) = 3j$
$b + c = (-i + 2j + k) + (-i + 2j - k) = -2i + 4j$
To find a vector perpendicular to both $a + b$ and $b + c$,we take their cross product:
$(a + b) \times (b + c) = (3j) \times (-2i + 4j) = -6(j \times i) + 12(j \times j)$
Since $j \times i = -k$ and $j \times j = 0$,we get:
$(a + b) \times (b + c) = -6(-k) + 0 = 6k$
The unit vector perpendicular to both is $\pm \frac{6k}{|6k|} = \pm k$.
Among the given options,$k$ is the correct unit vector.

(A) Given that $a, c, b$ form a right-handed system,the vector $c$ is defined by the cross product of $b$ and $a$ in the order $c = b \times a$.
Substituting the given vectors $a = xi + yj + zk$ and $b = j$:
$c = j \times (xi + yj + zk)$
Using the properties of the cross product of unit vectors ($j \times i = -k$,$j \times j = 0$,$j \times k = i$):
$c = x(j \times i) + y(j \times j) + z(j \times k)$
$c = x(-k) + y(0) + z(i)$
$c = zi - xk$.

(B) Let $A$ be the origin and let the position vectors of $B, C,$ and $D$ be $\vec{b}, \vec{c},$ and $\vec{d}$ respectively.
Then $\overrightarrow{AB} = \vec{b}, \overrightarrow{CD} = \vec{d} - \vec{c}, \overrightarrow{BC} = \vec{c} - \vec{b}, \overrightarrow{AD} = \vec{d}, \overrightarrow{CA} = -\vec{c},$ and $\overrightarrow{BD} = \vec{d} - \vec{b}$.
Now, consider the expression:
$|\overrightarrow{AB} \times \overrightarrow{CD} + \overrightarrow{BC} \times \overrightarrow{AD} + \overrightarrow{CA} \times \overrightarrow{BD}|$
$= |\vec{b} \times (\vec{d} - \vec{c}) + (\vec{c} - \vec{b}) \times \vec{d} - \vec{c} \times (\vec{d} - \vec{b})|$
$= |\vec{b} \times \vec{d} - \vec{b} \times \vec{c} + \vec{c} \times \vec{d} - \vec{b} \times \vec{d} - \vec{c} \times \vec{d} + \vec{c} \times \vec{b}|$
$= |-\vec{b} \times \vec{c} + \vec{c} \times \vec{b}|$
$= |-\vec{b} \times \vec{c} - \vec{b} \times \vec{c}| = |-2(\vec{b} \times \vec{c})| = 2|\vec{b} \times \vec{c}|$
Since the area of $\Delta ABC$ is $\Delta = \frac{1}{2}|\vec{b} \times \vec{c}|$, we have $|\vec{b} \times \vec{c}| = 2\Delta$.
Therefore, the expression equals $2(2\Delta) = 4\Delta$.

(B) Given equations are:
$r \times a = b \times a \implies r \times a - b \times a = 0 \implies (r - b) \times a = 0$
$r \times b = a \times b \implies r \times b + b \times a = 0 \implies (r + a) \times b = 0$
Adding the two original equations:
$r \times a + r \times b = b \times a + a \times b$
$r \times (a + b) = b \times a - b \times a = 0$
Since $r \times (a + b) = 0$,$r$ must be parallel to $(a + b)$,so $r = k(a + b)$ for some scalar $k$.
Substituting $r = k(a + b)$ into the first equation:
$k(a + b) \times a = b \times a$
$k(a \times a + b \times a) = b \times a$
$k(0 + b \times a) = b \times a$
$k(b \times a) = b \times a$
Since $a$ is not perpendicular to $b$ and $a \ne \lambda b$,$b \times a \ne 0$,so $k = 1$.
Therefore,$r = a + b$.

(A) Let $\vec{a} = 2i - j + k$ and $\vec{b} = 3i + 4j - k$.
$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 2 & -1 & 1 \\ 3 & 4 & -1 \end{vmatrix} = i(1 - 4) - j(-2 - 3) + k(8 + 3) = -3i + 5j + 11k$.
The magnitude of this vector is $|\vec{a} \times \vec{b}| = \sqrt{(-3)^2 + 5^2 + 11^2} = \sqrt{9 + 25 + 121} = \sqrt{155}$.
Therefore,the unit vector perpendicular to both is $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-3i + 5j + 11k}{\sqrt{155}}$.

(B) The position vectors are $\vec{A} = i + j + k$,$\vec{B} = 2i + 3j - 4k$,and $\vec{C} = 7i + 4j + 9k$.
First,we find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \vec{B} - \vec{A} = (2-1)i + (3-1)j + (-4-1)k = i + 2j - 5k$.
$\vec{AC} = \vec{C} - \vec{A} = (7-1)i + (4-1)j + (9-1)k = 6i + 3j + 8k$.
The vector perpendicular to the plane is given by the cross product $\vec{n} = \vec{AB} \times \vec{AC}$:
$\vec{n} = \begin{vmatrix} i & j & k \\ 1 & 2 & -5 \\ 6 & 3 & 8 \end{vmatrix} = i(16 - (-15)) - j(8 - (-30)) + k(3 - 12) = 31i - 38j - 9k$.
The magnitude of this vector is $|\vec{n}| = \sqrt{31^2 + (-38)^2 + (-9)^2} = \sqrt{961 + 1444 + 81} = \sqrt{2486}$.
The unit vector perpendicular to the plane is $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{31i - 38j - 9k}{\sqrt{2486}}$.

(B) The unit vector perpendicular to the plane of triangle $ABC$ is given by the formula: $\hat{n} = \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{|\overrightarrow{AB} \times \overrightarrow{AC}|}$.
Given the position vectors $a, b, c$ for vertices $A, B, C$,we have $\overrightarrow{AB} = b - a$ and $\overrightarrow{AC} = c - a$.
Calculating the cross product: $(b - a) \times (c - a) = b \times c - b \times a - a \times c + a \times a$.
Since $a \times a = 0$ and $-b \times a = a \times b$ and $-a \times c = c \times a$,we get: $(b - a) \times (c - a) = a \times b + b \times c + c \times a$.
Therefore,the unit vector is $\frac{a \times b + b \times c + c \times a}{|a \times b + b \times c + c \times a|}$.

(B) In $\Delta ABC$,the vectors representing the sides are $BC = a$,$CA = b$,and $AB = c$.
By the triangle law of vector addition,the sum of vectors along the perimeter of a triangle in order is zero: $a + b + c = 0$.
Taking the cross product with $a$ on both sides: $a \times (a + b + c) = a \times 0 \implies a \times a + a \times b + a \times c = 0$.
Since $a \times a = 0$,we get $a \times b = c \times a$ (as $a \times c = -c \times a$).
Similarly,taking the cross product with $b$: $b \times (a + b + c) = b \times 0 \implies b \times a + b \times b + b \times c = 0$.
Since $b \times b = 0$,we get $b \times a + b \times c = 0 \implies b \times c = a \times b$.
Combining these,we have $a \times b = b \times c = c \times a$.

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

(C) The vector perpendicular to both $a$ and $b$ is given by the cross product $a \times b$.
The unit vector perpendicular to the plane containing $a$ and $b$ is given by $\frac{a \times b}{|a \times b|}$.
First,calculate $a \times b = \begin{vmatrix} i & j & k \\ 2 & -6 & -3 \\ 4 & 3 & -1 \end{vmatrix}$.
$a \times b = i((-6)(-1) - (-3)(3)) - j((2)(-1) - (-3)(4)) + k((2)(3) - (-6)(4))$.
$a \times b = i(6 + 9) - j(-2 + 12) + k(6 + 24) = 15i - 10j + 30k$.
Next,calculate the magnitude $|a \times b| = \sqrt{15^2 + (-10)^2 + 30^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35$.
Therefore,the required unit vector is $\frac{15i - 10j + 30k}{35} = \frac{15}{35}i - \frac{10}{35}j + \frac{30}{35}k = \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k = \frac{3i - 2j + 6k}{7}$.

(A) Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.
First,we find the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(-1 - 3) + \hat{k}(2 - (-2)) = -4\hat{i} + 4\hat{j} + 4\hat{k}$.
The magnitude of the cross product is $|\vec{a} \times \vec{b}| = \sqrt{(-4)^2 + 4^2 + 4^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3}$.
The unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{-4\hat{i} + 4\hat{j} + 4\hat{k}}{4\sqrt{3}} = \pm \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
Comparing this with the given options,the correct option is $\frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$.

(C) Let the given vectors be $\vec{a} = i - j + k$ and $\vec{b} = 2i + 3j - k$.
The vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 2 & 3 & -1 \end{vmatrix}$
$= i((-1)(-1) - (1)(3)) - j((1)(-1) - (1)(2)) + k((1)(3) - (-1)(2))$
$= i(1 - 3) - j(-1 - 2) + k(3 + 2)$
$= -2i + 3j + 5k$
The magnitude of the cross product is $|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$.
The unit vector perpendicular to both vectors is $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-2i + 3j + 5k}{\sqrt{38}}$.

(B) The cross product $a \times b$ is calculated using the determinant of a matrix:
$a \times b = \begin{vmatrix} i & j & k \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{vmatrix}$
Expanding along the first row:
$= i((-3)(-2) - (-1)(4)) - j((2)(-2) - (-1)(1)) + k((2)(4) - (-3)(1))$
$= i(6 + 4) - j(-4 + 1) + k(8 + 3)$
$= i(10) - j(-3) + k(11)$
$= 10i + 3j + 11k$

(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 the vectors.
Given $|a| = 4$,$|b| = 2$,and $\theta = \frac{\pi}{6}$.
Substituting these values,we get $|a \times b| = 4 \times 2 \times \sin\left(\frac{\pi}{6}\right)$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $|a \times b| = 8 \times \frac{1}{2} = 4$.
Therefore,$|a \times b|^2 = 4^2 = 16$.

(A) The cross product $a \times b$ is calculated using the determinant of the matrix:
$a \times b = \begin{vmatrix} i & j & k \\ 2 & 4 & -5 \\ 1 & 2 & 3 \end{vmatrix}$
$= i(4 \times 3 - (-5) \times 2) - j(2 \times 3 - (-5) \times 1) + k(2 \times 2 - 4 \times 1)$
$= i(12 + 10) - j(6 + 5) + k(4 - 4)$
$= 22i - 11j + 0k$
Now,calculate the magnitude $|a \times b|$:
$|a \times b| = \sqrt{(22)^2 + (-11)^2 + (0)^2}$
$= \sqrt{484 + 121}$
$= \sqrt{605}$
$= \sqrt{121 \times 5} = 11\sqrt{5}$.

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

(B) The vertices of the triangle are $A(1, -1, 2)$,$B(2, 1, -1)$,and $C(3, -1, 2)$.
First,we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:
$\overrightarrow{AB} = (2-1)\hat{i} + (1-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 2\hat{j} - 3\hat{k}$
$\overrightarrow{AC} = (3-1)\hat{i} + (-1-(-1))\hat{j} + (2-2)\hat{k} = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i}$
The area of the triangle is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$.
Calculating the cross product:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - (-6)) + \hat{k}(0 - 4) = -6\hat{j} - 4\hat{k}$.
Now,find the magnitude:
$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$.
Therefore,the area of the triangle is $\frac{1}{2} \times 2\sqrt{13} = \sqrt{13}$ square units.

(B) The area of a triangle with vertices $A, B, C$ is given by $\Delta = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$.
First,find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:
$\overrightarrow{AB} = (2-1)i + (0 - (-1))j + (-1-2)k = i + j - 3k$
$\overrightarrow{AC} = (0-1)i + (2 - (-1))j + (1-2)k = -i + 3j - k$
Now,calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} i & j & k \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix}$
$= i(-1 - (-9)) - j(-1 - 3) + k(3 - (-1))$
$= i(8) - j(-4) + k(4) = 8i + 4j + 4k$
The magnitude of the cross product is:
$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{8^2 + 4^2 + 4^2} = \sqrt{64 + 16 + 16} = \sqrt{96} = 4\sqrt{6}$
Therefore,the area of the triangle is:
$\Delta = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} (4\sqrt{6}) = 2\sqrt{6}$.

(C) Let the vertices be $A(1, 2, 3)$,$B(2, 5, -1)$,and $C(-1, 1, 2)$.
The area of the triangle is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$.
First,find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:
$\overrightarrow{AB} = (2-1)\hat{i} + (5-2)\hat{j} + (-1-3)\hat{k} = \hat{i} + 3\hat{j} - 4\hat{k}$.
$\overrightarrow{AC} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-3)\hat{k} = -2\hat{i} - \hat{j} - \hat{k}$.
Now,calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -4 \\ -2 & -1 & -1 \end{vmatrix}$
$= \hat{i}(-3 - 4) - \hat{j}(-1 - 8) + \hat{k}(-1 + 6)$
$= -7\hat{i} + 9\hat{j} + 5\hat{k}$.
The magnitude of the cross product is $|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-7)^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155}$.
Therefore,the area of the triangle is $\frac{1}{2} \times \sqrt{155} = \frac{\sqrt{155}}{2}$ sq. unit.

(C) The area of a parallelogram with adjacent sides represented by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product: $\text{Area} = |\vec{a} \times \vec{b}|$.
Given $\vec{a} = 3i + 0j - k$ and $\vec{b} = i + 2j + 0k$.
The cross product is calculated as:
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 3 & 0 & -1 \\ 1 & 2 & 0 \end{vmatrix}$
$= i(0(0) - (-1)(2)) - j(3(0) - (-1)(1)) + k(3(2) - 0(1))$
$= i(0 + 2) - j(0 + 1) + k(6 - 0)$
$= 2i - j + 6k$.
The area is the magnitude of this vector:
$|\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (-1)^2 + (6)^2}$
$= \sqrt{4 + 1 + 36}$
$= \sqrt{41}$.

(B) The area of a parallelogram with diagonals $\vec{a}$ and $\vec{b}$ is given by the formula $\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}$
$= \hat{i}(1(4) - (-2)(-3)) - \hat{j}(3(4) - (-2)(1)) + \hat{k}(3(-3) - 1(1))$
$= \hat{i}(4 - 6) - \hat{j}(12 + 2) + \hat{k}(-9 - 1)$
$= -2\hat{i} - 14\hat{j} - 10\hat{k}$.
Now,calculate the magnitude of the cross product:
$|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2}$
$= \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3}$.
Finally,the area is:
$\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} (10\sqrt{3}) = 5\sqrt{3}$.

(D) The position vectors are $\vec{OA} = \vec{i} + \vec{j}$,$\vec{OB} = \vec{j} + \vec{k}$,and $\vec{OC} = \vec{k} + \vec{i}$.
First,find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \vec{OB} - \vec{OA} = (\vec{j} + \vec{k}) - (\vec{i} + \vec{j}) = -\vec{i} + \vec{k}$
$\vec{AC} = \vec{OC} - \vec{OA} = (\vec{k} + \vec{i}) - (\vec{i} + \vec{j}) = -\vec{j} + \vec{k}$
The vector area of $\Delta ABC$ is given by $\frac{1}{2}(\vec{AB} \times \vec{AC})$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 0 & 1 \\ 0 & -1 & 1 \end{vmatrix}$
$= \vec{i}(0 - (-1)) - \vec{j}(-1 - 0) + \vec{k}(1 - 0) = \vec{i} + \vec{j} + \vec{k}$
Thus,the vector area is $\frac{1}{2}(\vec{i} + \vec{j} + \vec{k})$.
Comparing this with $\pm \frac{1}{2} \vec{\alpha}$,we get $\vec{\alpha} = \vec{i} + \vec{j} + \vec{k}$.

(C) The area of a triangle $OAB$ with vertices at the origin $O$,$A$,and $B$ is given by the formula $\Delta = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}|$.
First,we calculate the cross product $\overrightarrow{OA} \times \overrightarrow{OB}$:
$\overrightarrow{OA} \times \overrightarrow{OB} = \begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 1 & 3 & 1 \end{vmatrix}$
$= i(2(1) - (-1)(3)) - j(3(1) - (-1)(1)) + k(3(3) - 2(1))$
$= i(2 + 3) - j(3 + 1) + k(9 - 2)$
$= 5i - 4j + 7k$.
Now,we find the magnitude of this vector:
$|\overrightarrow{OA} \times \overrightarrow{OB}| = \sqrt{5^2 + (-4)^2 + 7^2} = \sqrt{25 + 16 + 49} = \sqrt{90} = 3\sqrt{10}$.
Finally,the area of the triangle is:
$\Delta = \frac{1}{2} \times 3\sqrt{10} = \frac{3}{2}\sqrt{10}$.

(C) The area of a parallelogram with adjacent sides represented by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,Area $= |\vec{a} \times \vec{b}|$.
Given $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.
Calculating the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & 1 \end{vmatrix}$
$= \hat{i}(2(1) - 3(-2)) - \hat{j}(1(1) - 3(3)) + \hat{k}(1(-2) - 2(3))$
$= \hat{i}(2 + 6) - \hat{j}(1 - 9) + \hat{k}(-2 - 6)$
$= 8\hat{i} + 8\hat{j} - 8\hat{k}$.
Now,find the magnitude:
$|\vec{a} \times \vec{b}| = \sqrt{8^2 + 8^2 + (-8)^2} = \sqrt{64 + 64 + 64} = \sqrt{3 \times 64} = 8\sqrt{3}$.
Note: The original provided options were mathematically inconsistent with the calculation. Based on the standard calculation,the area is $8\sqrt{3}$.