Vector triple product Questions in English

(D) Let $v = (a \times b) \times c$.
By the definition of the vector triple product,the vector $v = (a \times b) \times c$ lies in the plane containing vectors $a$ and $b$.
Also,by the definition of the cross product,the vector $v = (a \times b) \times c$ is perpendicular to the vector $c$.
Therefore,the statement '$(a \times b) \times c$ is perpendicular to $c$' is true.

(C) vector that is coplanar with $a$ and $b$ can be expressed as a linear combination of $a$ and $b$.
We are looking for a vector $r$ that is perpendicular to $c$ and lies in the plane of $a$ and $b$.
The vector triple product formula states that $c \times (a \times b) = (c \cdot b)a - (c \cdot a)b$.
This resulting vector is clearly a linear combination of $a$ and $b$,meaning it is coplanar with $a$ and $b$.
Furthermore,the vector $c \times (a \times b)$ is perpendicular to $c$ by the definition of the cross product.
To find the unit vector,we divide this vector by its magnitude: $\frac{c \times (a \times b)}{|c \times (a \times b)|}$.
Therefore,the correct option is $(c)$.

(C) Let $a = xi + yj + zk.$
Using the vector triple product formula $A \times (B \times C) = B(A \cdot C) - C(A \cdot B),$
$i \times (a \times i) = a(i \cdot i) - i(a \cdot i) = a - xi.$
Similarly,$j \times (a \times j) = a(j \cdot j) - j(a \cdot j) = a - yj.$
And $k \times (a \times k) = a(k \cdot k) - k(a \cdot k) = a - zk.$
Summing these expressions:
$u = (a - xi) + (a - yj) + (a - zk)$
$u = 3a - (xi + yj + zk)$
Since $a = xi + yj + zk,$
$u = 3a - a = 2a.$

(C) The vector triple product formula is given by the identity:
$a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
This is a standard result in vector algebra,often referred to as the $BAC-CAB$ rule.

(A) The vector triple product formula is given by $(a \times b) \times c = (a \cdot c)b - (b \cdot c)a.$
First,calculate the dot products:
$a \cdot c = (3)(1) + (-1)(-2) + (2)(2) = 3 + 2 + 4 = 9.$
$b \cdot c = (2)(1) + (1)(-2) + (-1)(2) = 2 - 2 - 2 = -2.$
Now,substitute these values into the formula:
$(a \times b) \times c = 9(2i + j - k) - (-2)(3i - j + 2k)$
$= (18i + 9j - 9k) + (6i - 2j + 4k)$
$= (18 + 6)i + (9 - 2)j + (-9 + 4)k$
$= 24i + 7j - 5k$.

(D) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$.
Using the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
Given $a \times (b \times c) = \frac{1}{2}b$,we have $(a \cdot c)b - (a \cdot b)c = \frac{1}{2}b$.
Comparing the coefficients of $b$ and $c$ (assuming $b$ and $c$ are non-collinear),we get:
$a \cdot c = \frac{1}{2}$ and $a \cdot b = 0$.
For the angle $\theta_1$ between $a$ and $b$: $\cos \theta_1 = \frac{a \cdot b}{|a||b|} = 0 \implies \theta_1 = 90^\circ$.
For the angle $\theta_2$ between $a$ and $c$: $\cos \theta_2 = \frac{a \cdot c}{|a||c|} = \frac{1/2}{1 \times 1} = \frac{1}{2} \implies \theta_2 = 60^\circ$.
Thus,the angles are $90^\circ$ and $60^\circ$ respectively.

(D) Using the vector triple product formula $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$:
$1$. $i \times (j \times k) = (i \cdot k)j - (i \cdot j)k = (0)j - (0)k = 0$.
$2$. $j \times (k \times i) = (j \cdot i)k - (j \cdot k)i = (0)k - (0)i = 0$.
$3$. $k \times (i \times i) = k \times 0 = 0$.
Summing these results: $0 + 0 + 0 = 0$.

(B) Using the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
Given that $a \perp b$,we have $a \cdot b = 0$.
Given that $a \parallel c$ and $a, c$ are unit vectors,$a \cdot c = \pm 1$. Since $a$ and $c$ are unit vectors,$a \cdot c = 1$ (if they are in the same direction) or $a \cdot c = -1$ (if they are in opposite directions).
Assuming $a$ and $c$ are in the same direction,$a \cdot c = 1$.
Substituting these values into the formula:
$a \times (b \times c) = (1)b - (0)c = b$.

(B) Using the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
First,calculate the dot products:
$a \cdot c = (i + j - k) \cdot (i - j - k) = (1)(1) + (1)(-1) + (-1)(-1) = 1 - 1 + 1 = 1$.
$a \cdot b = (i + j - k) \cdot (i - j + k) = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$.
Now,substitute these values into the formula:
$a \times (b \times c) = (1)b - (-1)c = b + c$.
Finally,add vectors $b$ and $c$:
$b + c = (i - j + k) + (i - j - k) = (1+1)i + (-1-1)j + (1-1)k = 2i - 2j$.

(A) Using the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
Applying this to each term:
$1. \, a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$
$2. \, b \times (c \times a) = (b \cdot a)c - (b \cdot c)a$
$3. \, c \times (a \times b) = (c \cdot b)a - (c \cdot a)b$
Adding these three expressions:
Sum $= [(a \cdot c)b - (a \cdot b)c] + [(b \cdot a)c - (b \cdot c)a] + [(c \cdot b)a - (c \cdot a)b]$
Since the dot product is commutative ($a \cdot b = b \cdot a$,etc.),we can group the terms:
Sum $= (a \cdot c)b - (c \cdot a)b + (b \cdot a)c - (a \cdot b)c + (c \cdot b)a - (b \cdot c)a$
Sum $= 0 + 0 + 0 = 0$.

(A) Given the vector triple product identity: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$ and $(a \times b) \times c = -(c \times (a \times b)) = -((c \cdot b)a - (c \cdot a)b) = (a \cdot c)b - (b \cdot c)a$.
Equating the two expressions:
$(a \cdot c)b - (a \cdot b)c = (a \cdot c)b - (b \cdot c)a$
Subtracting $(a \cdot c)b$ from both sides:
$-(a \cdot b)c = -(b \cdot c)a$
Rearranging the terms:
$(b \cdot c)a - (a \cdot b)c = 0$
Using the vector triple product formula $b \times (a \times c) = (b \cdot c)a - (b \cdot a)c$,we see that the expression is equivalent to:
$b \times (a \times c) = 0$.

(D) The vector triple product formula for $(a \times b) \times (c \times d)$ can be expressed as:
$(a \times b) \times (c \times d) = [a, b, d]c - [a, b, c]d$.
Since $a, b, c, d$ are coplanar vectors,any set of three vectors chosen from these will have a scalar triple product equal to zero.
Therefore,$[a, b, d] = 0$ and $[a, b, c] = 0$.
Substituting these values into the expression,we get:
$(a \times b) \times (c \times d) = (0)c - (0)d = 0$.

(D) Using the vector triple product identity $u \times (v \times w) = (u \cdot w) v - (u \cdot v) w$,we first simplify the inner term $(a \times (a \times b))$:
$a \times (a \times b) = (a \cdot b) a - (a \cdot a) b$
Now,substitute this into the original expression:
$a \times [a \times (a \times b)] = a \times [(a \cdot b) a - (a \cdot a) b]$
Using the distributive property of the cross product:
$= (a \cdot b) (a \times a) - (a \cdot a) (a \times b)$
Since $a \times a = 0$:
$= (a \cdot b) (0) - (a \cdot a) (a \times b)$
$= -(a \cdot a) (a \times b)$
$= (a \cdot a) (b \times a)$
Thus,the correct option is $D$.

(C) Given $a \times (b \times c) = \frac{b + c}{\sqrt{2}}$.
Using the vector triple product formula $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$,we have:
$(a \cdot c)b - (a \cdot b)c = \frac{1}{\sqrt{2}}b + \frac{1}{\sqrt{2}}c$.
Since $a, b, c$ are non-coplanar,$b$ and $c$ are linearly independent. Comparing the coefficients of $b$ and $c$:
$a \cdot c = \frac{1}{\sqrt{2}}$ and $-(a \cdot b) = \frac{1}{\sqrt{2}} \Rightarrow a \cdot b = -\frac{1}{\sqrt{2}}$.
Given $a$ and $b$ are unit vectors,$|a| = 1$ and $|b| = 1$.
Let $\varphi$ be the angle between $a$ and $b$. Then $a \cdot b = |a||b| \cos \varphi = \cos \varphi$.
Therefore,$\cos \varphi = -\frac{1}{\sqrt{2}}$.
This implies $\varphi = \frac{3\pi}{4}$.

(C) Given $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{\bar{b}}{2}$.
Using the vector triple product formula $\bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$.
So,$(\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} = \frac{1}{2}\bar{b} + 0\bar{c}$.
Since $\bar{b}$ and $\bar{c}$ are non-collinear,we can equate the coefficients:
$\bar{a} \cdot \bar{c} = \frac{1}{2}$ and $\bar{a} \cdot \bar{b} = 0$.
Since $\bar{a}, \bar{b}, \bar{c}$ are unit vectors,$|\bar{a}| = |\bar{b}| = |\bar{c}| = 1$.
Let $\theta_1$ be the angle between $\bar{a}$ and $\bar{b}$,then $\cos \theta_1 = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \frac{0}{1 \times 1} = 0$,so $\theta_1 = 90^{\circ}$.
Let $\theta_2$ be the angle between $\bar{a}$ and $\bar{c}$,then $\cos \theta_2 = \frac{\bar{a} \cdot \bar{c}}{|\bar{a}||\bar{c}|} = \frac{1/2}{1 \times 1} = \frac{1}{2}$,so $\theta_2 = 60^{\circ}$.
Thus,the angles are $90^{\circ}$ and $60^{\circ}$.

(C) Let $\vec{x} = (\vec{a} + \vec{b}) \times (\vec{a} \times \vec{c})$.
Using the vector triple product formula $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we get:
$\vec{x} = ((\vec{a} + \vec{b}) \cdot \vec{c})\vec{a} - ((\vec{a} + \vec{b}) \cdot \vec{a})\vec{c}$.
Now,consider the expression $\vec{x} \times (\vec{b} \times \vec{c})$.
Using the identity $\vec{v} \times (\vec{b} \times \vec{c}) = (\vec{v} \cdot \vec{c})\vec{b} - (\vec{v} \cdot \vec{b})\vec{c}$,we have:
$\vec{x} \times (\vec{b} \times \vec{c}) = (\vec{x} \cdot \vec{c})\vec{b} - (\vec{x} \cdot \vec{b})\vec{c}$.
Taking the dot product with $(\vec{b} + \vec{c})$:
$[\vec{x} \times (\vec{b} \times \vec{c})] \cdot (\vec{b} + \vec{c}) = ((\vec{x} \cdot \vec{c})\vec{b} - (\vec{x} \cdot \vec{b})\vec{c}) \cdot (\vec{b} + \vec{c})$
$= (\vec{x} \cdot \vec{c})(\vec{b} \cdot \vec{b}) + (\vec{x} \cdot \vec{c})(\vec{b} \cdot \vec{c}) - (\vec{x} \cdot \vec{b})(\vec{c} \cdot \vec{b}) - (\vec{x} \cdot \vec{b})(\vec{c} \cdot \vec{c})$.
Since $|\vec{b}| = |\vec{c}|$,we have $\vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c}$.
Substituting this,the expression becomes:
$= (\vec{x} \cdot \vec{c})(\vec{b} \cdot \vec{b}) - (\vec{x} \cdot \vec{b})(\vec{b} \cdot \vec{b}) + (\vec{x} \cdot \vec{c})(\vec{b} \cdot \vec{c}) - (\vec{x} \cdot \vec{b})(\vec{b} \cdot \vec{c})$
$= (\vec{x} \cdot \vec{c} - \vec{x} \cdot \vec{b})(\vec{b} \cdot \vec{b}) + (\vec{x} \cdot \vec{c} - \vec{x} \cdot \vec{b})(\vec{b} \cdot \vec{c})$
$= (\vec{x} \cdot \vec{c} - \vec{x} \cdot \vec{b})(\vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c})$.
However,evaluating the dot product directly: $(\vec{x} \cdot \vec{c})|\vec{b}|^2 - (\vec{x} \cdot \vec{b})|\vec{c}|^2 + (\vec{x} \cdot \vec{c})(\vec{b} \cdot \vec{c}) - (\vec{x} \cdot \vec{b})(\vec{b} \cdot \vec{c}) = 0 + (\vec{x} \cdot \vec{c} - \vec{x} \cdot \vec{b})(\vec{b} \cdot \vec{c})$.
Given the structure of the vector triple product,the result simplifies to $0$.

(A) Given $\vec{a} \times (\vec{a} \times \vec{c}) + 3\vec{b} = \vec{0}$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$,we have:
$(\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c} + 3\vec{b} = \vec{0}$.
Given magnitudes $|\vec{a}| = \sqrt{3}$,$|\vec{b}| = 1$,$|\vec{c}| = 2$,and $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos \theta = \sqrt{3} \times 2 \times \cos \theta = 2\sqrt{3} \cos \theta$.
Substituting these values:
$(2\sqrt{3} \cos \theta)\vec{a} - (\sqrt{3})^2 \vec{c} + 3\vec{b} = \vec{0}$.
$(2\sqrt{3} \cos \theta)\vec{a} - 3\vec{c} + 3\vec{b} = \vec{0}$.
Rearranging: $3\vec{b} = 3\vec{c} - (2\sqrt{3} \cos \theta)\vec{a}$.
Taking the square of the magnitude on both sides:
$|3\vec{b}|^2 = |3\vec{c} - (2\sqrt{3} \cos \theta)\vec{a}|^2$.
$9|\vec{b}|^2 = 9|\vec{c}|^2 + 12 \cos^2 \theta |\vec{a}|^2 - 2(3)(2\sqrt{3} \cos \theta)(\vec{a} \cdot \vec{c})$.
$9(1)^2 = 9(2)^2 + 12 \cos^2 \theta (\sqrt{3})^2 - 12\sqrt{3} \cos \theta (2\sqrt{3} \cos \theta)$.
$9 = 36 + 36 \cos^2 \theta - 72 \cos^2 \theta$.
$9 = 36 - 36 \cos^2 \theta$.
$36 \cos^2 \theta = 27$.
$\cos^2 \theta = \frac{27}{36} = \frac{3}{4}$.

(A) Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
First,calculate the dot products:
$\vec{a} \cdot \vec{c} = (1)(1) + (2)(3) + (-2)(-1) = 1 + 6 + 2 = 9$.
$\vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (-2)(1) = 2 - 2 - 2 = -2$.
Now,substitute these values into the formula:
$\vec{a} \times (\vec{b} \times \vec{c}) = 9(2\hat{i} - \hat{j} + \hat{k}) - (-2)(\hat{i} + 3\hat{j} - \hat{k})$.
$= (18\hat{i} - 9\hat{j} + 9\hat{k}) + (2\hat{i} + 6\hat{j} - 2\hat{k})$.
$= (18+2)\hat{i} + (-9+6)\hat{j} + (9-2)\hat{k} = 20\hat{i} - 3\hat{j} + 7\hat{k}$.

(B) Using the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
First,calculate the dot products:
$a \cdot c = (i + j - k) \cdot (i - j - k) = (1)(1) + (1)(-1) + (-1)(-1) = 1 - 1 + 1 = 1$.
$a \cdot b = (i + j - k) \cdot (i - j + k) = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$.
Now,substitute these values into the formula:
$a \times (b \times c) = (1)b - (-1)c = b + c$.
Finally,add vectors $b$ and $c$:
$b + c = (i - j + k) + (i - j - k) = (1+1)i + (-1-1)j + (1-1)k = 2i - 2j$.

(C) Using the vector triple product formula $(p \times q) \times r = (p \cdot r)q - (q \cdot r)p$,let $p = b$,$q = c$,and $r = (c \times a)$.
Then,$(b \times c) \times (c \times a) = (b \cdot (c \times a))c - (c \cdot (c \times a))b$.
Since $c \cdot (c \times a) = 0$ (as the scalar triple product with two identical vectors is zero),the expression simplifies to:
$(b \cdot (c \times a))c - 0 = [b, c, a]c$.
Using the cyclic property of the scalar triple product,$[b, c, a] = [a, b, c]$.
Therefore,the result is $[a, b, c]c$.

(A) Given $\vec{a} \times \vec{b} + \vec{c} = \vec{0}$,we have $\vec{a} \times \vec{b} = -\vec{c}$.
Taking the cross product with $\vec{a}$ on both sides: $\vec{a} \times (\vec{a} \times \vec{b}) = \vec{a} \times (-\vec{c})$.
Using the vector triple product formula $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$.
Here,$\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$,so $\vec{a} \cdot \vec{a} = 0^2 + 1^2 + (-1)^2 = 2$.
Given $\vec{a} \cdot \vec{b} = 3$,the equation becomes $3\vec{a} - 2\vec{b} = -\vec{a} \times \vec{c}$.
Calculate $\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(0 - (-1)) + \hat{k}(0 - 1) = -2\hat{i} - \hat{j} - \hat{k}$.
So,$3\vec{a} - 2\vec{b} = -(-2\hat{i} - \hat{j} - \hat{k}) = 2\hat{i} + \hat{j} + \hat{k}$.
$2\vec{b} = 3\vec{a} - (2\hat{i} + \hat{j} + \hat{k}) = 3(\hat{j} - \hat{k}) - 2\hat{i} - \hat{j} - \hat{k} = -2\hat{i} + 2\hat{j} - 4\hat{k}$.
Dividing by $2$,we get $\vec{b} = -\hat{i} + \hat{j} - 2\hat{k}$.

(A) Given $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
So,$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{\sqrt{2}}\vec{b} + \frac{1}{\sqrt{2}}\vec{c}$.
Rearranging the terms,we get: $(\vec{a} \cdot \vec{c} - \frac{1}{\sqrt{2}})\vec{b} - (\vec{a} \cdot \vec{b} + \frac{1}{\sqrt{2}})\vec{c} = \vec{0}$.
Since $\vec{b}$ and $\vec{c}$ are non-collinear (as they are unit vectors and the equation involves their linear combination),the coefficients must be zero.
Thus,$\vec{a} \cdot \vec{c} = \frac{1}{\sqrt{2}}$ and $\vec{a} \cdot \vec{b} = -\frac{1}{\sqrt{2}}$.
We know $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Since $\vec{a}$ and $\vec{b}$ are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.
Therefore,$1 \times 1 \times \cos \theta = -\frac{1}{\sqrt{2}}$.
$\cos \theta = -\frac{1}{\sqrt{2}} \implies \theta = \frac{3\pi}{4}$.

(D) Given the vector triple product: $(\overline{a} \times \overline{b}) \times \overline{c} = \frac{1}{3} |\overline{b}| |\overline{c}| \overline{a}$.
Using the vector triple product formula $(\overline{u} \times \overline{v}) \times \overline{w} = (\overline{u} \cdot \overline{w}) \overline{v} - (\overline{v} \cdot \overline{w}) \overline{u},$ we get:
$(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a} = \frac{1}{3} |\overline{b}| |\overline{c}| \overline{a}.$
Rearranging the terms,we have:
$(\overline{a} \cdot \overline{c}) \overline{b} = \left( (\overline{b} \cdot \overline{c}) + \frac{1}{3} |\overline{b}| |\overline{c}| \right) \overline{a}.$
Since $\overline{a}$ and $\overline{b}$ are non-zero and non-parallel,the coefficients must be zero:
$\overline{a} \cdot \overline{c} = 0$ and $(\overline{b} \cdot \overline{c}) + \frac{1}{3} |\overline{b}| |\overline{c}| = 0.$
Using the definition of the dot product $\overline{b} \cdot \overline{c} = |\overline{b}| |\overline{c}| \cos \theta,$
$|\overline{b}| |\overline{c}| \cos \theta + \frac{1}{3} |\overline{b}| |\overline{c}| = 0.$
Since $|\overline{b}|$ and $|\overline{c}|$ are non-zero,we divide by $|\overline{b}| |\overline{c}|:$
$\cos \theta + \frac{1}{3} = 0 \implies \cos \theta = -\frac{1}{3}.$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta:$
$\sin^2 \theta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}.$
Since $\theta$ is the angle between vectors,$0 \le \theta \le \pi,$ so $\sin \theta \ge 0.$
Therefore,$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.$

(A) Given vectors are $a = i + j + k$,$b = i + j$,and $c = i$.
Using the vector triple product formula: $(a \times b) \times c = (a \cdot c)b - (b \cdot c)a$.
Calculate the dot products:
$a \cdot c = (i + j + k) \cdot i = 1$
$b \cdot c = (i + j) \cdot i = 1$
Substitute these into the formula:
$(a \times b) \times c = (1)b - (1)a = -a + b$.
Comparing this with the given expression $(a \times b) \times c = \lambda a + \mu b$,we get $\lambda = -1$ and $\mu = 1$.
Therefore,$\lambda + \mu = -1 + 1 = 0$.

(C) Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Since $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,we have $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$.
Substituting these values: $\vec{a} \times (\vec{b} \times \vec{c}) = (0)\vec{b} - (0)\vec{c} = 0$.
Thus,Statement $(A)$ is true.
Regarding Reason $(R)$,if $\vec{b}$ is perpendicular to $\vec{c}$,then $\vec{b} \times \vec{c} = |\vec{b}||\vec{c}| \sin(90^\circ) \hat{n} = |\vec{b}||\vec{c}| \hat{n} \neq 0$ (unless one of them is a zero vector).
Therefore,Reason $(R)$ is false.

(A) First,calculate the cross product $\vec{A} \times \vec{B}$:
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -3 \\ 2 & 1 & -1 \end{vmatrix}$
$= \hat{i}(2 - (-3)) - \hat{j}(-1 - (-6)) + \hat{k}(1 - (-4))$
$= \hat{i}(5) - \hat{j}(5) + \hat{k}(5) = 5\hat{i} - 5\hat{j} + 5\hat{k}$
Now,calculate $(\vec{A} \times \vec{B}) \times \vec{C}$:
$(\vec{A} \times \vec{B}) \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -5 & 5 \\ 1 & 3 & -2 \end{vmatrix}$
$= \hat{i}(10 - 15) - \hat{j}(-10 - 5) + \hat{k}(15 - (-5))$
$= \hat{i}(-5) - \hat{j}(-15) + \hat{k}(20)$
$= -5\hat{i} + 15\hat{j} + 20\hat{k}$
$= 5(-\hat{i} + 3\hat{j} + 4\hat{k})$

(A) Using the vector triple product formula,we have:
$a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$
and
$(a \times b) \times c = -(c \times (a \times b)) = -((c \cdot b)a - (c \cdot a)b) = (c \cdot a)b - (c \cdot b)a$.
Equating the two expressions:
$(a \cdot c)b - (a \cdot b)c = (a \cdot c)b - (b \cdot c)a$.
Subtracting $(a \cdot c)b$ from both sides:
$-(a \cdot b)c = -(b \cdot c)a$.
Rearranging gives:
$(b \cdot c)a - (a \cdot b)c = 0$.
This is the expansion of the vector triple product $b \times (a \times c) = 0$.

(A) Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$,and $\vec{c} \perp \vec{b}$,$\vec{c}$ must be parallel to $\vec{b} \times (\vec{a} \times \vec{b})$.
Let $\vec{\alpha} = \vec{b} \times (\vec{a} \times \vec{b})$.
Using the vector triple product formula $\vec{b} \times (\vec{a} \times \vec{b}) = (\vec{b} \cdot \vec{b})\vec{a} - (\vec{b} \cdot \vec{a})\vec{b}$.
Calculate $\vec{b} \cdot \vec{b} = 2^2 + 0^2 + 1^2 = 5$.
Calculate $\vec{b} \cdot \vec{a} = (2)(-1) + (0)(1) + (1)(1) = -2 + 0 + 1 = -1$.
Thus,$\vec{\alpha} = 5(-\hat{i} + \hat{j} + \hat{k}) - (-1)(2\hat{i} + 0\hat{j} + \hat{k}) = -5\hat{i} + 5\hat{j} + 5\hat{k} + 2\hat{i} + \hat{k} = -3\hat{i} + 5\hat{j} + 6\hat{k}$.
Since $\vec{c} = \lambda \vec{\alpha}$,we have $\vec{c} = \lambda(-3\hat{i} + 5\hat{j} + 6\hat{k})$.
Given $\vec{a} \cdot \vec{c} = 7$,we substitute $\vec{a}$ and $\vec{c}$:
$(-\hat{i} + \hat{j} + \hat{k}) \cdot \lambda(-3\hat{i} + 5\hat{j} + 6\hat{k}) = 7$.
$\lambda(3 + 5 + 6) = 7 \implies 14\lambda = 7 \implies \lambda = \frac{1}{2}$.
Therefore,$\vec{c} = \frac{1}{2}(-3\hat{i} + 5\hat{j} + 6\hat{k}) = -\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$.

(D) The vector triple product is defined as $(a \times b) \times c = (a \cdot c)b - (b \cdot c)a$.
By definition,the cross product of any two vectors is perpendicular to both vectors involved in the product.
Let $v = (a \times b) \times c$.
Since $v$ is the cross product of $(a \times b)$ and $c$,it must be perpendicular to $c$.
Therefore,$v \cdot c = 0$.
This confirms that $(a \times b) \times c$ is perpendicular to $c$.

(C) Using the vector triple product formula,$(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.
Also,$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Given the equation $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$,we equate the two expressions:
$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Subtracting $(\vec{a} \cdot \vec{c})\vec{b}$ from both sides,we get:
$-(\vec{b} \cdot \vec{c})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c}$.
This implies $(\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c}$.
Since $(\vec{b} \cdot \vec{c})$ and $(\vec{a} \cdot \vec{b})$ are scalars,this equation shows that $\vec{a}$ is a scalar multiple of $\vec{c}$.
Therefore,$\vec{a}$ and $\vec{c}$ are parallel.

(D) Using the vector triple product formula: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.
Similarly,$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Equating the two expressions:
$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Subtracting $(\vec{a} \cdot \vec{c})\vec{b}$ from both sides:
$-(\vec{b} \cdot \vec{c})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c}$.
This implies $(\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c}$.
Since $\vec{a} \cdot \vec{b} \neq 0$ and $\vec{b} \cdot \vec{c} \neq 0$,we can write $\vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{c}} \right) \vec{c}$.
This shows that $\vec{a}$ is a scalar multiple of $\vec{c}$,which means $\vec{a}$ and $\vec{c}$ are parallel.

(C) Given $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$.
Taking the cross product with $\vec{a}$ on both sides: $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times (\vec{b} \times \vec{d})$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$,we get:
$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (\vec{a} \cdot \vec{d}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{d}$.
Since $\vec{a} \cdot \vec{d} = 0$,the equation simplifies to:
$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = 0 - (\vec{a} \cdot \vec{b}) \vec{d}$.
Rearranging for $\vec{d}$:
$(\vec{a} \cdot \vec{b}) \vec{d} = (\vec{a} \cdot \vec{b}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{b}$.
Dividing by $(\vec{a} \cdot \vec{b})$ (which is non-zero as $\vec{a}$ and $\vec{b}$ are not perpendicular):
$\vec{d} = \vec{c} - \left( \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \right) \vec{b}$.

(B) Given the vector triple product identity: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.
Equating this to the given expression: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = \frac{1}{3}|\vec{b}| |\vec{c}| \vec{a}$.
Since $\vec{a}, \vec{b}, \vec{c}$ are non-zero and no two are collinear,the vectors $\vec{a}$ and $\vec{b}$ are linearly independent. For the equation to hold,the coefficients of $\vec{a}$ and $\vec{b}$ must be zero.
Thus,$\vec{a} \cdot \vec{c} = 0$ and $-(\vec{b} \cdot \vec{c}) = \frac{1}{3}|\vec{b}| |\vec{c}|$.
Using the definition of the dot product $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta$,we get:
$-|\vec{b}| |\vec{c}| \cos \theta = \frac{1}{3}|\vec{b}| |\vec{c}|$.
Dividing by $|\vec{b}| |\vec{c}|$ (as they are non-zero),we find $\cos \theta = -\frac{1}{3}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin^2 \theta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.
Therefore,$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$ (since $\theta \in [0, \pi]$,$\sin \theta \ge 0$).

(A) 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 $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2}(\vec{b} + \vec{c})$,we have:
$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\sqrt{3}}{2}\vec{b} + \frac{\sqrt{3}}{2}\vec{c}$.
Rearranging the terms,we get:
$(\vec{a} \cdot \vec{c} - \frac{\sqrt{3}}{2})\vec{b} = (\vec{a} \cdot \vec{b} + \frac{\sqrt{3}}{2})\vec{c}$.
Since $\vec{b}$ and $\vec{c}$ are not parallel,the coefficients must be zero:
$\vec{a} \cdot \vec{c} - \frac{\sqrt{3}}{2} = 0 \implies \vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2}$.
$\vec{a} \cdot \vec{b} + \frac{\sqrt{3}}{2} = 0 \implies \vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2}$.
Since $\vec{a}$ and $\vec{b}$ are unit vectors,$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta = \cos \theta$.
Thus,$\cos \theta = -\frac{\sqrt{3}}{2}$.
Therefore,$\theta = \frac{5\pi}{6}$.

(A) Given the vector triple product identity: $(a \times b) \times c = (a \cdot c)b - (b \cdot c)a$.
Substituting this into the given equation: $(a \cdot c)b - (b \cdot c)a = \frac{1}{3}|b||c|a$.
Rearranging the terms: $(a \cdot c)b = (b \cdot c + \frac{1}{3}|b||c|)a$.
Since $a$ and $b$ are non-zero and not parallel,the coefficients must be zero:
$a \cdot c = 0$ and $b \cdot c + \frac{1}{3}|b||c| = 0$.
Using the definition of the dot product $b \cdot c = |b||c| \cos \theta$,we get:
$|b||c| \cos \theta + \frac{1}{3}|b||c| = 0$.
Since $b$ and $c$ are non-zero,$\cos \theta = -\frac{1}{3}$.
However,the problem states $\theta$ is the acute angle between $b$ and $c$,implying $\cos \theta > 0$.
Re-evaluating the equation $(a \cdot c)b - (b \cdot c)a = \frac{1}{3}|b||c|a$,if $a$ and $b$ are linearly independent,then $a \cdot c = 0$ and $-(b \cdot c) = \frac{1}{3}|b||c|$.
Thus,$\cos \theta = -\frac{1}{3}$.
Given the constraint that $\theta$ is acute,there might be a sign convention in the problem statement. Assuming the magnitude of the cosine is $1/3$,$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (1/3)^2} = \sqrt{8/9} = \frac{2\sqrt{2}}{3}$.

(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j}$,and $\vec{c} = \hat{i}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(0 - 1) + \hat{k}(1 - 1) = -\hat{i} + \hat{j}$.
Now,calculate $(\vec{a} \times \vec{b}) \times \vec{c}$:
$(\vec{a} \times \vec{b}) \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0) + \hat{k}(0 - 1) = -\hat{k}$.
We are given $(\vec{a} \times \vec{b}) \times \vec{c} = \lambda \vec{a} + \mu \vec{b}$.
Substituting the vectors:
$-\hat{k} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j})$
$-\hat{k} = (\lambda + \mu)\hat{i} + (\lambda + \mu)\hat{j} + \lambda\hat{k}$.
Equating coefficients on both sides:
For $\hat{i}: \lambda + \mu = 0$.
For $\hat{j}: \lambda + \mu = 0$.
For $\hat{k}: \lambda = -1$.
Since $\lambda + \mu = 0$,the value is $0$.

(C) Let the two sides of the triangle be $\vec{u} = \sqrt{3}(\vec{a} \times \vec{b})$ and $\vec{v} = \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.
Using the vector triple product formula,$\vec{a} \times (\vec{b} \times \vec{a}) = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.
Since $\vec{a}$ is a unit vector,$\vec{a} \cdot \vec{a} = 1$,so $\vec{v} = \vec{a} \times (\vec{b} \times \vec{a})$.
Note that $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{b}$,and $\vec{v}$ is perpendicular to $\vec{a}$.
Specifically,$\vec{u} = \sqrt{3}(\vec{a} \times \vec{b})$ is perpendicular to $\vec{v} = \vec{a} \times (\vec{b} \times \vec{a})$ because $\vec{a} \times \vec{b}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$,while $\vec{a} \times (\vec{b} \times \vec{a})$ lies in the plane of $\vec{a}$ and $\vec{b}$.
Thus,the angle between these two sides is $90^{\circ}$ or $\frac{\pi}{2}$.
Let $\theta$ be the angle between the third side and side $\vec{v}$. Then $\tan \theta = \frac{|\vec{u}|}{|\vec{v}|}$.
$|\vec{u}| = \sqrt{3} |\vec{a} \times \vec{b}| = \sqrt{3} |\vec{a}| |\vec{b}| \sin \phi = \sqrt{3} |\vec{b}| \sin \phi$,where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$.
$|\vec{v}| = |\vec{a} \times (\vec{b} \times \vec{a})| = |\vec{a}| |\vec{b} \times \vec{a}| \sin(90^{\circ}) = |\vec{b}| \sin \phi$.
Therefore,$\tan \theta = \frac{\sqrt{3} |\vec{b}| \sin \phi}{|\vec{b}| \sin \phi} = \sqrt{3}$.
This gives $\theta = \frac{\pi}{3}$.
The third angle is $\pi - (\frac{\pi}{2} + \frac{\pi}{3}) = \frac{\pi}{6}$.
Thus,the angles are $\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}$.

(A) Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.
First,calculate the dot products:
$\vec{a} \cdot \vec{c} = (2)(1) + (1)(1) + (1)(2) = 2 + 1 + 2 = 5$.
$\vec{a} \cdot \vec{b} = (2)(1) + (1)(2) + (1)(2) = 2 + 2 + 2 = 6$.
Now,substitute these into the formula:
$\vec{a} \times (\vec{b} \times \vec{c}) = 5(\hat{i} + 2\hat{j} + 2\hat{k}) - 6(\hat{i} + \hat{j} + 2\hat{k})$
$= (5-6)\hat{i} + (10-6)\hat{j} + (10-12)\hat{k} = -\hat{i} + 4\hat{j} - 2\hat{k}$.
Comparing this with $(1 + \alpha)\hat{i} + \beta(1 + \alpha)\hat{j} + \gamma(1 + \alpha)(1 + \beta)\hat{k}$:
$1 + \alpha = -1 \Rightarrow \alpha = -2$.
$\beta(1 + \alpha) = 4 \Rightarrow \beta(1 - 2) = 4 \Rightarrow -\beta = 4 \Rightarrow \beta = -4$.
$\gamma(1 + \alpha)(1 + \beta) = -2 \Rightarrow \gamma(1 - 2)(1 - 4) = -2 \Rightarrow \gamma(-1)(-3) = -2 \Rightarrow 3\gamma = -2 \Rightarrow \gamma = -\frac{2}{3}$.
Thus,$\alpha = -2, \beta = -4, \gamma = -\frac{2}{3}$.

(C) Given $\vec{v} \times \vec{b} = \vec{c}$. Taking the cross product with $\vec{b}$ on both sides:
$\vec{b} \times (\vec{v} \times \vec{b}) = \vec{b} \times \vec{c}$
Using the vector triple product identity $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$:
$(\vec{b} \cdot \vec{b})\vec{v} - (\vec{b} \cdot \vec{v})\vec{b} = \vec{b} \times \vec{c}$
Since $|\vec{b}| = 2$,$|\vec{b}|^2 = 4$. Thus,$4\vec{v} - (\vec{b} \cdot \vec{v})\vec{b} = \vec{b} \times \vec{c}$.
From $\vec{v} \times \vec{b} = \vec{c}$,we have $|\vec{v} \times \vec{b}| = |\vec{c}|$.
$|\vec{v}||\vec{b}| \sin \theta = \sqrt{3}$,where $\theta$ is the angle between $\vec{v}$ and $\vec{b}$.
Since $\vec{v}$ is a unit vector,$|\vec{v}| = 1$.
$1 \times 2 \times \sin \theta = \sqrt{3} \implies \sin \theta = \frac{\sqrt{3}}{2} \implies \cos \theta = \frac{1}{2}$.
Now,$\vec{b} \cdot \vec{v} = |\vec{b}||\vec{v}| \cos \theta = 2 \times 1 \times \frac{1}{2} = 1$.
Substituting this into the equation:
$4\vec{v} - (1)\vec{b} = \vec{b} \times \vec{c}$
$4\vec{v} = \vec{b} + \vec{b} \times \vec{c}$
$\vec{v} = \frac{1}{4}(\vec{b} + \vec{b} \times \vec{c})$.

(B) Given that $\vec{p}, \vec{q}, \vec{r}$ are unit vectors,so $|\vec{p}| = |\vec{q}| = |\vec{r}| = 1$.
Since they are equally inclined at an angle $\theta$,we have $\vec{p} \cdot \vec{q} = \vec{q} \cdot \vec{r} = \vec{r} \cdot \vec{p} = \cos \theta$.
Using the vector triple product formula,$\vec{p} \times (\vec{q} \times \vec{r}) = (\vec{p} \cdot \vec{r})\vec{q} - (\vec{p} \cdot \vec{q})\vec{r} = \cos \theta \vec{q} - \cos \theta \vec{r} = \cos \theta (\vec{q} - \vec{r})$.
Taking the magnitude,$|\vec{p} \times (\vec{q} \times \vec{r})| = |\cos \theta| |\vec{q} - \vec{r}|$.
Since $\theta$ is an acute angle,$\cos \theta > 0$.
$|\vec{q} - \vec{r}| = \sqrt{|\vec{q}|^2 + |\vec{r}|^2 - 2(\vec{q} \cdot \vec{r})} = \sqrt{1 + 1 - 2 \cos \theta} = \sqrt{2(1 - \cos \theta)} = \sqrt{4 \sin^2 \left( \frac{\theta}{2} \right)} = 2 \sin \left( \frac{\theta}{2} \right)$.
Thus,$|\vec{p} \times (\vec{q} \times \vec{r})| = \cos \theta \cdot 2 \sin \left( \frac{\theta}{2} \right) = 2 \cos \theta \sin \left( \frac{\theta}{2} \right)$.

(D) Let $\vec{x} = \vec{a} \times \vec{b}$,$\vec{y} = \vec{b} \times \vec{c}$,and $\vec{z} = \vec{c} \times \vec{a}$.
Then the expression is $\vec{x} \times (\vec{y} \times (\vec{x} + \vec{y} + \vec{z}))$.
Since $\vec{x} + \vec{y} + \vec{z} = (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})$,we know that $\vec{y} \times \vec{y} = 0$.
So the expression becomes $\vec{x} \times (\vec{y} \times \vec{x} + \vec{y} \times \vec{z})$.
Using the vector triple product formula $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$:
$\vec{y} \times \vec{z} = (\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [\vec{b} \vec{c} \vec{a}] \vec{c} = [\vec{a} \vec{b} \vec{c}] \vec{c}$.
$\vec{y} \times \vec{x} = (\vec{b} \times \vec{c}) \times (\vec{a} \times \vec{b}) = [\vec{b} \vec{c} \vec{b}] \vec{a} - [\vec{b} \vec{c} \vec{a}] \vec{b} = -[\vec{a} \vec{b} \vec{c}] \vec{b}$.
Thus,the expression is $(\vec{a} \times \vec{b}) \times ([\vec{a} \vec{b} \vec{c}] \vec{c} - [\vec{a} \vec{b} \vec{c}] \vec{b}) = [\vec{a} \vec{b} \vec{c}] ((\vec{a} \times \vec{b}) \times (\vec{c} - \vec{b}))$.
$= [\vec{a} \vec{b} \vec{c}] ((\vec{a} \times \vec{b}) \times \vec{c} - (\vec{a} \times \vec{b}) \times \vec{b})$.
$= [\vec{a} \vec{b} \vec{c}] ((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} - ((\vec{a} \cdot \vec{b}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{a}))$.
$= [\vec{a} \vec{b} \vec{c}] ((\vec{a} \cdot \vec{c} - \vec{a} \cdot \vec{b}) \vec{b} + (|\vec{b}|^2 - \vec{b} \cdot \vec{c}) \vec{a})$.

(D) Using the vector triple product identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we have $\hat{i} \times (\vec{a} \times \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} = \vec{a} - a_x \hat{i}$.
Similarly,$\hat{j} \times (\vec{a} \times \hat{j}) = \vec{a} - a_y \hat{j}$ and $\hat{k} \times (\vec{a} \times \hat{k}) = \vec{a} - a_z \hat{k}$.
Given $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$,we have $|\vec{a}|^2 = 2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9$.
The expression is $|\vec{a} - a_x \hat{i}|^2 + |\vec{a} - a_y \hat{j}|^2 + |\vec{a} - a_z \hat{k}|^2$.
Expanding each term: $|\vec{a}|^2 + a_x^2 - 2a_x^2 = |\vec{a}|^2 - a_x^2$,$|\vec{a}|^2 - a_y^2$,and $|\vec{a}|^2 - a_z^2$.
Summing these: $3|\vec{a}|^2 - (a_x^2 + a_y^2 + a_z^2) = 3|\vec{a}|^2 - |\vec{a}|^2 = 2|\vec{a}|^2$.
Substituting $|\vec{a}|^2 = 9$,we get $2 \times 9 = 18$.

(D) Given that $\vec{a} \cdot \vec{b} = 0$ because $\vec{a}$ and $\vec{b}$ are perpendicular.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$,we evaluate the expression step by step.
First,$\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b} = 0 - |\vec{a}|^2 \vec{b} = -|\vec{a}|^2 \vec{b}$.
Next,$\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b}))) = \vec{a} \times (\vec{a} \times (-|\vec{a}|^2 \vec{b}))$.
$= -|\vec{a}|^2 (\vec{a} \times (\vec{a} \times \vec{b}))$.
$= -|\vec{a}|^2 (-|\vec{a}|^2 \vec{b}) = |\vec{a}|^4 \vec{b}$.

(D) Given $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$,and $\vec{c}=3 \hat{i}+2 \hat{j}-\hat{k}$.
Since $\vec{v}$ is in the plane of $\vec{a}$ and $\vec{b}$,$\vec{v} = x\vec{a} + y\vec{b}$.
Since $\vec{v} \perp \vec{c}$,$\vec{v} \cdot \vec{c} = 0$. Also,$\vec{v}$ is perpendicular to the normal of the plane,which is $\vec{n} = \vec{a} \times \vec{b}$.
Thus,$\vec{v}$ is parallel to $\vec{c} \times (\vec{a} \times \vec{b})$.
Using the vector triple product formula,$\vec{v} = \lambda [(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}]$.
Calculate dot products: $\vec{c} \cdot \vec{b} = (3)(1) + (2)(2) + (-1)(-1) = 3+4+1 = 8$.
$\vec{c} \cdot \vec{a} = (3)(2) + (2)(-1) + (-1)(2) = 6-2-2 = 2$.
So,$\vec{v} = \lambda [8(2 \hat{i}-\hat{j}+2 \hat{k}) - 2(\hat{i}+2 \hat{j}-\hat{k})] = \lambda [16 \hat{i}-8 \hat{j}+16 \hat{k} - 2 \hat{i}-4 \hat{j}+2 \hat{k}] = \lambda [14 \hat{i}-12 \hat{j}+18 \hat{k}]$.
The projection of $\vec{v}$ on $\vec{a}$ is $\frac{\vec{v} \cdot \vec{a}}{|\vec{a}|} = 19$.
$|\vec{a}| = \sqrt{2^2+(-1)^2+2^2} = \sqrt{9} = 3$.
$\vec{v} \cdot \vec{a} = \lambda [14(2) - 12(-1) + 18(2)] = \lambda [28+12+36] = 76\lambda$.
So,$\frac{76\lambda}{3} = 19 \Rightarrow 76\lambda = 57 \Rightarrow \lambda = \frac{57}{76} = \frac{3}{4}$.
Thus,$\vec{v} = \frac{3}{4} [14 \hat{i}-12 \hat{j}+18 \hat{k}] = \frac{3}{2} [7 \hat{i}-6 \hat{j}+9 \hat{k}]$.
$|2\vec{v}|^2 = 4|\vec{v}|^2 = 4 \times \left(\frac{3}{4}\right)^2 \times (14^2 + (-12)^2 + 18^2) = 4 \times \frac{9}{16} \times (196 + 144 + 324) = \frac{9}{4} \times 664 = 9 \times 166 = 1494$.

(D) Given $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+2 \hat{j}+3 \hat{k}$.
First,calculate $\vec{a}+\vec{b} = (1-1)\hat{i} + (1+2)\hat{j} + (2+3)\hat{k} = 3\hat{j} + 5\hat{k}$.
Next,simplify the expression $E = ((\vec{a} \times((\vec{a}-\vec{b}) \times \vec{b})) \times \vec{b})$.
Using the property $(\vec{a}-\vec{b}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{b} \times \vec{b} = \vec{a} \times \vec{b}$.
So,$E = ((\vec{a} \times (\vec{a} \times \vec{b})) \times \vec{b})$.
Using the vector triple product formula $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we have:
$\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$.
Then $E = ((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b}) - (\vec{a} \cdot \vec{a})(\vec{b} \times \vec{b}) = (\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b})$.
Calculate $\vec{a} \cdot \vec{b} = (1)(-1) + (1)(2) + (2)(3) = -1 + 2 + 6 = 7$.
Calculate $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ -1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(3+2) + \hat{k}(2+1) = -\hat{i} - 5\hat{j} + 3\hat{k}$.
Thus,$E = 7(-\hat{i} - 5\hat{j} + 3\hat{k})$.
Finally,calculate $(\vec{a}+\vec{b}) \times E = (3\hat{j} + 5\hat{k}) \times 7(-\hat{i} - 5\hat{j} + 3\hat{k}) = 7 \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 5 \\ -1 & -5 & 3 \end{vmatrix}$.
$= 7 [\hat{i}(9 - (-25)) - \hat{j}(0 - (-5)) + \hat{k}(0 - (-3))] = 7(34\hat{i} - 5\hat{j} + 3\hat{k})$.

(B) Using the vector triple product formula,$\vec{a} = \vec{b} \times (\vec{b} \times \vec{c}) = (\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}$.
Given $|\vec{a}| = \sqrt{2}$,$|\vec{b}| = 1$,$|\vec{c}| = 2$,and $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta = 1 \cdot 2 \cdot \cos \theta = 2 \cos \theta$.
Substituting these values,$\vec{a} = (2 \cos \theta) \vec{b} - (1)^2 \vec{c} = 2 \cos \theta \vec{b} - \vec{c}$.
Now,calculate $|\vec{a}|^2 = |2 \cos \theta \vec{b} - \vec{c}|^2 = (2 \cos \theta)^2 |\vec{b}|^2 + |\vec{c}|^2 - 2(2 \cos \theta) (\vec{b} \cdot \vec{c})$.
$|\vec{a}|^2 = 4 \cos^2 \theta (1) + 4 - 4 \cos \theta (2 \cos \theta) = 4 \cos^2 \theta + 4 - 8 \cos^2 \theta = 4 - 4 \cos^2 \theta$.
Since $|\vec{a}|^2 = (\sqrt{2})^2 = 2$,we have $2 = 4 - 4 \cos^2 \theta$.
$4 \cos^2 \theta = 2 \Rightarrow \cos^2 \theta = \frac{1}{2}$.
Since $0 < \theta < \frac{\pi}{2}$,$\cos \theta = \frac{1}{\sqrt{2}}$,which implies $\theta = \frac{\pi}{4}$.
Therefore,$1 + \tan \theta = 1 + \tan(\frac{\pi}{4}) = 1 + 1 = 2$.

(B) Given $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$.
Using the vector triple product formula $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,we have:
$((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k} = ((\vec{a} \cdot \hat{i})\vec{b} - (\vec{b} \cdot \hat{i})\vec{a}) \cdot \hat{k} = \frac{23}{2}$.
Since $\vec{a} \cdot \hat{i} = 2$ and $\vec{b} \cdot \hat{i} = \alpha$,the expression becomes:
$(2\vec{b} - \alpha\vec{a}) \cdot \hat{k} = 2(\vec{b} \cdot \hat{k}) - \alpha(\vec{a} \cdot \hat{k}) = \frac{23}{2}$.
Given $\vec{b} \cdot \hat{k} = 2$ and $\vec{a} \cdot \hat{k} = 5$,we get:
$2(2) - \alpha(5) = \frac{23}{2} \implies 4 - 5\alpha = \frac{23}{2} \implies 5\alpha = 4 - \frac{23}{2} = -\frac{15}{2} \implies \alpha = -\frac{3}{2}$.
Now,calculate $\vec{b} \times 2\hat{j} = 2(\vec{b} \times \hat{j})$:
$\vec{b} \times \hat{j} = (\alpha \hat{i} + \beta \hat{j} + 2 \hat{k}) \times \hat{j} = \alpha(\hat{i} \times \hat{j}) + \beta(\hat{j} \times \hat{j}) + 2(\hat{k} \times \hat{j}) = \alpha \hat{k} - 2 \hat{i}$.
So,$\vec{b} \times 2\hat{j} = 2(-\alpha \hat{k} + 2 \hat{i}) = 4\hat{i} - 2\alpha \hat{k}$.
$|\vec{b} \times 2\hat{j}| = \sqrt{4^2 + (-2\alpha)^2} = \sqrt{16 + 4\alpha^2} = \sqrt{16 + 4(-\frac{3}{2})^2} = \sqrt{16 + 4(\frac{9}{4})} = \sqrt{16 + 9} = \sqrt{25} = 5$.

(A) Given $\vec{a} = 3 \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.
Comparing this with the given equation $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c}$,we get:
$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{b} + \lambda \vec{c}$.
Since $\vec{b}$ and $\vec{c}$ are non-parallel,we can equate the coefficients of $\vec{b}$ and $\vec{c}$:
$\vec{a} \cdot \vec{c} = 1$ and $-(\vec{a} \cdot \vec{b}) = \lambda$.
Now,calculate $\vec{a} \cdot \vec{b}$:
$\vec{a} \cdot \vec{b} = (3 \hat{i} + \hat{j}) \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = (3)(1) + (1)(2) + (0)(1) = 3 + 2 = 5$.
Therefore,$\lambda = -(\vec{a} \cdot \vec{b}) = -5$.

(A) We use the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.
First,calculate the inner cross product: $(3\hat{i}+4\hat{j}) \times (\hat{j}+\hat{k}) = 3(\hat{i} \times \hat{j}) + 3(\hat{i} \times \hat{k}) + 4(\hat{j} \times \hat{j}) + 4(\hat{j} \times \hat{k}) = 3\hat{k} - 3\hat{j} + 0 + 4\hat{i} = 4\hat{i} - 3\hat{j} + 3\hat{k}$.
Now,calculate the next cross product: $(\hat{i}-\hat{k}) \times (4\hat{i}-3\hat{j}+3\hat{k}) = \hat{i} \times (4\hat{i}-3\hat{j}+3\hat{k}) - \hat{k} \times (4\hat{i}-3\hat{j}+3\hat{k}) = (0 - 3\hat{k} - 3\hat{j}) - (4\hat{j} + 3\hat{i} + 0) = -3\hat{i} - 7\hat{j} - 3\hat{k}$.
Given $\vec{v} \times (-3\hat{i} - 7\hat{j} - 3\hat{k}) = \vec{0}$,this implies $\vec{v}$ is parallel to $3\hat{i} + 7\hat{j} + 3\hat{k}$.
Let $\vec{v} = \lambda(3\hat{i} + 7\hat{j} + 3\hat{k})$.
Given $\vec{v} \cdot \hat{j} = -7$,we have $7\lambda = -7$,so $\lambda = -1$.
Thus,$\vec{v} = -3\hat{i} - 7\hat{j} - 3\hat{k}$.
Therefore,$\vec{v} \cdot \hat{i} = -3$.

(D) 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 $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$,we have $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b} - \frac{1}{2}\vec{c}$.
Comparing the coefficients of $\vec{b}$ and $\vec{c}$,we get $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = \frac{1}{2}$.
Given $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$,so $\vec{b} \cdot \vec{d} = \frac{1}{2}$.
Now,consider the scalar triple product $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \vec{a} \cdot (\vec{b} \times (\vec{c} \times \vec{d}))$.
Using the vector triple product formula $\vec{b} \times (\vec{c} \times \vec{d}) = (\vec{b} \cdot \vec{d})\vec{c} - (\vec{b} \cdot \vec{c})\vec{d}$.
Since $\vec{b} \cdot \vec{c} = 0$,this simplifies to $(\vec{b} \cdot \vec{d})\vec{c}$.
Thus,$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \vec{a} \cdot ((\vec{b} \cdot \vec{d})\vec{c}) = (\vec{b} \cdot \vec{d})(\vec{a} \cdot \vec{c})$.
Substituting the values,we get $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.