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

(D) The total force $\vec{F}$ is the sum of the individual forces:
$\vec{F} = (4\hat{i} + \hat{j} - 3\hat{k}) + (3\hat{i} + \hat{j} - \hat{k}) = 7\hat{i} + 2\hat{j} - 4\hat{k}$.
The displacement vector $\vec{d}$ is the difference between the final position vector and the initial position vector:
$\vec{d} = (5\hat{i} + 4\hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 4\hat{i} + 2\hat{j} - 2\hat{k}$.
The work done $W$ is the dot product of the total force and the displacement:
$W = \vec{F} \cdot \vec{d} = (7\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (4\hat{i} + 2\hat{j} - 2\hat{k})$.
$W = (7 \times 4) + (2 \times 2) + (-4 \times -2) = 28 + 4 + 8 = 40$ units.

(A) Let the position vectors of the points be $\vec{A} = \vec{l} - 2\vec{m} + 3\vec{n}$,$\vec{B} = 2\vec{l} + \lambda\vec{m} - 4\vec{n}$,and $\vec{C} = -7\vec{m} + 10\vec{n}$.
For the points to be collinear,the vectors $\vec{AB}$ and $\vec{AC}$ must be parallel,meaning $\vec{AB} = k\vec{AC}$ for some scalar $k$.
$\vec{AB} = \vec{B} - \vec{A} = (2\vec{l} + \lambda\vec{m} - 4\vec{n}) - (\vec{l} - 2\vec{m} + 3\vec{n}) = \vec{l} + (\lambda + 2)\vec{m} - 7\vec{n}$.
$\vec{AC} = \vec{C} - \vec{A} = (-7\vec{m} + 10\vec{n}) - (\vec{l} - 2\vec{m} + 3\vec{n}) = -\vec{l} - 5\vec{m} + 7\vec{n}$.
Since $\vec{AB} = k\vec{AC}$,we have $\vec{l} + (\lambda + 2)\vec{m} - 7\vec{n} = k(-\vec{l} - 5\vec{m} + 7\vec{n})$.
Comparing the coefficients of $\vec{l}, \vec{m}, \vec{n}$:
For $\vec{l}$: $1 = -k \implies k = -1$.
For $\vec{n}$: $-7 = 7k \implies k = -1$.
For $\vec{m}$: $\lambda + 2 = -5k$.
Substituting $k = -1$ into the equation for $\vec{m}$:
$\lambda + 2 = -5(-1) = 5$.
$\lambda = 5 - 2 = 3$.

(B) Given that $(\bar{a} + 2\bar{b})$ and $(5\bar{a} - 4\bar{b})$ are perpendicular,their dot product is zero:
$(\bar{a} + 2\bar{b}) \cdot (5\bar{a} - 4\bar{b}) = 0$
Expanding the dot product:
$5|\bar{a}|^2 - 4(\bar{a} \cdot \bar{b}) + 10(\bar{a} \cdot \bar{b}) - 8|\bar{b}|^2 = 0$
Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$:
$5(1)^2 + 6(\bar{a} \cdot \bar{b}) - 8(1)^2 = 0$
$5 + 6(\bar{a} \cdot \bar{b}) - 8 = 0$
$6(\bar{a} \cdot \bar{b}) = 3$
$\bar{a} \cdot \bar{b} = \frac{3}{6} = \frac{1}{2}$
Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$. Then $\cos \theta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \frac{1/2}{1 \times 1} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^o$.

(B) Let the given vector be $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.
The line is given by $\vec{r} = (3\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$.
The direction vector of the line is $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The projection of vector $\vec{a}$ on the line is the projection of $\vec{a}$ onto its direction vector $\vec{b}$,which is given by the formula $\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(1) + (1)(2) + (1)(3) = 1 + 2 + 3 = 6$.
Next,calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
Therefore,the projection is $\frac{6}{\sqrt{14}}$.

(C) The work done $W$ is defined as the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{r}$.
Mathematically,$W = \vec{F} \cdot \vec{r}$.
The dot product of two vectors always results in a scalar quantity.
Therefore,work done is a scalar quantity.
Since Assertion $(A)$ is true and Reason $(R)$ states that work done is not a scalar (which is false),the correct option is $C$.

(D) The resultant force $\vec{F}$ is given by $\vec{F} = \vec{F_1} + \vec{F_2} = (2\hat{i} - 5\hat{j} + 6\hat{k}) + (-\hat{i} + 2\hat{j} - \hat{k}) = \hat{i} - 3\hat{j} + 5\hat{k}$.
The displacement vector $\vec{d}$ is given by $\vec{d} = \vec{PQ} = \vec{OQ} - \vec{OP} = (6\hat{i} + \hat{j} + 3\hat{k}) - (4\hat{i} - 3\hat{j} + 2\hat{k}) = 2\hat{i} + 4\hat{j} + \hat{k}$.
The work done $W$ is the dot product of the resultant force and the displacement vector: $W = \vec{F} \cdot \vec{d} = (\hat{i} - 3\hat{j} + 5\hat{k}) \cdot (2\hat{i} + 4\hat{j} + \hat{k})$.
Calculating the dot product: $W = (1)(2) + (-3)(4) + (5)(1) = 2 - 12 + 5 = -5$ units.

(B) Two vectors are perpendicular if their dot product is $0$.
Given that $(a + \lambda b)$ is perpendicular to $(a - \lambda b)$,we have:
$(a + \lambda b) \cdot (a - \lambda b) = 0$
Using the property of the dot product $(u + v) \cdot (u - v) = |u|^2 - |v|^2$,we get:
$|a|^2 - |\lambda b|^2 = 0$
$|a|^2 - \lambda^2 |b|^2 = 0$
Substitute the given values $|a| = 3$ and $|b| = 4$:
$3^2 - \lambda^2 (4^2) = 0$
$9 - 16\lambda^2 = 0$
$16\lambda^2 = 9$
$\lambda^2 = \frac{9}{16}$
$\lambda = \pm \frac{3}{4}$

(C) The work done $W$ is defined as the scalar (dot) product of the force vector $\vec{F}$ and the displacement vector $\vec{d}$.
$W = \vec{F} \cdot \vec{d}$
$W = (2\hat{i} - \hat{j} - \hat{k}) \cdot (3\hat{i} + 2\hat{j} - 5\hat{k})$
Using the property of dot product $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and others as $0$:
$W = (2 \times 3) + (-1 \times 2) + (-1 \times -5)$
$W = 6 - 2 + 5$
$W = 9$ units.

(NONE) The resultant force $\vec{F}$ is the sum of the two force vectors.
First,find the unit vectors in the given directions:
$\hat{u}_1 = \frac{-\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{(-1)^2 + (-2)^2 + 2^2}} = \frac{-\hat{i} - 2\hat{j} + 2\hat{k}}{3}$
$\hat{u}_2 = \frac{2\hat{i} - 3\hat{j} - 6\hat{k}}{\sqrt{2^2 + (-3)^2 + (-6)^2}} = \frac{2\hat{i} - 3\hat{j} - 6\hat{k}}{7}$
Now,calculate the force vectors:
$\vec{F}_1 = 6 \hat{u}_1 = 2(-\hat{i} - 2\hat{j} + 2\hat{k}) = -2\hat{i} - 4\hat{j} + 4\hat{k}$
$\vec{F}_2 = 7 \hat{u}_2 = 1(2\hat{i} - 3\hat{j} - 6\hat{k}) = 2\hat{i} - 3\hat{j} - 6\hat{k}$
Resultant force $\vec{F} = \vec{F}_1 + \vec{F}_2 = (-2+2)\hat{i} + (-4-3)\hat{j} + (4-6)\hat{k} = 0\hat{i} - 7\hat{j} - 2\hat{k}$
Displacement vector $\vec{d} = \vec{PQ} = (5-2)\hat{i} + (-1 - (-1))\hat{j} + (1 - (-3))\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k}$
Work done $W = \vec{F} \cdot \vec{d} = (0\hat{i} - 7\hat{j} - 2\hat{k}) \cdot (3\hat{i} + 0\hat{j} + 4\hat{k}) = (0)(3) + (-7)(0) + (-2)(4) = -8$ units.

(D) Given that $|\vec{a} + \vec{b}| = \sqrt{3}$.
Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = 3$.
$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 3$.
Since $|\vec{a}| = 1$ and $|\vec{b}| = 1$,we have $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 3$.
$2(\vec{a} \cdot \vec{b}) = 1$,so $\vec{a} \cdot \vec{b} = 1/2$ $(1)$.
Now,consider the given equation $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} \times \vec{b})$.
Taking the dot product with $\vec{b}$ on both sides:
$(\vec{c} - \vec{a} - 2\vec{b}) \cdot \vec{b} = 3(\vec{a} \times \vec{b}) \cdot \vec{b}$.
$\vec{c} \cdot \vec{b} - \vec{a} \cdot \vec{b} - 2(\vec{b} \cdot \vec{b}) = 3[\vec{a} \vec{b} \vec{b}]$.
Since the scalar triple product $[\vec{a} \vec{b} \vec{b}] = 0$ (as two vectors are identical),we have:
$\vec{c} \cdot \vec{b} - 1/2 - 2(1) = 0$.
$\vec{c} \cdot \vec{b} = 1/2 + 2 = 5/2$.

(D) Given vectors are $a = i + 2j - 3k$ and $b = 3i - j + 2k$.
First,calculate the sum and difference of the vectors:
$a + b = (1+3)i + (2-1)j + (-3+2)k = 4i + j - k$
$a - b = (1-3)i + (2-(-1))j + (-3-2)k = -2i + 3j - 5k$
Let $u = a + b$ and $v = a - b$. The angle $\theta$ between $u$ and $v$ is given by $\cos \theta = \frac{u \cdot v}{|u| |v|}$.
Calculate the dot product $u \cdot v = (4)(-2) + (1)(3) + (-1)(-5) = -8 + 3 + 5 = 0$.
Since the dot product is $0$,the vectors $u$ and $v$ are perpendicular.
Therefore,$\theta = 90^o$.

(A) Two vectors are perpendicular if and only if their dot product is zero.
Given $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$.
$\vec{a} \cdot \vec{b} = 0$
$(1)(1) + (-2x)(3x) + (-3y)(2y) = 0$
$1 - 6x^2 - 6y^2 = 0$
$6x^2 + 6y^2 = 1$
$x^2 + y^2 = \frac{1}{6}$
This is the equation of a circle with center at the origin $(0, 0)$ and radius $\frac{1}{\sqrt{6}}$.
Therefore,the locus of the point $(x, y)$ is a circle.

(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
The angle between them is $2\theta$.
We know that $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
$|\vec{a} - \vec{b}|^2 = 1^2 + 1^2 - 2|\vec{a}||\vec{b}| \cos(2\theta) = 2 - 2 \cos(2\theta)$.
Using the trigonometric identity $1 - \cos(2\theta) = 2 \sin^2(\theta)$,we get:
$|\vec{a} - \vec{b}|^2 = 2(1 - \cos(2\theta)) = 2(2 \sin^2(\theta)) = 4 \sin^2(\theta)$.
Taking the square root on both sides,$|\vec{a} - \vec{b}| = 2|\sin(\theta)|$.
Given the condition $|\vec{a} - \vec{b}| < 1$,we have $2|\sin(\theta)| < 1$,which implies $|\sin(\theta)| < 1/2$.
Since $0 \le \theta \le \pi$,$\sin(\theta)$ is non-negative,so $\sin(\theta) < 1/2$.
This inequality holds for $\theta \in [0, \pi/6) \cup (5\pi/6, \pi]$.

(C) Let $\theta$ be the angle between the unit vectors $\vec{a}$ and $\vec{b}$. Since they are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.
We know that $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 1 + 1 + 2\cos\theta = 2(1 + \cos\theta) = 4\cos^2(\frac{\theta}{2})$.
Thus,$|\vec{a} + \vec{b}| = 2\cos(\frac{\theta}{2})$ (assuming $\cos(\frac{\theta}{2}) \ge 0$ for $0 \le \theta \le \pi$).
Similarly,$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 1 + 1 - 2\cos\theta = 2(1 - \cos\theta) = 4\sin^2(\frac{\theta}{2})$.
Thus,$|\vec{a} - \vec{b}| = 2\sin(\frac{\theta}{2})$.
Let $f(\theta) = |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}| = 2(\cos(\frac{\theta}{2}) + \sin(\frac{\theta}{2}))$.
Using the identity $\cos x + \sin x = \sqrt{2}\sin(x + \frac{\pi}{4})$,we get $f(\theta) = 2\sqrt{2}\sin(\frac{\theta}{2} + \frac{\pi}{4})$.
The maximum value of $\sin(\frac{\theta}{2} + \frac{\pi}{4})$ is $1$ (when $\frac{\theta}{2} + \frac{\pi}{4} = \frac{\pi}{2}$,i.e.,$\theta = \frac{\pi}{2}$).
Therefore,the maximum value is $2\sqrt{2} \times 1 = 2\sqrt{2}$.

(A) Given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
We need the angle between $\vec{b}$ and $\vec{c}$,so we write $\vec{b} + \vec{c} = -\vec{a}$.
Squaring both sides,we get $|\vec{b} + \vec{c}|^2 = |-\vec{a}|^2$.
$|\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c}) = |\vec{a}|^2$.
Substituting the given magnitudes: $5^2 + 3^2 + 2(\vec{b} \cdot \vec{c}) = 7^2$.
$25 + 9 + 2(\vec{b} \cdot \vec{c}) = 49$.
$34 + 2(\vec{b} \cdot \vec{c}) = 49$.
$2(\vec{b} \cdot \vec{c}) = 15 \implies \vec{b} \cdot \vec{c} = \frac{15}{2}$.
Let $\theta$ be the angle between $\vec{b}$ and $\vec{c}$.
Then $\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} = \frac{15/2}{5 \times 3} = \frac{15/2}{15} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,$\theta = 60^o$.

(C) Let the sides of the parallelogram be $\vec{u} = 5\vec{a} + 2\vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$.
The diagonals of the parallelogram are given by $\vec{d_1} = \vec{u} + \vec{v}$ and $\vec{d_2} = \vec{u} - \vec{v}$.
$\vec{d_1} = (5\vec{a} + 2\vec{b}) + (\vec{a} - 3\vec{b}) = 6\vec{a} - \vec{b}$.
$\vec{d_2} = (5\vec{a} + 2\vec{b}) - (\vec{a} - 3\vec{b}) = 4\vec{a} + 5\vec{b}$.
Given $|\vec{a}| = 2\sqrt{2}$,$|\vec{b}| = 3$,and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{4}) = (2\sqrt{2})(3)(\frac{1}{\sqrt{2}}) = 6$.
$|\vec{d_1}|^2 = |6\vec{a} - \vec{b}|^2 = 36|\vec{a}|^2 + |\vec{b}|^2 - 12(\vec{a} \cdot \vec{b}) = 36(8) + 9 - 12(6) = 288 + 9 - 72 = 225$.
$|\vec{d_1}| = \sqrt{225} = 15$.
$|\vec{d_2}|^2 = |4\vec{a} + 5\vec{b}|^2 = 16|\vec{a}|^2 + 25|\vec{b}|^2 + 40(\vec{a} \cdot \vec{b}) = 16(8) + 25(9) + 40(6) = 128 + 225 + 240 = 593$.
$|\vec{d_2}| = \sqrt{593}$.
Since $\sqrt{593} > 15$,the length of the longer diagonal is $\sqrt{593}$.

(C) The dot product of two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is given by $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$.
Given $\vec{a} = 2\hat{i} + 2\hat{j} - 1\hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$.
Substituting the values,we get $\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2)$.
$\vec{a} \cdot \vec{b} = 12 - 6 - 2$.
$\vec{a} \cdot \vec{b} = 4$.

(C) Given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
Taking the dot product of the sum with itself: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{0} \cdot \vec{0}$.
This expands to $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
Substituting the given magnitudes $|\vec{a}| = 1, |\vec{b}| = 2, |\vec{c}| = 3$:
$1^2 + 2^2 + 3^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
$1 + 4 + 9 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
$14 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -14$.
Therefore,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -7$.

(D) Given vectors are $\vec{u} = \hat{i} + \hat{k}$ and $\vec{v} = \hat{i} - \hat{j} + a\hat{k}$.
The angle $\theta$ between two vectors is given by $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.
Here,$\theta = \pi/3$,so $\cos(\pi/3) = 1/2$.
The dot product $\vec{u} \cdot \vec{v} = (1)(1) + (0)(-1) + (1)(a) = 1 + a$.
The magnitudes are $|\vec{u}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$ and $|\vec{v}| = \sqrt{1^2 + (-1)^2 + a^2} = \sqrt{2 + a^2}$.
Substituting these into the formula: $\frac{1}{2} = \frac{1 + a}{\sqrt{2} \sqrt{2 + a^2}}$.
Squaring both sides: $\frac{1}{4} = \frac{(1 + a)^2}{2(2 + a^2)} \implies \frac{1}{2} = \frac{1 + 2a + a^2}{2 + a^2}$.
Cross-multiplying: $2 + a^2 = 2(1 + 2a + a^2) = 2 + 4a + 2a^2$.
Rearranging terms: $a^2 + 4a = 0 \implies a(a + 4) = 0$.
Thus,$a = 0$ or $a = -4$. Given the options,the correct value is $a = 0$.

(C) In a parallelogram $ABCD$,the diagonals $AC$ and $DB$ bisect each other at the same point $M$.
Therefore,$M$ is the midpoint of $AC$ as well.
The position vector of $M$ is given by $\vec{OM} = \frac{\vec{OA} + \vec{OC}}{2}$.
Given $\vec{OA} = 3\hat{i} + 3\hat{j} + 5\hat{k}$ and $\vec{OC} = \hat{i} - 5\hat{j} - 5\hat{k}$.
$\vec{OM} = \frac{(3+1)\hat{i} + (3-5)\hat{j} + (5-5)\hat{k}}{2} = \frac{4\hat{i} - 2\hat{j} + 0\hat{k}}{2} = 2\hat{i} - \hat{j}$.
The projection of $\vec{OM}$ on $\vec{OC}$ is given by $\frac{\vec{OM} \cdot \vec{OC}}{|\vec{OC}|}$.
$\vec{OM} \cdot \vec{OC} = (2)(1) + (-1)(-5) + (0)(-5) = 2 + 5 + 0 = 7$.
$|\vec{OC}| = \sqrt{1^2 + (-5)^2 + (-5)^2} = \sqrt{1 + 25 + 25} = \sqrt{51}$.
Thus,the projection is $\frac{7}{\sqrt{51}}$.

(B) The vector component of $a$ along the direction of $b$ is given by the formula: $\text{Component} = \frac{(a \cdot b)b}{|b|^2}$.
First,calculate the dot product $a \cdot b$:
$a \cdot b = (4i + 6j) \cdot (3j + 4k) = (4 \times 0) + (6 \times 3) + (0 \times 4) = 18$.
Next,calculate the magnitude squared of $b$:
$|b|^2 = 3^2 + 4^2 = 9 + 16 = 25$.
Substitute these values into the formula:
$\text{Component} = \frac{18}{25}(3j + 4k)$.

(A) Let the position vectors of the vertices be $\vec{A} = 2\hat{i} + 4\hat{j} - \hat{k}$,$\vec{B} = 4\hat{i} + 5\hat{j} + \hat{k}$,and $\vec{C} = 3\hat{i} + 6\hat{j} - 3\hat{k}$.
First,find the vectors representing the sides:
$\vec{AB} = \vec{B} - \vec{A} = (4-2)\hat{i} + (5-4)\hat{j} + (1-(-1))\hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$.
$\vec{BC} = \vec{C} - \vec{B} = (3-4)\hat{i} + (6-5)\hat{j} + (-3-1)\hat{k} = -\hat{i} + \hat{j} - 4\hat{k}$.
$\vec{AC} = \vec{C} - \vec{A} = (3-2)\hat{i} + (6-4)\hat{j} + (-3-(-1))\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$.
Now,check the dot products to identify the right angle:
$\vec{AB} \cdot \vec{BC} = (2)(-1) + (1)(1) + (2)(-4) = -2 + 1 - 8 = -9 \neq 0$.
$\vec{BA} \cdot \vec{BC} = -\vec{AB} \cdot \vec{BC} = 9 \neq 0$.
$\vec{AB} \cdot \vec{AC} = (2)(1) + (1)(2) + (2)(-2) = 2 + 2 - 4 = 0$.
Since $\vec{AB} \cdot \vec{AC} = 0$,the vectors $\vec{AB}$ and $\vec{AC}$ are perpendicular.
Therefore,$\angle A = 90^{\circ}$.

(B) The distance $d$ of a point $P$ with position vector $\vec{p}$ from a line passing through point $A$ with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is given by the formula: $d = \frac{|(\vec{p} - \vec{a}) \times \vec{b}|}{|\vec{b}|}$.
Here,$\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}$,$\vec{p} = -\hat{i} + 2\hat{j} + 6\hat{k}$,and $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.
First,calculate $\vec{p} - \vec{a} = (-1 - 2)\hat{i} + (2 - 3)\hat{j} + (6 - (-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$.
Next,calculate the cross product $(\vec{p} - \vec{a}) \times \vec{b}$:
$(\vec{p} - \vec{a}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -1 & 10 \\ 6 & 3 & -4 \end{vmatrix} = \hat{i}(4 - 30) - \hat{j}(12 - 60) + \hat{k}(-9 - (-6)) = -26\hat{i} + 48\hat{j} - 3\hat{k}$.
The magnitude $|(\vec{p} - \vec{a}) \times \vec{b}| = \sqrt{(-26)^2 + 48^2 + (-3)^2} = \sqrt{676 + 2304 + 9} = \sqrt{2989} = \sqrt{49 \times 61} = 7\sqrt{61}$ (Wait,let's recheck the calculation).
Re-calculating: $\vec{p}-\vec{a} = -3\hat{i}-\hat{j}+10\hat{k}$. Cross product: $\hat{i}(4-30) - \hat{j}(12-60) + \hat{k}(-9+6) = -26\hat{i} + 48\hat{j} - 3\hat{k}$. Magnitude is $\sqrt{676+2304+9} = \sqrt{2989}$.
Let's re-verify the vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$. Magnitude $|\vec{b}| = \sqrt{36+9+16} = \sqrt{61}$.
Distance $d = \frac{\sqrt{2989}}{\sqrt{61}} = \sqrt{\frac{2989}{61}} = \sqrt{49} = 7$.

(D) Given points are $A = (k, 1, -1)$,$B = (2k, 0, 2)$,and $C = (2 + 2k, k, 1)$.
The vector $\vec{AB} = B - A = (2k - k, 0 - 1, 2 - (-1)) = (k, -1, 3)$.
The vector $\vec{BC} = C - B = (2 + 2k - 2k, k - 0, 1 - 2) = (2, k, -1)$.
Since $AB \perp BC$,the dot product $\vec{AB} \cdot \vec{BC} = 0$.
$(k)(2) + (-1)(k) + (3)(-1) = 0$.
$2k - k - 3 = 0$.
$k - 3 = 0$.
Therefore,$k = 3$.

(A) The vector $\vec{AB} = (3-2, 5-3, -3-(-1)) = (1, 2, -2)$.
The vector $\vec{CD} = (3-1, 5-2, 7-3) = (2, 3, 4)$.
The projection of $\vec{AB}$ on $\vec{CD}$ is given by the formula $\frac{\vec{AB} \cdot \vec{CD}}{|\vec{CD}|}$.
Calculate the dot product: $\vec{AB} \cdot \vec{CD} = (1)(2) + (2)(3) + (-2)(4) = 2 + 6 - 8 = 0$.
Since the dot product is $0$,the projection of $AB$ on $CD$ is $0$.

(A) Given the position vectors of vertices $A, B, C$ are $\vec{A} = 2\hat{i}-\hat{j}+\hat{k}$,$\vec{B} = \hat{i}-3\hat{j}-5\hat{k}$,and $\vec{C} = a\hat{i}-3\hat{j}+\hat{k}$.
Since $m\angle C = 90^\circ$,the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ must be perpendicular,meaning their dot product is zero.
First,calculate $\overrightarrow{CA} = \vec{A} - \vec{C} = (2-a)\hat{i} + 2\hat{j} + 0\hat{k}$.
Next,calculate $\overrightarrow{CB} = \vec{B} - \vec{C} = (1-a)\hat{i} + 0\hat{j} - 6\hat{k}$.
Now,set the dot product $\overrightarrow{CA} \cdot \overrightarrow{CB} = 0$:
$((2-a)\hat{i} + 2\hat{j}) \cdot ((1-a)\hat{i} - 6\hat{k}) = 0$.
$(2-a)(1-a) + (2)(0) + (0)(-6) = 0$.
$(2-a)(1-a) = 0$.
This gives $a = 2$ or $a = 1$.

(D) Since $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar,their scalar triple product is zero:
$[\vec{a} \vec{b} \vec{c}] = 0$
$\begin{vmatrix} \alpha & 2 & \beta \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = 0$
$\alpha(1-0) - 2(1-0) + \beta(1-0) = 0 \Rightarrow \alpha - 2 + \beta = 0 \Rightarrow \alpha + \beta = 2 \dots (i)$
Since $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$,$\vec{a}$ must be proportional to the sum of the unit vectors along $\vec{b}$ and $\vec{c}$:
$\hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{(1, 1, 0)}{\sqrt{2}}, \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{(0, 1, 1)}{\sqrt{2}}$
$\vec{a} = k(\hat{b} + \hat{c}) = \frac{k}{\sqrt{2}}(1, 2, 1)$
Comparing this with $\vec{a} = (\alpha, 2, \beta)$,we get $\frac{k}{\sqrt{2}} = 1$,so $k = \sqrt{2}$.
Thus,$\alpha = 1$ and $\beta = 1$.

(D) Given vectors are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$.
Since the vectors are mutually orthogonal,their dot products are zero: $\vec{a} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{c} = 0$.
For $\vec{a} \cdot \vec{c} = 0$: $(\hat{i} - \hat{j} + 2\hat{k}) \cdot (\alpha\hat{i} + \hat{j} + \beta\hat{k}) = \alpha - 1 + 2\beta = 0 \implies \alpha + 2\beta = 1$ (Equation $1$).
For $\vec{b} \cdot \vec{c} = 0$: $(2\hat{i} + 4\hat{j} + \hat{k}) \cdot (\alpha\hat{i} + \hat{j} + \beta\hat{k}) = 2\alpha + 4 + \beta = 0 \implies 2\alpha + \beta = -4$ (Equation $2$).
From Equation $2$,$\beta = -4 - 2\alpha$. Substituting this into Equation $1$:
$\alpha + 2(-4 - 2\alpha) = 1 \implies \alpha - 8 - 4\alpha = 1 \implies -3\alpha = 9 \implies \alpha = -3$.
Substituting $\alpha = -3$ into $\beta = -4 - 2\alpha$: $\beta = -4 - 2(-3) = -4 + 6 = 2$.
Thus,$(\alpha, \beta) = (-3, 2)$.

(C) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,we have $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
Since $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular,their dot product is zero: $\vec{c} \cdot \vec{d} = 0$.
$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$.
Expanding the dot product: $5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$.
Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$,and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$:
$5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0$.
$6(\vec{a} \cdot \vec{b}) - 3 = 0$.
$6(\vec{a} \cdot \vec{b}) = 3$.
$\vec{a} \cdot \vec{b} = \frac{3}{6} = \frac{1}{2}$.
We know $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$1 \cdot 1 \cdot \cos \theta = \frac{1}{2}$.
$\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$.

(B) Let $E$ be the foot of the perpendicular from vertex $B$ to the side $AD$. The vector $\vec{AE}$ is the projection of $\vec{AB}$ onto $\vec{AD}$.
Thus,$\vec{AE} = \text{proj}_{\vec{p}} \vec{q} = \left( \frac{\vec{q} \cdot \vec{p}}{\vec{p} \cdot \vec{p}} \right) \vec{p}$.
In $\triangle ABE$,by the triangle law of vector addition,we have $\vec{AB} + \vec{BE} = \vec{AE}$.
Given $\vec{AB} = \vec{q}$ and $\vec{BE} = \vec{r}$,we have $\vec{q} + \vec{r} = \vec{AE}$.
Therefore,$\vec{r} = \vec{AE} - \vec{q}$.
Substituting the expression for $\vec{AE}$,we get $\vec{r} = \left( \frac{\vec{q} \cdot \vec{p}}{\vec{p} \cdot \vec{p}} \right) \vec{p} - \vec{q}$.

(D) Since $\vec{u}$ is coplanar with $\vec{a}$ and $\vec{b}$,we can write $\vec{u} = x\vec{a} + y\vec{b}$.
Given $\vec{u} \perp \vec{a}$,so $\vec{u} \cdot \vec{a} = 0$.
$(x\vec{a} + y\vec{b}) \cdot \vec{a} = 0 \implies x|\vec{a}|^2 + y(\vec{a} \cdot \vec{b}) = 0$.
Calculate $|\vec{a}|^2 = 2^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14$.
Calculate $\vec{a} \cdot \vec{b} = (2)(0) + (3)(1) + (-1)(1) = 3 - 1 = 2$.
So,$14x + 2y = 0 \implies y = -7x$.
Thus,$\vec{u} = x\vec{a} - 7x\vec{b} = x(\vec{a} - 7\vec{b})$.
$vec{a} - 7\vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) - 7(\hat{j} + \hat{k}) = 2\hat{i} - 4\hat{j} - 8\hat{k}$.
Given $\vec{u} \cdot \vec{b} = 24$,so $x(\vec{a} - 7\vec{b}) \cdot \vec{b} = 24$.
$(\vec{a} - 7\vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} - 7|\vec{b}|^2 = 2 - 7(1^2 + 1^2) = 2 - 14 = -12$.
So,$x(-12) = 24 \implies x = -2$.
Therefore,$\vec{u} = -2(2\hat{i} - 4\hat{j} - 8\hat{k}) = -4\hat{i} + 8\hat{j} + 16\hat{k}$.
$|\vec{u}|^2 = (-4)^2 + 8^2 + 16^2 = 16 + 64 + 256 = 336$.

(C) Given that $|a| = 3, |b| = 4, |c| = 5$.
Since $a$ and $(b + c)$ are perpendicular,$a \cdot (b + c) = 0 \implies a \cdot b + a \cdot c = 0$ .....$(i)$
Since $b$ and $(c + a)$ are perpendicular,$b \cdot (c + a) = 0 \implies b \cdot c + b \cdot a = 0$ .....$(ii)$
Since $c$ and $(a + b)$ are perpendicular,$c \cdot (a + b) = 0 \implies c \cdot a + c \cdot b = 0$ .....$(iii)$
Adding equations $(i), (ii),$ and $(iii)$,we get:
$2(a \cdot b + b \cdot c + c \cdot a) = 0 \implies a \cdot b + b \cdot c + c \cdot a = 0$.
Now,the modulus of $a + b + c$ is given by:
$|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$
$|a + b + c|^2 = 3^2 + 4^2 + 5^2 + 2(0)$
$|a + b + c|^2 = 9 + 16 + 25 = 50$
$|a + b + c| = \sqrt{50} = 5\sqrt{2}$.

(C) Let the two vectors forming the sides of the parallelogram be $u = 5a + 2b$ and $v = a - 3b$.
The diagonals of the parallelogram are given by $d_1 = u + v$ and $d_2 = u - v$.
$d_1 = (5a + 2b) + (a - 3b) = 6a - b$.
$d_2 = (5a + 2b) - (a - 3b) = 4a + 5b$.
We calculate the squared magnitudes:
$|d_1|^2 = |6a - b|^2 = 36|a|^2 + |b|^2 - 12(a \cdot b)$.
Given $|a| = 2\sqrt{2}$,$|b| = 3$,and $\theta = \frac{\pi}{4}$,$a \cdot b = |a||b|\cos(\frac{\pi}{4}) = (2\sqrt{2})(3)(\frac{1}{\sqrt{2}}) = 6$.
$|d_1|^2 = 36(8) + 9 - 12(6) = 288 + 9 - 72 = 225$,so $|d_1| = 15$.
$|d_2|^2 = |4a + 5b|^2 = 16|a|^2 + 25|b|^2 + 40(a \cdot b)$.
$|d_2|^2 = 16(8) + 25(9) + 40(6) = 128 + 225 + 240 = 593$.
$|d_2| = \sqrt{593}$.
Comparing the two,$\sqrt{593} > 15$,so the longer diagonal is $\sqrt{593}$.

(D) Given that $|a| = |b| = |c| = 2$ and the angle between any two vectors is $60^\circ$.
Thus,$a \cdot b = b \cdot c = c \cdot a = |a||b| \cos 60^\circ = 2 \times 2 \times \frac{1}{2} = 2$.
Now,consider the dot product:
$(2a + 3b - 5c) \cdot (4a - 6b + 10c) = 8(a \cdot a) - 12(a \cdot b) + 20(a \cdot c) + 12(b \cdot a) - 18(b \cdot b) + 30(b \cdot c) - 20(c \cdot a) + 30(c \cdot b) - 50(c \cdot c)$.
Since $a \cdot b = b \cdot a$,$a \cdot c = c \cdot a$,and $b \cdot c = c \cdot b$,the expression simplifies to:
$8|a|^2 - 18|b|^2 - 50|c|^2 + (-12 + 12)(a \cdot b) + (20 - 20)(a \cdot c) + (30 + 30)(b \cdot c)$.
$= 8(2)^2 - 18(2)^2 - 50(2)^2 + 60(b \cdot c)$.
$= 8(4) - 18(4) - 50(4) + 60(2)$.
$= 32 - 72 - 200 + 120$.
$= -120$.

(A) Let $c = (c_1, c_2, c_3).$ Since $|a| = |b| = |c|,$ we have $|c|^2 = |a|^2 = |b|^2 = 1^2 + 1^2 = 2.$ Thus,$c_1^2 + c_2^2 + c_3^2 = 2.$
Given that the vectors are taken pairwise at equal angles $\varphi,$ we have $\cos \varphi = \frac{a \cdot b}{|a||b|} = \frac{(i+j) \cdot (j+k)}{2} = \frac{1}{2}.$
Since the angle between $a$ and $c$ is $\varphi,$ we have $\frac{a \cdot c}{|a||c|} = \frac{1}{2} \implies \frac{c_1 + c_2}{2} = \frac{1}{2} \implies c_1 + c_2 = 1.$
Similarly,for $b$ and $c,$ $\frac{b \cdot c}{|b||c|} = \frac{1}{2} \implies \frac{c_2 + c_3}{2} = \frac{1}{2} \implies c_2 + c_3 = 1.$
From these,$c_1 = 1 - c_2$ and $c_3 = 1 - c_2.$
Substituting into $c_1^2 + c_2^2 + c_3^2 = 2,$ we get $(1 - c_2)^2 + c_2^2 + (1 - c_2)^2 = 2.$
$1 - 2c_2 + c_2^2 + c_2^2 + 1 - 2c_2 + c_2^2 = 2 \implies 3c_2^2 - 4c_2 = 0.$
This gives $c_2(3c_2 - 4) = 0,$ so $c_2 = 0$ or $c_2 = \frac{4}{3}.$
If $c_2 = 0,$ then $c_1 = 1$ and $c_3 = 1,$ giving $c = (1, 0, 1).$
If $c_2 = \frac{4}{3},$ then $c_1 = -\frac{1}{3}$ and $c_3 = -\frac{1}{3},$ giving $c = (-\frac{1}{3}, \frac{4}{3}, -\frac{1}{3}).$
Since $(1, 0, 1)$ is an option,the correct answer is $A$.

(A) The position vectors of $R$ and $S$ are $\frac{3p + 2q}{5}$ and $3p - 2q$ respectively.
Therefore,$\overrightarrow{OR} = \frac{3p + 2q}{5}$ and $\overrightarrow{OS} = 3p - 2q.$
Since $\overrightarrow{OR} \perp \overrightarrow{OS},$ their dot product is zero,so $\overrightarrow{OR} \cdot \overrightarrow{OS} = 0.$
$\left( \frac{3p + 2q}{5} \right) \cdot (3p - 2q) = 0$
$9|p|^2 - 6(p \cdot q) + 6(q \cdot p) - 4|q|^2 = 0$
$9|p|^2 - 4|q|^2 = 0$
$9|p|^2 = 4|q|^2$
Given $|p| = p$ and $|q| = q,$ we get $9p^2 = 4q^2.$

(D) Let the unit vector be $\vec{u} = x\,i + y\,j$. Since it is a unit vector,$x^2 + y^2 = 1$.
Given the angle with $\vec{a} = i + j$ is $45^{\circ}$,we have $\vec{u} \cdot \vec{a} = |\vec{u}| |\vec{a}| \cos(45^{\circ})$.
$(x\,i + y\,j) \cdot (i + j) = (1)(\sqrt{1^2 + 1^2}) \frac{1}{\sqrt{2}} \Rightarrow x + y = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$.
Given the angle with $\vec{b} = 3i - 4j$ is $60^{\circ}$,we have $\vec{u} \cdot \vec{b} = |\vec{u}| |\vec{b}| \cos(60^{\circ})$.
$(x\,i + y\,j) \cdot (3i - 4j) = (1)(\sqrt{3^2 + (-4)^2}) \frac{1}{2} \Rightarrow 3x - 4y = 5 \cdot \frac{1}{2} = 2.5$.
We have the system: $x + y = 1$ and $3x - 4y = 2.5$.
From the first equation,$y = 1 - x$. Substituting into the second: $3x - 4(1 - x) = 2.5 \Rightarrow 3x - 4 + 4x = 2.5 \Rightarrow 7x = 6.5 \Rightarrow x = \frac{6.5}{7} = \frac{13}{14}$.
Then $y = 1 - \frac{13}{14} = \frac{1}{14}$.
Check the unit vector condition: $x^2 + y^2 = (\frac{13}{14})^2 + (\frac{1}{14})^2 = \frac{169 + 1}{196} = \frac{170}{196} \neq 1$.
Since no option satisfies these conditions,the correct answer is $(d)$.

(C) Let the two vectors be $\vec{a} = cxi - 6j + 3k$ and $\vec{b} = xi + 2j + 2cxk$.
Since the vectors make an obtuse angle, their dot product must be negative, i.e., $\vec{a} \cdot \vec{b} < 0$.
Calculating the dot product: $\vec{a} \cdot \vec{b} = (cx)(x) + (-6)(2) + (3)(2cx) = cx^2 - 12 + 6cx$.
We require $cx^2 + 6cx - 12 < 0$ for all real $x$.
For a quadratic expression $Ax^2 + Bx + C < 0$ to hold for all $x$, we must have $A < 0$ and the discriminant $D = B^2 - 4AC < 0$.
Here, $A = c$, $B = 6c$, and $C = -12$.
Condition $1$: $c < 0$.
Condition $2$: $D = (6c)^2 - 4(c)(-12) < 0
\Rightarrow 36c^2 + 48c < 0
\Rightarrow 12c(3c + 4) < 0$.
Since $c < 0$, we divide by $12c$ (which reverses the inequality sign): $3c + 4 > 0
\Rightarrow 3c > -4
\Rightarrow c > -\frac{4}{3}$.
Combining $c < 0$ and $c > -\frac{4}{3}$, we get $-\frac{4}{3} < c < 0$.

(A) Given $b = 3j + 4k$ and $a = i + j$.
Since $b_1$ is parallel to $a$,we can write $b_1 = \lambda a = \lambda(i + j) = \lambda i + \lambda j$.
We know that $b = b_1 + b_2$,where $b_2$ is perpendicular to $a$.
Thus,$b_2 = b - b_1 = (0 - \lambda)i + (3 - \lambda)j + 4k = -\lambda i + (3 - \lambda)j + 4k$.
Since $b_2 \perp a$,their dot product must be zero: $b_2 \cdot a = 0$.
$(-\lambda i + (3 - \lambda)j + 4k) \cdot (i + j) = 0$.
$-\lambda + (3 - \lambda) = 0$.
$3 - 2\lambda = 0 \Rightarrow \lambda = \frac{3}{2}$.
Therefore,$b_1 = \frac{3}{2}(i + j)$.

(C) Given $|u| = 1, |v| = 2, |w| = 3$. Since $v \perp w$,we have $v \cdot w = 0$.
Let the projection of $v$ along $u$ be equal to the projection of $w$ along $u$. This implies $\frac{v \cdot u}{|u|} = \frac{w \cdot u}{|u|}$.
Since $|u| = 1$,we have $v \cdot u = w \cdot u$,which means $(v - w) \cdot u = 0$.
We want to find $|u - v + w|$. Let $X = u - (v - w)$.
Then $|X|^2 = |u - (v - w)|^2 = |u|^2 + |v - w|^2 - 2u \cdot (v - w)$.
Since $u \cdot (v - w) = 0$,we have $|X|^2 = |u|^2 + |v - w|^2$.
$|v - w|^2 = |v|^2 + |w|^2 - 2(v \cdot w) = 2^2 + 3^2 - 2(0) = 4 + 9 = 13$.
Thus,$|u - v + w|^2 = 1^2 + 13 = 14$.
Therefore,$|u - v + w| = \sqrt{14}$.

(C) The unit vector in the direction of $\vec{a} = 5\hat{i} + 3\hat{j} + 4\hat{k}$ is $\hat{a} = \frac{5\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{25 + 9 + 16}} = \frac{5\hat{i} + 3\hat{j} + 4\hat{k}}{5\sqrt{2}}$.
The unit vector in the direction of $\vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k}$ is $\hat{b} = \frac{3\hat{i} + 4\hat{j} - 5\hat{k}}{\sqrt{9 + 16 + 25}} = \frac{3\hat{i} + 4\hat{j} - 5\hat{k}}{5\sqrt{2}}$.
The forces are $\vec{F_1} = 3\hat{a} = \frac{3(5\hat{i} + 3\hat{j} + 4\hat{k})}{5\sqrt{2}}$ and $\vec{F_2} = 2\hat{b} = \frac{2(3\hat{i} + 4\hat{j} - 5\hat{k})}{5\sqrt{2}}$.
The net force $\vec{F} = \vec{F_1} + \vec{F_2} = \frac{1}{5\sqrt{2}} [(15\hat{i} + 9\hat{j} + 12\hat{k}) + (6\hat{i} + 8\hat{j} - 10\hat{k})] = \frac{21\hat{i} + 17\hat{j} + 2\hat{k}}{5\sqrt{2}}$.
The displacement vector $\vec{d} = (3 - 1)\hat{i} + (3 - (-1))\hat{j} + (1 - (-1))\hat{k} = 2\hat{i} + 4\hat{j} + 2\hat{k}$.
Work done $W = \vec{F} \cdot \vec{d} = \frac{1}{5\sqrt{2}} (21\hat{i} + 17\hat{j} + 2\hat{k}) \cdot (2\hat{i} + 4\hat{j} + 2\hat{k})$.
$W = \frac{1}{5\sqrt{2}} (21 \times 2 + 17 \times 4 + 2 \times 2) = \frac{42 + 68 + 4}{5\sqrt{2}} = \frac{114}{5\sqrt{2}}$.
Rationalizing the denominator: $W = \frac{114\sqrt{2}}{5 \times 2} = \frac{57\sqrt{2}}{5}$ units.

(D) Given $|a \times b| = 4$ and $|a \cdot b| = 2$.
We know that $|a \times b| = |a| |b| \sin \theta = 4$ and $|a \cdot b| = |a| |b| |\cos \theta| = 2$.
Squaring both equations,we get:
$|a|^2 |b|^2 \sin^2 \theta = 16$
$|a|^2 |b|^2 \cos^2 \theta = 4$
Adding these two equations:
$|a|^2 |b|^2 (\sin^2 \theta + \cos^2 \theta) = 16 + 4$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$|a|^2 |b|^2 = 20$.

(C) Let the origin be the circumcenter $O$ of the equilateral triangle $ABC.$ Let the position vectors of $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ and $P$ be $\vec{r}.$
Since $O$ is the circumcenter,$|\vec{a}| = |\vec{b}| = |\vec{c}| = R,$ where $R$ is the circumradius.
For an equilateral triangle of side $s=1,$ the circumradius $R = \frac{s}{\sqrt{3}} = \frac{1}{\sqrt{3}}.$
Since $P$ is on the circumcircle,$|\vec{r}| = R = \frac{1}{\sqrt{3}}.$
We need to calculate $S = |\vec{r}-\vec{a}|^2 + |\vec{r}-\vec{b}|^2 + |\vec{r}-\vec{c}|^2.$
Expanding this,we get:
$S = (\vec{r}-\vec{a}) \cdot (\vec{r}-\vec{a}) + (\vec{r}-\vec{b}) \cdot (\vec{r}-\vec{b}) + (\vec{r}-\vec{c}) \cdot (\vec{r}-\vec{c})$
$S = 3|\vec{r}|^2 + (|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) - 2\vec{r} \cdot (\vec{a}+\vec{b}+\vec{c}).$
For an equilateral triangle centered at the origin,$\vec{a}+\vec{b}+\vec{c} = \vec{0}.$
Also,$|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = R^2 = \frac{1}{3}$ and $|\vec{r}|^2 = R^2 = \frac{1}{3}.$
Substituting these values:
$S = 3(\frac{1}{3}) + (\frac{1}{3} + \frac{1}{3} + \frac{1}{3}) - 2\vec{r} \cdot \vec{0}$
$S = 1 + 1 - 0 = 2.$

(D) Given that $OACB$ is a parallelogram with $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$.
Since $AC$ is parallel to $OB$,the vector $\overrightarrow{AC} = \overrightarrow{OB} = \vec{b}$.
The position vector of $M$ lies on $AC$. Since $M$ is on $AC$,we can write $\overrightarrow{OM} = \vec{a} + \lambda \vec{b}$ for some scalar $\lambda$.
Then $\overrightarrow{BM} = \overrightarrow{OM} - \overrightarrow{OB} = \vec{a} + \lambda \vec{b} - \vec{b} = \vec{a} + (\lambda - 1)\vec{b}$.
Since $BM \perp AC$,we have $\overrightarrow{BM} \cdot \overrightarrow{AC} = 0$.
$(\vec{a} + (\lambda - 1)\vec{b}) \cdot \vec{b} = 0$.
$\vec{a} \cdot \vec{b} + (\lambda - 1)|\vec{b}|^2 = 0$.
Given $\vec{a} \cdot \vec{b} = 1$ and $|\vec{b}| = 2$,so $|\vec{b}|^2 = 4$.
$1 + (\lambda - 1)(4) = 0 \implies 4(\lambda - 1) = -1 \implies \lambda - 1 = -\frac{1}{4} \implies \lambda = \frac{3}{4}$.
Now,$\overrightarrow{BM} = \vec{a} - \frac{1}{4}\vec{b}$.
$|\overrightarrow{BM}|^2 = |\vec{a} - \frac{1}{4}\vec{b}|^2 = |\vec{a}|^2 + \frac{1}{16}|\vec{b}|^2 - \frac{2}{4}(\vec{a} \cdot \vec{b})$.
$|\overrightarrow{BM}|^2 = 4 + \frac{4}{16} - \frac{1}{2} = 4 + \frac{1}{4} - \frac{1}{2} = \frac{16 + 1 - 2}{4} = \frac{15}{4}$.
Therefore,$|\overrightarrow{BM}| = \frac{\sqrt{15}}{2}$.

(C) Given that $\vec a \cdot (\vec b + \vec c) = 0$,$\vec b \cdot (\vec c + \vec a) = 0$,and $\vec c \cdot (\vec a + \vec b) = 0$.
This implies:
$(i) \quad \vec a \cdot \vec b + \vec a \cdot \vec c = 0$
$(ii) \quad \vec b \cdot \vec c + \vec b \cdot \vec a = 0$
$(iii) \quad \vec c \cdot \vec a + \vec c \cdot \vec b = 0$
Adding these three equations,we get $2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a) = 0$,which means $\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a = 0$.
Now,$|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a)$.
Substituting the given magnitudes $|\vec a| = 3, |\vec b| = 4, |\vec c| = 5$:
$|\vec a + \vec b + \vec c|^2 = 3^2 + 4^2 + 5^2 + 2(0) = 9 + 16 + 25 = 50$.
Therefore,$|\vec a + \vec b + \vec c| = \sqrt{50} = 5 \sqrt{2}$.

(B) We check the orthogonality of the set $\{\vec{a}, \vec{b_1}, \vec{c_2}\}$.
First,$\vec{a} \cdot \vec{b_1} = \vec{a} \cdot \left( \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} \right) = \vec{a} \cdot \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} |\vec{a}|^2 = \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} = 0$.
Thus,$\vec{a}$ and $\vec{b_1}$ are orthogonal.
Next,$\vec{a} \cdot \vec{c_2} = \vec{a} \cdot \left( \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1} \right) = \vec{a} \cdot \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} |\vec{a}|^2 - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2} (\vec{a} \cdot \vec{b_1}) = \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{a} - 0 = 0$.
Thus,$\vec{a}$ and $\vec{c_2}$ are orthogonal.
Finally,$\vec{b_1} \cdot \vec{c_2} = \vec{b_1} \cdot \left( \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1} \right) = \vec{b_1} \cdot \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} (\vec{b_1} \cdot \vec{a}) - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2} |\vec{b_1}|^2 = \vec{b_1} \cdot \vec{c} - 0 - \vec{c} \cdot \vec{b_1} = 0$.
Thus,$\vec{b_1}$ and $\vec{c_2}$ are orthogonal.
Since all pairs are orthogonal,$\{\vec{a}, \vec{b_1}, \vec{c_2}\}$ is a set of mutually orthogonal vectors.

(C) Given that $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are unit vectors.
We have $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$.
Using the identity $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) = 1$.
Since $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are unit vectors,the maximum value of $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ is $1$,which occurs when $\vec{a} \times \vec{b} = \vec{c} \times \vec{d}$.
This implies that the vectors are coplanar in specific configurations.
Given $\vec{a} \cdot \vec{c} = \frac{1}{2}$,the angle between $\vec{a}$ and $\vec{c}$ is $60^{\circ}$.
If $\vec{b}$ and $\vec{d}$ were parallel,then $\vec{a} \times \vec{b}$ and $\vec{c} \times \vec{d}$ would not satisfy the dot product condition of $1$ unless specific constraints are met.
By analyzing the geometry,if $\vec{b}$ and $\vec{d}$ were parallel,the scalar triple products would vanish,contradicting the result.
Thus,$\vec{b}$ and $\vec{d}$ must be non-parallel.
Therefore,option $C$ is correct.

(A) Given $\vec{a} \times \vec{b} = \vec{d} \times \vec{b}$,we can write $(\vec{a} - \vec{d}) \times \vec{b} = 0$.
This implies that $\vec{a} - \vec{d}$ is parallel to $\vec{b}$,so $\vec{a} - \vec{d} = \lambda \vec{b}$ for some scalar $\lambda$.
Thus,$\vec{d} = \vec{a} - \lambda \vec{b}$.
Taking the dot product with $\vec{c}$ on both sides,we get $\vec{d} \cdot \vec{c} = \vec{a} \cdot \vec{c} - \lambda (\vec{b} \cdot \vec{c})$.
Given $\vec{d} \cdot \vec{c} = 8$,$\vec{a} \cdot \vec{c} = (2)(1) + (3)(1) + (1)(1) = 6$,and $\vec{b} \cdot \vec{c} = (1)(1) + (-1)(1) + (1)(1) = 1$.
Substituting these values: $8 = 6 - \lambda(1) \Rightarrow \lambda = -2$.
Therefore,$\vec{d} = \vec{a} + 2\vec{b} = (2\hat{i} + 3\hat{j} + \hat{k}) + 2(\hat{i} - \hat{j} + \hat{k}) = 4\hat{i} + \hat{j} + 3\hat{k}$.
Finally,$\vec{d} \cdot \vec{b} = (4\hat{i} + \hat{j} + 3\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = (4)(1) + (1)(-1) + (3)(1) = 4 - 1 + 3 = 6$.

(A) Given that $|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and the area of the parallelogram is $|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin \theta = 4 \sin \theta$.
The dot product is $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta = 4 \cos \theta$.
The given equation is $|\overrightarrow{a} \times \overrightarrow{b}| + \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{2} |\overrightarrow{a}| |\overrightarrow{b}|$.
Substituting the values,we get $4 \sin \theta + 4 \cos \theta = \sqrt{2} (2)(2) = 4\sqrt{2}$.
Dividing by $4$,we get $\sin \theta + \cos \theta = \sqrt{2}$.
Multiplying by $\frac{1}{\sqrt{2}}$,we get $\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = 1$,which is $\sin(\theta + \frac{\pi}{4}) = 1$.
Thus,$\theta + \frac{\pi}{4} = \frac{\pi}{2}$,which implies $\theta = \frac{\pi}{4}$.
The area of the parallelogram is $|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin \theta = (2)(2) \sin(\frac{\pi}{4}) = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2}$ square units.

(C) Given that $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors,we have $|\vec{a}| = 1$,$|\vec{b}| = 1$,and $\vec{a} \cdot \vec{b} = 0$.
Given $\vec{c} = \vec{a} + \vec{b}$.
We know the scalar triple product identity $(\vec{u} \times \vec{v}) \cdot (\vec{w} \times \vec{z}) = (\vec{u} \cdot \vec{w})(\vec{v} \cdot \vec{z}) - (\vec{u} \cdot \vec{z})(\vec{v} \cdot \vec{w})$.
Applying this to each term:
$1$. $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b}) = 0 \cdot (\vec{b} \cdot \vec{c}) - (\vec{a} \cdot (\vec{a} + \vec{b}))(1) = -(\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b}) = -(1 + 0) = -1$.
$2$. $(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{c})(\vec{c} \cdot \vec{a}) - (\vec{b} \cdot \vec{a})(\vec{c} \cdot \vec{c}) = (\vec{b} \cdot (\vec{a} + \vec{b}))(\vec{c} \cdot \vec{a}) - 0 = (\vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b})(\vec{c} \cdot \vec{a}) = (0 + 1)(\vec{c} \cdot \vec{a}) = \vec{c} \cdot \vec{a} = (\vec{a} + \vec{b}) \cdot \vec{a} = 1 + 0 = 1$.
$3$. $(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{a})(\vec{a} \cdot \vec{b}) - (\vec{c} \cdot \vec{b})(\vec{a} \cdot \vec{a}) = (\vec{c} \cdot \vec{a})(0) - ((\vec{a} + \vec{b}) \cdot \vec{b})(1) = -(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}) = -(0 + 1) = -1$.
Summing these values: $-1 + 1 - 1 = -1$.