Basic , Modulus and Algebra of vectors Questions in English

(A) The perimeter of a triangle is the sum of the magnitudes of its sides.
Let the sides be $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\vec{b} = 4\hat{i} - 3\hat{j} - 5\hat{k}$,and $\vec{c} = 7\hat{i} + \hat{j}$.
Magnitude of side $\vec{a}$ is $|\vec{a}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.
Magnitude of side $\vec{b}$ is $|\vec{b}| = \sqrt{4^2 + (-3)^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$.
Magnitude of side $\vec{c}$ is $|\vec{c}| = \sqrt{7^2 + 1^2 + 0^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$.
Perimeter = $|\vec{a}| + |\vec{b}| + |\vec{c}| = 5\sqrt{2} + 5\sqrt{2} + 5\sqrt{2} = 15\sqrt{2}$.
Since $15\sqrt{2} = \sqrt{225 \times 2} = \sqrt{450}$,the correct option is $A$.

(D) The side of the square is given by the vector $\vec{a} = 3i + 4j + 5k.$
The magnitude of the side is $|\vec{a}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}.$
The area of a square is given by the square of its side length,which is $|\vec{a}|^2.$
Area $= (\sqrt{50})^2 = 50.$

(A) Given vector $a = 2i + 2j - k$.
First,calculate the magnitude of vector $a$:
$|a| = \sqrt{(2)^2 + (2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
We are given the equation $|xa| = 1$.
Using the property of modulus $|xa| = |x||a|$,we have:
$|x| \cdot |a| = 1$
$|x| \cdot 3 = 1$
$|x| = \frac{1}{3}$
Therefore,$x = \pm \frac{1}{3}$.

(C) vector $\vec{v} = x\,i + y\,j$ is a unit vector if its magnitude $|\vec{v}| = \sqrt{x^2 + y^2} = 1$,which implies $x^2 + y^2 = 1$.
For option $A$: $(\cos \theta)^2 + (-\sin \theta)^2 = \cos^2 \theta + \sin^2 \theta = 1$.
For option $B$: $(\sin \theta)^2 + (\cos \theta)^2 = \sin^2 \theta + \cos^2 \theta = 1$.
For option $C$: $(\sin 2\theta)^2 + (-\cos \theta)^2 = \sin^2 2\theta + \cos^2 \theta$. This is not equal to $1$ for all values of $\theta$ (e.g.,at $\theta = \frac{\pi}{4}$,$\sin^2(\frac{\pi}{2}) + \cos^2(\frac{\pi}{4}) = 1^2 + (\frac{1}{\sqrt{2}})^2 = 1 + 0.5 = 1.5 \neq 1$).
For option $D$: $(\cos 2\theta)^2 + (-\sin 2\theta)^2 = \cos^2 2\theta + \sin^2 2\theta = 1$.
Therefore,the vector in option $C$ is not a unit vector for all values of $\theta$.

(C) Let the angle between vectors $a$ and $b$ be $\theta$. The unit vectors along $a$ and $b$ are $\hat{a} = \frac{a}{|a|}$ and $\hat{b} = \frac{b}{|b|}$.
The vector that bisects the angle between $a$ and $b$ is given by the sum of their unit vectors,which is $\hat{a} + \hat{b} = \frac{a}{|a|} + \frac{b}{|b|}$.
Given that $a + b$ is the bisector,it must be parallel to $\hat{a} + \hat{b}$.
Thus,$a + b = k \left( \frac{a}{|a|} + \frac{b}{|b|} \right)$ for some scalar $k$.
Comparing the coefficients of $a$ and $b$,we get $1 = \frac{k}{|a|}$ and $1 = \frac{k}{|b|}$.
This implies $|a| = k$ and $|b| = k$,therefore $|a| = |b|$.
Hence,$a$ and $b$ are equal in magnitude.

(B) Given vectors are $a = i + 2j + 2k$ and $b = 3i + 6j + 2k.$
First,calculate the magnitude of vector $b$:
$|b| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7.$
Next,find the unit vector in the direction of $a$,denoted as $\hat{a}$:
$|a| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.$
$\hat{a} = \frac{a}{|a|} = \frac{i + 2j + 2k}{3}.$
The required vector has magnitude $|b|$ and direction $\hat{a}$,so it is $|b|\hat{a}$:
$7 \times \left( \frac{i + 2j + 2k}{3} \right) = \frac{7}{3}(i + 2j + 2k).$
Thus,the correct option is $B$.

(D) Given vectors are $p = 7i - 2j + 3k$ and $q = 3i + j + 5k$.
First,calculate $2q = 2(3i + j + 5k) = 6i + 2j + 10k$.
Now,find $p - 2q = (7i - 2j + 3k) - (6i + 2j + 10k)$.
$p - 2q = (7 - 6)i + (-2 - 2)j + (3 - 10)k = i - 4j - 7k$.
The magnitude is given by $|p - 2q| = \sqrt{(1)^2 + (-4)^2 + (-7)^2}$.
$|p - 2q| = \sqrt{1 + 16 + 49} = \sqrt{66}$.

(C) Given $a = i$ and $|b| = 1$. The angle between $a$ and $b$ is $120^\circ$.
Since $a$ is along the $x$-axis,we can represent $b$ as $b = \cos(120^\circ)i + \sin(120^\circ)j$.
Calculating the components: $b = -\frac{1}{2}i + \frac{\sqrt{3}}{2}j$.
Now,find the vector $a + b$: $a + b = i + (-\frac{1}{2}i + \frac{\sqrt{3}}{2}j) = \frac{1}{2}i + \frac{\sqrt{3}}{2}j$.
Since $|a+b| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$,the vector $\frac{1}{2}i + \frac{\sqrt{3}}{2}j$ is already a unit vector.

(C) Let the position vectors of the vertices be $\vec{A} = 6i + 4j + 5k$,$\vec{B} = 4i + 5j + 6k$,and $\vec{C} = 5i + 6j + 4k$.
To find the lengths of the sides,we calculate the magnitudes of the vectors $\vec{AB}$,$\vec{BC}$,and $\vec{CA}$.
$\vec{AB} = \vec{B} - \vec{A} = (4-6)i + (5-4)j + (6-5)k = -2i + j + k$.
The length $AB = |\vec{AB}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
$\vec{BC} = \vec{C} - \vec{B} = (5-4)i + (6-5)j + (4-6)k = i + j - 2k$.
The length $BC = |\vec{BC}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
$\vec{CA} = \vec{A} - \vec{C} = (6-5)i + (4-6)j + (5-4)k = i - 2j + k$.
The length $CA = |\vec{CA}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
Since $AB = BC = CA = \sqrt{6}$,the triangle is an equilateral triangle.

(A) Let the vertices of the triangle be $A(1, 1, 1)$,$B(5, 3, -3)$,and $C(2, 5, 9)$.
The side lengths are the magnitudes of the vectors representing the sides:
$AB = \sqrt{(5-1)^2 + (3-1)^2 + (-3-1)^2} = \sqrt{4^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
$BC = \sqrt{(2-5)^2 + (5-3)^2 + (9-(-3))^2} = \sqrt{(-3)^2 + 2^2 + 12^2} = \sqrt{9 + 4 + 144} = \sqrt{157}$.
$CA = \sqrt{(1-2)^2 + (1-5)^2 + (1-9)^2} = \sqrt{(-1)^2 + (-4)^2 + (-8)^2} = \sqrt{1 + 16 + 64} = \sqrt{81} = 9$.
The perimeter of the triangle is the sum of the side lengths:
Perimeter $= AB + BC + CA = 6 + \sqrt{157} + 9 = 15 + \sqrt{157}$.

(B) Given position vectors are $\vec{OA} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{OB} = 2\hat{i} - \hat{j} + \hat{k}$.
We know that $\overrightarrow{AB} = \vec{OB} - \vec{OA}$.
$\overrightarrow{AB} = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - \hat{k})$
$\overrightarrow{AB} = (2-1)\hat{i} + (-1-1)\hat{j} + (1-(-1))\hat{k}$
$\overrightarrow{AB} = \hat{i} - 2\hat{j} + 2\hat{k}$
Now,the magnitude $|\overrightarrow{AB}| = \sqrt{(1)^2 + (-2)^2 + (2)^2}$
$|\overrightarrow{AB}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Thus,the correct option is $B$.

(B) Since the forces $a, b$ and $c$ are mutually perpendicular,they can be represented along the $x, y$ and $z$ axes respectively.
Let the vectors be $\vec{a} = 2\hat{i}$,$\vec{b} = 10\hat{j}$,and $\vec{c} = 11\hat{k}$.
The resultant vector is $\vec{R} = \vec{a} + \vec{b} + \vec{c} = 2\hat{i} + 10\hat{j} + 11\hat{k}$.
The magnitude of the resultant is given by $|\vec{R}| = \sqrt{(2)^2 + (10)^2 + (11)^2}$.
$|\vec{R}| = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.

(A) The unit vectors $i, j, k$ represent the standard basis vectors along the $x, y,$ and $z$ axes respectively in a three-dimensional Cartesian coordinate system.
By definition,these vectors are mutually perpendicular to each other,meaning the dot product of any two distinct vectors from this set is zero (e.g.,$i \cdot j = 0, j \cdot k = 0, k \cdot i = 0$).
Vectors that are mutually perpendicular are termed as orthogonal vectors.
Therefore,the system of vectors $i, j, k$ is orthogonal.

(D) Let the given vectors be $\vec{a} = i + j + k$,$\vec{b} = -i + j + k$,$\vec{c} = i - j + k$,and $\vec{d} = i + j - k$.
The resultant vector $\vec{R} = \vec{a} + \vec{b} + \vec{c} + \vec{d} = (1-1+1+1)i + (1+1-1+1)j + (1+1+1-1)k = 2i + 2j + 2k$.
The magnitude of the resultant vector is $|\vec{R}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$.
The direction cosines $(l, m, n)$ are given by $\frac{x}{|\vec{R}|}, \frac{y}{|\vec{R}|}, \frac{z}{|\vec{R}|}$.
Thus,$l = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$m = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,and $n = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Therefore,the direction cosines are $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.

(A) The position vectors are $\vec{OP} = 5i + 4j + ak$ and $\vec{OQ} = -i + 2j - 2k$.
The vector $\vec{PQ} = \vec{OQ} - \vec{OP} = (-1 - 5)i + (2 - 4)j + (-2 - a)k = -6i - 2j - (a + 2)k$.
The distance between $P$ and $Q$ is given by the magnitude of $\vec{PQ}$,which is $7$.
$|\vec{PQ}| = \sqrt{(-6)^2 + (-2)^2 + (-(a + 2))^2} = 7$.
Squaring both sides,we get $36 + 4 + (a + 2)^2 = 49$.
$40 + (a + 2)^2 = 49$.
$(a + 2)^2 = 9$.
$a + 2 = \pm 3$.
Case $1$: $a + 2 = 3 \Rightarrow a = 1$.
Case $2$: $a + 2 = -3 \Rightarrow a = -5$.
Thus,the values of $a$ are $-5$ and $1$.

(A) zero vector is defined as a vector whose magnitude is $0$.
By definition,a zero vector has an arbitrary or indeterminate direction.
Therefore,it is often said to have any direction,as it does not point in any specific direction in space.
Thus,the correct option is $A$.

(C) Let $\vec{a} = l\hat{i} + m\hat{j} + n\hat{k}$,where $l^2 + m^2 + n^2 = 1$.
Since $\vec{a}$ makes an angle $\frac{\pi}{4}$ with the $z$-axis,$n = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Thus,$l^2 + m^2 + (\frac{1}{\sqrt{2}})^2 = 1 \implies l^2 + m^2 = \frac{1}{2} \dots (i)$.
Given that $\vec{a} + \hat{i} + \hat{j}$ is a unit vector,we have $|(l+1)\hat{i} + (m+1)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}| = 1$.
Squaring both sides,$(l+1)^2 + (m+1)^2 + (\frac{1}{\sqrt{2}})^2 = 1^2$.
$l^2 + 2l + 1 + m^2 + 2m + 1 + \frac{1}{2} = 1$.
$(l^2 + m^2) + 2(l+m) + 2.5 = 1$.
Substituting $(i)$,$\frac{1}{2} + 2(l+m) + 2.5 = 1 \implies 2(l+m) = -2 \implies l+m = -1$.
Since $l^2 + m^2 = \frac{1}{2}$ and $l+m = -1$,we have $(l+m)^2 = l^2 + m^2 + 2lm = 1 \implies \frac{1}{2} + 2lm = 1 \implies 2lm = \frac{1}{2} \implies lm = \frac{1}{4}$.
Solving $l+m = -1$ and $lm = \frac{1}{4}$ gives $l = m = -\frac{1}{2}$.
Therefore,$\vec{a} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.

(B) force is defined as a vector that acts at a specific point on a body.
Since the effect of a force depends on its point of application,it is classified as a localised vector (or bound vector).

(B) The position vectors are $\vec{OA} = i + 3j - 7k$ and $\vec{OB} = 5i - 2j + 4k$.
The vector $\overrightarrow{AB}$ is given by $\vec{OB} - \vec{OA} = (5-1)i + (-2-3)j + (4 - (-7))k = 4i - 5j + 11k$.
The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{4^2 + (-5)^2 + 11^2} = \sqrt{16 + 25 + 121} = \sqrt{162}$.
The direction cosine along the $y$-axis is the component of the unit vector along the $y$-direction,which is the coefficient of $j$ divided by the magnitude of the vector.
Direction cosine along $y$-axis $= \frac{-5}{\sqrt{162}}$.

(C) The vector is given by $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
To find the direction cosines,we first calculate the magnitude of the vector $\vec{a}$:
$|\vec{a}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$.
The direction cosine in the direction of the positive $x$-axis is given by the ratio of the $x$-component to the magnitude of the vector:
$\cos \alpha = \frac{a_x}{|\vec{a}|} = \frac{3}{\sqrt{50}}$.
Thus,the correct option is $C$.

(C) Let the position vectors of the points be $\overrightarrow{OA} = 2i + 3j + 4k,$ $\overrightarrow{OB} = 3i + 4j + 2k,$ and $\overrightarrow{OC} = 4i + 2j + 3k.$
First,we find the vectors representing the sides of the triangle:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3-2)i + (4-3)j + (2-4)k = i + j - 2k.$
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (4-3)i + (2-4)j + (3-2)k = i - 2j + k.$
$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2-4)i + (3-2)j + (4-3)k = -2i + j + k.$
Next,we calculate the lengths of the sides:
$|\overrightarrow{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}.$
$|\overrightarrow{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}.$
$|\overrightarrow{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}.$
Since $|\overrightarrow{AB}| = |\overrightarrow{BC}| = |\overrightarrow{CA}| = \sqrt{6},$ all three sides are equal in length.
Therefore,the points form an equilateral triangle.

(B) Let $P, Q,$ and $R$ be points with position vectors $\vec{p} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\vec{q} = \beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$,and $\vec{r} = \gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$.
The distance between $P$ and $Q$ is $|\vec{q} - \vec{p}| = \sqrt{(\beta - \alpha)^2 + (\gamma - \beta)^2 + (\alpha - \gamma)^2}$.
The distance between $Q$ and $R$ is $|\vec{r} - \vec{q}| = \sqrt{(\gamma - \beta)^2 + (\alpha - \gamma)^2 + (\beta - \alpha)^2}$.
The distance between $R$ and $P$ is $|\vec{p} - \vec{r}| = \sqrt{(\alpha - \gamma)^2 + (\beta - \alpha)^2 + (\gamma - \beta)^2}$.
Since $|\vec{PQ}| = |\vec{QR}| = |\vec{RP}|$,the points form an equilateral triangle.

(B) Given that $|a| = 3$,$|b| = 4$,and $|a + b| = 5$.
Using the parallelogram law of vectors,we have the identity:
$|a + b|^2 + |a - b|^2 = 2(|a|^2 + |b|^2)$
Substituting the given values:
$5^2 + |a - b|^2 = 2(3^2 + 4^2)$
$25 + |a - b|^2 = 2(9 + 16)$
$25 + |a - b|^2 = 2(25)$
$25 + |a - b|^2 = 50$
$|a - b|^2 = 50 - 25 = 25$
$|a - b| = \sqrt{25} = 5$.
Thus,the correct option is $B$.

(B) Let $x(a + b) + y(a - b) = 0$ for some scalars $x$ and $y$.
This can be rewritten as $(x + y)a + (x - y)b = 0$.
Since $a$ and $b$ are non-collinear,they are linearly independent.
Therefore,the coefficients must be zero:
$x + y = 0$ and $x - y = 0$.
Adding these equations gives $2x = 0$,so $x = 0$.
Substituting $x = 0$ into $x + y = 0$ gives $y = 0$.
Since the only solution is $x = 0$ and $y = 0$,the vectors $a + b$ and $a - b$ are linearly independent.

(B) Let the vector be $\vec{A} = 3\hat{i} - 4\hat{j} + 5\hat{k}$.
The magnitude of the vector is $|\vec{A}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.
The direction cosines $(l, m, n)$ are given by $\frac{x}{|\vec{A}|}, \frac{y}{|\vec{A}|}, \frac{z}{|\vec{A}|}$.
Thus,the direction cosines are $\frac{3}{5\sqrt{2}}, \frac{-4}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}$.
Simplifying the third term,$\frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,the direction cosines are $\frac{3}{5\sqrt{2}}, \frac{-4}{5\sqrt{2}}, \frac{1}{\sqrt{2}}$.

(D) Given the position vectors of $A$ and $B$ are $\overrightarrow{OA} = 2i - 9j - 4k$ and $\overrightarrow{OB} = 6i - 3j + 8k$.
The vector $\overrightarrow{AB}$ is given by $\overrightarrow{OB} - \overrightarrow{OA}$.
$\overrightarrow{AB} = (6 - 2)i + (-3 - (-9))j + (8 - (-4))k$
$\overrightarrow{AB} = 4i + 6j + 12k$
The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{4^2 + 6^2 + 12^2}$.
$|\overrightarrow{AB}| = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$.

(D) Given the position vectors of points $P$ and $Q$ are $\vec{p} = i + 3j - 7k$ and $\vec{q} = 5i - 2j + 4k$.
The vector $\overrightarrow{PQ}$ is given by $\vec{q} - \vec{p}$.
$\overrightarrow{PQ} = (5 - 1)i + (-2 - 3)j + (4 - (-7))k$
$\overrightarrow{PQ} = 4i - 5j + 11k$
The magnitude $|\overrightarrow{PQ}|$ is calculated as $\sqrt{(4)^2 + (-5)^2 + (11)^2}$.
$|\overrightarrow{PQ}| = \sqrt{16 + 25 + 121} = \sqrt{162}$.

(C) vector $\vec{v}$ is a unit vector if its magnitude is $1$,i.e.,$|\vec{v}| = 1$.
Given that $m\vec{a}$ is a unit vector,we have $|m\vec{a}| = 1$.
Using the property of the modulus of a scalar multiple of a vector,$|m\vec{a}| = |m| |\vec{a}|$.
Therefore,$|m| |\vec{a}| = 1$.
Since $m$ is a non-zero scalar,we can write $|m| = \frac{1}{|\vec{a}|}$.
Assuming $m$ is positive,$m = \frac{1}{|\vec{a}|}$.
Thus,option $(C)$ is the correct condition.

(C) Let the position vectors be $\vec{a} = 2i + j - k$,$\vec{b} = 3i - 2j + k$,and $\vec{c} = i + 4j - 3k$.
Calculate the vectors $\vec{AB}$ and $\vec{BC}$:
$\vec{AB} = \vec{b} - \vec{a} = (3-2)i + (-2-1)j + (1-(-1))k = i - 3j + 2k$.
$\vec{BC} = \vec{c} - \vec{b} = (1-3)i + (4-(-2))j + (-3-1)k = -2i + 6j - 4k$.
Observe that $\vec{BC} = -2(i - 3j + 2k) = -2\vec{AB}$.
Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$,the vectors $\vec{AB}$ and $\vec{BC}$ are parallel.
Because they share a common point $B$,the points $A, B$,and $C$ must be collinear.

(D) Given,position vectors of $A, B$,and $C$ are $\vec{a} = 7j + 10k$,$\vec{b} = -i + 6j + 6k$,and $\vec{c} = -4i + 9j + 6k$.
First,we calculate the vectors representing the sides:
$\vec{AB} = \vec{b} - \vec{a} = (-i + 6j + 6k) - (7j + 10k) = -i - j - 4k$
$|\vec{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$
$\vec{BC} = \vec{c} - \vec{b} = (-4i + 9j + 6k) - (-i + 6j + 6k) = -3i + 3j$
$|\vec{BC}| = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$\vec{AC} = \vec{c} - \vec{a} = (-4i + 9j + 6k) - (7j + 10k) = -4i + 2j - 4k$
$|\vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Since $|\vec{AB}| = |\vec{BC}| = \sqrt{18}$,the triangle is isosceles.
Also,$|\vec{AB}|^2 + |\vec{BC}|^2 = 18 + 18 = 36 = |\vec{AC}|^2$.
By the converse of the Pythagorean theorem,the triangle is right-angled at $B$.
Therefore,the triangle is right-angled and isosceles.

(B) Let the forces be $\vec{F_1} = 6$ along $\overrightarrow{AB}$,$\vec{F_2} = 5$ along $\overrightarrow{CA}$,and $\vec{F_3} = 2\sqrt{2}$ along $\overrightarrow{BC}$.
Taking $A$ as the origin $(0,0)$,$\overrightarrow{AB}$ is along the $x$-axis and $\overrightarrow{AC}$ is along the $y$-axis.
Since $\triangle ABC$ is an isosceles right-angled triangle,$\angle B = \angle C = 45^\circ$.
Vector $\vec{F_1} = 6\hat{i}$.
Vector $\vec{F_2} = 5\hat{j}$.
Vector $\vec{F_3}$ acts along $\overrightarrow{BC}$. The direction of $\overrightarrow{BC}$ makes an angle of $180^\circ - 45^\circ = 135^\circ$ with the $x$-axis.
So,$\vec{F_3} = 2\sqrt{2}(\cos 135^\circ \hat{i} + \sin 135^\circ \hat{j}) = 2\sqrt{2}(-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}) = -2\hat{i} + 2\hat{j}$.
The resultant force $\vec{R} = \vec{F_1} + \vec{F_2} + \vec{F_3} = (6 - 2)\hat{i} + (5 + 2)\hat{j} = 4\hat{i} + 7\hat{j}$.
Wait,re-evaluating the direction of $\overrightarrow{CA}$ and $\overrightarrow{BC}$ from the diagram:
$\overrightarrow{AB}$ is along $+x$ direction: $\vec{F_1} = 6\hat{i}$.
$\overrightarrow{CA}$ is along $-y$ direction: $\vec{F_2} = 5(-\hat{j}) = -5\hat{j}$.
$\overrightarrow{BC}$ direction: $\vec{BC} = \vec{AC} - \vec{AB}$. If $AB=AC=a$,then $\vec{B}=(a,0), \vec{C}=(0,a)$. $\vec{BC} = (-a, a)$. Unit vector is $\frac{1}{\sqrt{2}}(-\hat{i} + \hat{j})$.
$\vec{F_3} = 2\sqrt{2} \cdot \frac{1}{\sqrt{2}}(-\hat{i} + \hat{j}) = -2\hat{i} + 2\hat{j}$.
Resultant $\vec{R} = (6-2)\hat{i} + (-5+2)\hat{j} = 4\hat{i} - 3\hat{j}$.
Magnitude $|R| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

(B) Let the center of the regular hexagon be $O$. In a regular hexagon,the sum of vectors from the center to the vertices is zero,i.e.,$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} = \vec{0}$.
Alternatively,using the property of a regular hexagon,we have $\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO}$ and $\overrightarrow{AC} + \overrightarrow{AE} = 2\overrightarrow{AO} + \overrightarrow{AD}$ is not directly helpful. Let us use the property that $\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO}$ is incorrect.
Correct approach: Let the origin be $A$. Then $\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{AO} + \overrightarrow{OF} = 2\overrightarrow{AO} + (\overrightarrow{OB} + \overrightarrow{OF}) = 2\overrightarrow{AO} + \vec{0} = 2\overrightarrow{AO} = \overrightarrow{AD}$.
Similarly,$\overrightarrow{AC} + \overrightarrow{AE} = 2\overrightarrow{AD}$.
Thus,$\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} = (\overrightarrow{AB} + \overrightarrow{AF}) + (\overrightarrow{AC} + \overrightarrow{AE}) + \overrightarrow{AD} = \overrightarrow{AD} + 2\overrightarrow{AD} + \overrightarrow{AD} = 4\overrightarrow{AD}$.
Wait,let's re-evaluate: In a regular hexagon $ABCDEF$,$\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO}$. Also $\overrightarrow{AC} + \overrightarrow{AE} = 2\overrightarrow{AD}$ is not correct.
Let $A$ be the origin. $\overrightarrow{AB} = \vec{b}$,$\overrightarrow{AF} = \vec{f}$. $\overrightarrow{AC} = \vec{b} + \vec{c}$ where $\vec{c}$ is vector $BC$.
Actually,the simplest way: $\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO}$. $\overrightarrow{AC} + \overrightarrow{AE} = 2\overrightarrow{AD}$. Sum $= \overrightarrow{AO} + 2\overrightarrow{AD} + \overrightarrow{AD} = \overrightarrow{AO} + 3\overrightarrow{AD}$. Since $\overrightarrow{AD} = 2\overrightarrow{AO}$,this is $0.5\overrightarrow{AD} + 3\overrightarrow{AD} = 3.5\overrightarrow{AD}$.
Let's use the property: $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} = 3\overrightarrow{AD}$ is a standard result for a regular hexagon where $\overrightarrow{AD}$ is the main diagonal. Thus $\lambda = 3$.

(D) In parallelogram $ABCD$,let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AD} = \vec{b}$. Then $\overrightarrow{BC} = \overrightarrow{AD} = \vec{b}$ and $\overrightarrow{DC} = \overrightarrow{AB} = \vec{a}$.
Since $P$ is the midpoint of $BC$,$\overrightarrow{BP} = \frac{1}{2}\overrightarrow{BC} = \frac{1}{2}\vec{b}$.
Thus,$\overrightarrow{AP} = \overrightarrow{AB} + \overrightarrow{BP} = \vec{a} + \frac{1}{2}\vec{b}$.
Since $Q$ is the midpoint of $CD$,$\overrightarrow{DQ} = \frac{1}{2}\overrightarrow{DC} = \frac{1}{2}\vec{a}$.
Thus,$\overrightarrow{AQ} = \overrightarrow{AD} + \overrightarrow{DQ} = \vec{b} + \frac{1}{2}\vec{a}$.
Adding these two vectors:
$\overrightarrow{AP} + \overrightarrow{AQ} = (\vec{a} + \frac{1}{2}\vec{b}) + (\vec{b} + \frac{1}{2}\vec{a}) = \frac{3}{2}\vec{a} + \frac{3}{2}\vec{b} = \frac{3}{2}(\vec{a} + \vec{b})$.
Since $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b}$,we have:
$\overrightarrow{AP} + \overrightarrow{AQ} = \frac{3}{2}\overrightarrow{AC}$.

(C) Given that $\overrightarrow{PQ} = \overrightarrow{AP} + \overrightarrow{PB} + \overrightarrow{PC}$.
By the triangle law of vector addition,$\overrightarrow{AP} + \overrightarrow{PB} = \overrightarrow{AB}$.
Substituting this into the given equation,we get $\overrightarrow{AB} + \overrightarrow{PC} = \overrightarrow{PQ}$.
Rearranging the terms,we have $\overrightarrow{AB} = \overrightarrow{PQ} - \overrightarrow{PC}$.
Since $\overrightarrow{PQ} - \overrightarrow{PC} = \overrightarrow{PQ} + \overrightarrow{CP} = \overrightarrow{CQ}$,we get $\overrightarrow{AB} = \overrightarrow{CQ}$.
Since one pair of opposite sides is equal and parallel $(\overrightarrow{AB} = \overrightarrow{CQ})$,the quadrilateral $ABQC$ is a parallelogram.

(D) From the triangle law of vector addition in $\triangle ADC$,we have $\vec{AD} + \vec{DC} = \vec{AC}$. Given $\vec{AD} = b$ and $\vec{DC} = v$,we have $\vec{AC} = b + v$. However,looking at the triangle $\triangle ADC$,the vector $v$ is directed from $C$ to $D$,so $\vec{DC} = -v$. Thus,$\vec{AC} = b - v$.
From the triangle law in $\triangle ABD$,$\vec{AB} + \vec{BD} = \vec{AD}$. Given $\vec{AB} = a$ and $\vec{BD} = w$,we have $a + w = b$,so $w = b - a$.
Given $x - w = v$,we have $x = v + w$.
Substituting the vectors from the figure:
In $\triangle ADC$,$\vec{AC} = \vec{AD} + \vec{DC}$. If we define the vectors based on the directions shown:
$w = \vec{BD}$,$v = \vec{CD}$.
From $\triangle ABD$,$\vec{AD} = \vec{AB} + \vec{BD} = a + w$.
From $\triangle ADC$,$\vec{AC} = \vec{AD} + \vec{DC} = (a + w) + v$.
Given the equation $x = v + w$,and observing the geometry,the correct vector sum is $x = a + b + c$.

(A) vector $\vec{r}$ is coplanar with two non-collinear vectors $\vec{a}$ and $\vec{b}$ if and only if it can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$.
That is,$\vec{r} = x\vec{a} + y\vec{b}$ for some scalars $x$ and $y$.
Since the provided options do not explicitly include the general form $x\vec{a} + y\vec{b}$ as a standard choice,and option $(B)$ $\vec{a} + \vec{b}$ is a specific case of this linear combination (where $x=1, y=1$),the most appropriate general answer is the linear combination form. However,if the question implies a specific vector from the list,$\vec{a} + \vec{b}$ is indeed coplanar. Given the structure,we select the linear combination form.

(C) In a parallelogram $ABCD$,by the triangle law of vector addition,we have $\overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AD}$.
Therefore,$\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB}$.
Substituting the given vectors: $\overrightarrow{BD} = (i + 2j + 3k) - (2i + 4j - 5k) = -i - 2j + 8k$.
The magnitude of $\overrightarrow{BD}$ is $|\overrightarrow{BD}| = \sqrt{(-1)^2 + (-2)^2 + 8^2} = \sqrt{1 + 4 + 64} = \sqrt{69}$.
The unit vector in the direction of $\overrightarrow{BD}$ is given by $\frac{\overrightarrow{BD}}{|\overrightarrow{BD}|} = \frac{-i - 2j + 8k}{\sqrt{69}} = \frac{1}{\sqrt{69}}(-i - 2j + 8k)$.

(D) Given vectors are $a = 2i + 5j$ and $b = 2i - j.$
First,calculate the sum $a + b = (2i + 5j) + (2i - j) = 4i + 4j.$
Next,find the magnitude of the resultant vector $|a + b| = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}.$
The unit vector along $a + b$ is given by $\frac{a + b}{|a + b|} = \frac{4i + 4j}{4\sqrt{2}} = \frac{4(i + j)}{4\sqrt{2}} = \frac{i + j}{\sqrt{2}}.$

(A) Let the vector to be added be $b$.
According to the problem,$a + b = i$.
Substituting the given vector $a = 3i + 4j - 2k$,we get:
$3i + 4j - 2k + b = i$
$b = i - (3i + 4j - 2k)$
$b = i - 3i - 4j + 2k$
$b = - 2i - 4j + 2k$
Thus,the required vector is $- 2i - 4j + 2k$.

(C) The resultant vector $R$ is given by the sum of the vectors $a$,$b$,and $c$:
$R = a + b + c$
$R = (i + 2j + 3k) + (-i + 2j + k) + (3i + j)$
$R = (1 - 1 + 3)i + (2 + 2 + 1)j + (3 + 1 + 0)k$
$R = 3i + 5j + 4k$
Now,find the magnitude of the resultant vector $|R|$:
$|R| = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$
The unit vector along the resultant is given by $\hat{R} = \frac{R}{|R|}$:
$\hat{R} = \frac{3i + 5j + 4k}{5\sqrt{2}}$

(B) In a regular hexagon $ABCDEF$,we can express the vector $\overrightarrow{AE}$ using the triangle law of vector addition.
Consider the path $A \to C \to D \to E$. Thus,$\overrightarrow{AE} = \overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DE}$.
In a regular hexagon,opposite sides are parallel and equal in magnitude. Therefore,$\overrightarrow{CD} = \overrightarrow{AF}$ and $\overrightarrow{DE} = -\overrightarrow{AB}$ (since $\overrightarrow{AB} = -\overrightarrow{ED} = \overrightarrow{DE}$ is incorrect,rather $\overrightarrow{ED} = \overrightarrow{AB}$,so $\overrightarrow{DE} = -\overrightarrow{AB}$).
Substituting these into the expression,we get $\overrightarrow{AE} = \overrightarrow{AC} + \overrightarrow{AF} - \overrightarrow{AB}$.

(C) Given expression: $3\overrightarrow{OD} + \overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC}$
We can rewrite the expression by splitting $3\overrightarrow{OD}$ into three parts:
$= \overrightarrow{OD} + \overrightarrow{DA} + \overrightarrow{OD} + \overrightarrow{DB} + \overrightarrow{OD} + \overrightarrow{DC}$
Using the triangle law of vector addition,$\overrightarrow{OX} + \overrightarrow{XY} = \overrightarrow{OY}$:
$\overrightarrow{OD} + \overrightarrow{DA} = \overrightarrow{OA}$
$\overrightarrow{OD} + \overrightarrow{DB} = \overrightarrow{OB}$
$\overrightarrow{OD} + \overrightarrow{DC} = \overrightarrow{OC}$
Substituting these into the expression:
$= \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$
Thus,the correct option is $C$.

(B) Let $-2a + 3b - c = xp + yq + zr$.
Substituting the expressions for $p, q,$ and $r$:
$-2a + 3b - c = x(2a - 3b) + y(a - 2b + c) + z(-3a + b + 2c)$
$-2a + 3b - c = (2x + y - 3z)a + (-3x - 2y + z)b + (y + 2z)c$.
Comparing the coefficients of $a, b,$ and $c$:
$1) 2x + y - 3z = -2$
$2) -3x - 2y + z = 3$
$3) y + 2z = -1$
From $(3)$,$y = -1 - 2z$.
Substituting into $(1)$ and $(2)$:
$2x + (-1 - 2z) - 3z = -2 \implies 2x - 5z = -1$
$-3x - 2(-1 - 2z) + z = 3 \implies -3x + 2 + 4z + z = 3 \implies -3x + 5z = 1$
Adding the two equations: $(2x - 5z) + (-3x + 5z) = -1 + 1 \implies -x = 0 \implies x = 0$.
Then $5z = 1 \implies z = \frac{1}{5}$.
$y = -1 - 2(\frac{1}{5}) = -1 - \frac{2}{5} = -\frac{7}{5}$.
Thus,$-2a + 3b - c = 0p - \frac{7}{5}q + \frac{1}{5}r = \frac{-7q + r}{5}$.

(A) Given vectors are $a = 2i + j - 8k$ and $b = i + 3j - 4k$.
First,calculate the sum of the vectors $a + b$:
$a + b = (2i + i) + (j + 3j) + (-8k - 4k) = 3i + 4j - 12k$.
Now,find the magnitude of the resulting vector $|a + b|$:
$|a + b| = \sqrt{(3)^2 + (4)^2 + (-12)^2}$
$|a + b| = \sqrt{9 + 16 + 144}$
$|a + b| = \sqrt{169}$
$|a + b| = 13$.
Thus,the magnitude of $a + b$ is $13$.

(B) Given that $A, B, C, D, E$ are five coplanar points.
We need to evaluate the sum: $\overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC} + \overrightarrow{AE} + \overrightarrow{BE} + \overrightarrow{CE}$.
By using the triangle law of vector addition,we know that $\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$.
Rearranging the terms in the given expression:
$= (\overrightarrow{DA} + \overrightarrow{AE}) + (\overrightarrow{DB} + \overrightarrow{BE}) + (\overrightarrow{DC} + \overrightarrow{CE})$
Applying the triangle law to each group:
$= \overrightarrow{DE} + \overrightarrow{DE} + \overrightarrow{DE}$
$= 3\,\overrightarrow{DE}$.

(C) Given vectors are $a = 3i - 2j + k$,$b = 2i - 4j - 3k$,and $c = -i + 2j + 2k$.
To find $a + b + c$,we add the corresponding components of $i$,$j$,and $k$:
$a + b + c = (3 + 2 - 1)i + (-2 - 4 + 2)j + (1 - 3 + 2)k$
$a + b + c = (4)i + (-4)j + (0)k$
$a + b + c = 4i - 4j$.

(B) The resultant force $R$ is the sum of all the given vectors:
$R = (\overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE}) + (\overrightarrow{CB} + \overrightarrow{DB} + \overrightarrow{EB})$
By rearranging the terms using the commutative property of vector addition:
$R = (\overrightarrow{AC} + \overrightarrow{CB}) + (\overrightarrow{AD} + \overrightarrow{DB}) + (\overrightarrow{AE} + \overrightarrow{EB})$
Using the triangle law of vector addition,where $\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$:
$\overrightarrow{AC} + \overrightarrow{CB} = \overrightarrow{AB}$
$\overrightarrow{AD} + \overrightarrow{DB} = \overrightarrow{AB}$
$\overrightarrow{AE} + \overrightarrow{EB} = \overrightarrow{AB}$
Substituting these into the expression for $R$:
$R = \overrightarrow{AB} + \overrightarrow{AB} + \overrightarrow{AB} = 3\overrightarrow{AB}$.

(A) Let the two forces be $P$ and $Q$,where $Q > P$.
Given that the sum of the forces is $P + Q = 18 \ N$.
Let the resultant $R = 12 \ N$ be perpendicular to the smaller force $P$.
The angle between the resultant $R$ and the force $P$ is $90^o$.
The formula for the direction of the resultant is given by $\tan \alpha = \frac{Q \sin \theta}{P + Q \cos \theta}$,where $\theta$ is the angle between the forces.
Since the resultant is perpendicular to $P$,$\tan 90^o = \infty$,which implies $P + Q \cos \theta = 0$,or $\cos \theta = -\frac{P}{Q}$.
Using the magnitude formula for the resultant: $R^2 = P^2 + Q^2 + 2PQ \cos \theta$.
Substituting $\cos \theta = -\frac{P}{Q}$:
$12^2 = P^2 + Q^2 + 2PQ(-\frac{P}{Q}) = P^2 + Q^2 - 2P^2 = Q^2 - P^2$.
$144 = (Q - P)(Q + P)$.
Since $Q + P = 18$,we have $144 = (Q - P) \times 18$,so $Q - P = 8$.
Solving the system $Q + P = 18$ and $Q - P = 8$:
Adding the equations: $2Q = 26 \Rightarrow Q = 13 \ N$.
Subtracting the equations: $2P = 10 \Rightarrow P = 5 \ N$.
Thus,the magnitudes of the two forces are $13 \ N$ and $5 \ N$.

(A) Let the given vectors be $\vec{a} = 2i + 4j - 5k$ and $\vec{b} = i + 2j + 3k$.
The resultant vector $\vec{R}$ is the sum of $\vec{a}$ and $\vec{b}$:
$\vec{R} = \vec{a} + \vec{b} = (2+1)i + (4+2)j + (-5+3)k = 3i + 6j - 2k$.
The magnitude of the resultant vector is $|\vec{R}| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
The unit vector $\hat{R}$ parallel to $\vec{R}$ is given by $\hat{R} = \frac{\vec{R}}{|\vec{R}|} = \frac{3i + 6j - 2k}{7} = \frac{1}{7}(3i + 6j - 2k)$.

(A) The centroid of a triangle is the point where the three medians intersect.
If the position vectors of the vertices $A, B,$ and $C$ are given by $\vec{a}, \vec{b},$ and $\vec{c}$ respectively,the position vector of the centroid $G$ is calculated as the arithmetic mean of the position vectors of the vertices.
Therefore,the position vector of the centroid $G$ is given by $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}.$
Thus,the correct option is $A$.