Basic , Modulus and Algebra of vectors Questions in English

(B) Let the direction ratios of the two lines be $\vec{a} = (3, 0, 2)$ and $\vec{b} = (0, 2, k)$.
The formula for the cosine of the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $|\cos \theta| = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the given values: $|\cos \theta| = \frac{|(3)(0) + (0)(2) + (2)(k)|}{\sqrt{3^2 + 0^2 + 2^2} \sqrt{0^2 + 2^2 + k^2}} = \frac{|2k|}{\sqrt{13} \sqrt{4 + k^2}}$.
Given $|\cos \theta| = \frac{6}{13}$, we have $\frac{6}{13} = \frac{|2k|}{\sqrt{13} \sqrt{4 + k^2}}$.
Squaring both sides: $\frac{36}{169} = \frac{4k^2}{13(4 + k^2)}$.
Simplifying: $\frac{9}{13} = \frac{k^2}{4 + k^2}$.
$9(4 + k^2) = 13k^2 \Rightarrow 36 + 9k^2 = 13k^2 \Rightarrow 4k^2 = 36 \Rightarrow k^2 = 9$.
Thus, $k = \pm 3$.

(A) The position vector of point $B$ is $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$.
Since point $A$ lies on the line passing through $B$ and parallel to $\vec{v} = 2 \hat{i} - \hat{j} + 2 \hat{k}$, the vector $\overrightarrow{B A}$ can be written as $\overrightarrow{B A} = t \vec{v} = t(2 \hat{i} - \hat{j} + 2 \hat{k})$ for some scalar $t$.
Given $|\overrightarrow{B A}| = 18$, we have $|t| \sqrt{2^2 + (-1)^2 + 2^2} = 18$.
$|t| \sqrt{4 + 1 + 4} = 18 \Rightarrow |t| \sqrt{9} = 18 \Rightarrow 3|t| = 18 \Rightarrow |t| = 6$.
Thus, $t = 6$ or $t = -6$.
The position vector of $A$ is $\vec{a} = \vec{b} + \overrightarrow{B A} = (3 \hat{i} + \hat{j} - \hat{k}) + t(2 \hat{i} - \hat{j} + 2 \hat{k})$.
For $t = 6$: $\vec{a} = (3 + 12) \hat{i} + (1 - 6) \hat{j} + (-1 + 12) \hat{k} = 15 \hat{i} - 5 \hat{j} + 11 \hat{k}$.
For $t = -6$: $\vec{a} = (3 - 12) \hat{i} + (1 + 6) \hat{j} + (-1 - 12) \hat{k} = -9 \hat{i} + 7 \hat{j} - 13 \hat{k}$.
Comparing with the given options, the correct position vector is $-9 \hat{i} + 7 \hat{j} - 13 \hat{k}$.

(B) Let vector $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.
Since the vector $v = 19 \hat{i} + 22 \hat{j} + 5 \hat{k}$ bisects the angle between $a$ and $b = 6 \hat{i} + 8 \hat{j}$, it must be proportional to the sum of their unit vectors:
$\lambda \left( \frac{a}{|a|} + \frac{b}{|b|} \right) = v$
Given $|b| = \sqrt{6^2 + 8^2} = 10$, so $\frac{b}{|b|} = \frac{6 \hat{i} + 8 \hat{j}}{10} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$.
Thus, $\frac{a}{|a|} = \frac{19 \hat{i} + 22 \hat{j} + 5 \hat{k}}{\lambda} - (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) = (\frac{19}{\lambda} - \frac{3}{5}) \hat{i} + (\frac{22}{\lambda} - \frac{4}{5}) \hat{j} + \frac{5}{\lambda} \hat{k}$.
Since $\frac{a}{|a|}$ is a unit vector, its magnitude squared is $1$:
$(\frac{19}{\lambda} - \frac{3}{5})^2 + (\frac{22}{\lambda} - \frac{4}{5})^2 + (\frac{5}{\lambda})^2 = 1$.
Expanding this: $\frac{361}{\lambda^2} - \frac{114}{5\lambda} + \frac{9}{25} + \frac{484}{\lambda^2} - \frac{176}{5\lambda} + \frac{16}{25} + \frac{25}{\lambda^2} = 1$.
$\frac{870}{\lambda^2} - \frac{290}{5\lambda} + 1 = 1 \Rightarrow \frac{870}{\lambda^2} = \frac{58}{\lambda} \Rightarrow \lambda = \frac{870}{58} = 15$.
Substituting $\lambda = 15$ back, we get $\frac{a}{|a|} = (\frac{19}{15} - \frac{9}{15}) \hat{i} + (\frac{22}{15} - \frac{12}{15}) \hat{j} + \frac{5}{15} \hat{k} = \frac{10}{15} \hat{i} + \frac{10}{15} \hat{j} + \frac{5}{15} \hat{k} = \frac{1}{3}(2 \hat{i} + 2 \hat{j} + \hat{k})$.
Therefore, the correct option is $(b)$.

(C) Given vectors are $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$.
Sum of the vectors is $\vec{s} = \vec{a} + \vec{b} = (2-2) \hat{i} + (-2+5) \hat{j} + (5-3) \hat{k} = 0 \hat{i} + 3 \hat{j} + 2 \hat{k} = 3 \hat{j} + 2 \hat{k}$.
Since the $\hat{i}$ component of the resulting vector $\vec{s}$ is $0$, the vector lies in the $YZ$-plane.
$A$ vector that lies in a plane is parallel to that plane.
Therefore, the vector is parallel to the $YZ$-plane.

(D) For a point $P$ defined by $P = xa + yb + zc$ to lie in the plane containing non-collinear points $a, b$ and $c$, the sum of the coefficients must be equal to $1$ if the points are represented as position vectors relative to an origin, or the expression must be of the form $P = x a + y b + (1 - x - y) c$.
However, if the expression is given as a linear combination $ka + 2b + 3c$ representing a point in the plane, it implies that the vector is normalized such that the sum of coefficients is $1$, or it represents a point where the sum of coefficients is $0$ for the vector to be a linear combination of vectors in the plane.
Given the standard form for a point in the plane of $a, b, c$ is $P = x a + y b + z c$ where $x + y + z = 1$.
Here, $x = k, y = 2, z = 3$.
Thus, $k + 2 + 3 = 1$.
$k + 5 = 1$.
$k = 1 - 5 = -4$.

(C) Let $\vec{a}$ and $\vec{b}$ be two unit vectors, so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
Given that their sum is a unit vector, $|\vec{a} + \vec{b}| = 1$.
Squaring both sides, we get $|\vec{a} + \vec{b}|^2 = 1^2$.
Using the identity $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$, we have $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1$.
This simplifies to $2 + 2(\vec{a} \cdot \vec{b}) = 1$, which gives $2(\vec{a} \cdot \vec{b}) = -1$.
Now, we need to find the magnitude of their difference, $|\vec{a} - \vec{b}|$.
We know that $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
Substituting the known values, $|\vec{a} - \vec{b}|^2 = 1 + 1 - (-1) = 1 + 1 + 1 = 3$.
Therefore, $|\vec{a} - \vec{b}| = \sqrt{3}$ units.

(A) Let the unit vector be $\vec{r} = x\hat{i} + y\hat{j}$. Since it is a unit vector, $x^2 + y^2 = 1$.
Given $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 3\hat{i} - 4\hat{j}$.
The angle between $\vec{r}$ and $\vec{a}$ is $45^{\circ}$, so $\vec{r} \cdot \vec{a} = |\vec{r}||\vec{a}| \cos 45^{\circ} = 1 \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$.
Thus, $x + y = 1 \implies y = 1 - x$.
The angle between $\vec{r}$ and $\vec{b}$ is $60^{\circ}$, so $\vec{r} \cdot \vec{b} = |\vec{r}||\vec{b}| \cos 60^{\circ} = 1 \cdot \sqrt{3^2 + (-4)^2} \cdot \frac{1}{2} = 5 \cdot \frac{1}{2} = \frac{5}{2}$.
Thus, $3x - 4y = \frac{5}{2} \implies 6x - 8y = 5$.
Substituting $y = 1 - x$ into the second equation: $6x - 8(1 - x) = 5 \implies 6x - 8 + 8x = 5 \implies 14x = 13 \implies x = \frac{13}{14}$.
Then $y = 1 - \frac{13}{14} = \frac{1}{14}$.
Therefore, the unit vector is $\frac{13}{14}\hat{i} + \frac{1}{14}\hat{j}$.

(B) Given that $O$ is the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$.
Since $R$ lies on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, we have $\vec{OR} = \frac{1}{5}\vec{OP} = \frac{1}{5}\vec{a}$.
Given $\vec{OQ} = 5\vec{RM}$, we have $\vec{RM} = \frac{1}{5}\vec{OQ} = \frac{1}{5}\vec{b}$.
Since $\vec{RM} = \vec{OM} - \vec{OR}$, we can write $\vec{OM} = \vec{RM} + \vec{OR} = \frac{1}{5}\vec{b} + \frac{1}{5}\vec{a}$.
Now, $\vec{PM} = \vec{OM} - \vec{OP} = (\frac{1}{5}\vec{a} + \frac{1}{5}\vec{b}) - \vec{a} = \frac{1}{5}\vec{b} + \frac{1}{5}\vec{a} - \frac{5}{5}\vec{a} = \frac{1}{5}(\vec{b} - 4\vec{a})$.

(B) Given that $O$ is the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$.
Since $\vec{OP} = 5\vec{OR}$, we have $\vec{OR} = \frac{1}{5}\vec{a}$.
Given $\vec{OQ} = 5\vec{RM}$, we have $\vec{RM} = \frac{1}{5}\vec{b}$.
We know that $\vec{RM} = \vec{OM} - \vec{OR}$, so $\vec{OM} = \vec{OR} + \vec{RM}$.
Substituting the values, $\vec{OM} = \frac{1}{5}\vec{a} + \frac{1}{5}\vec{b}$.
Now, $\vec{PM} = \vec{OM} - \vec{OP} = (\frac{1}{5}\vec{a} + \frac{1}{5}\vec{b}) - \vec{a}$.
$\vec{PM} = \frac{1}{5}\vec{b} + (\frac{1}{5} - 1)\vec{a} = \frac{1}{5}\vec{b} - \frac{4}{5}\vec{a} = \frac{1}{5}(\vec{b} - 4\vec{a})$.

(B) $1$. Expression $(A)$ $(a \times b) \cdot c$ represents the scalar triple product, which is meaningful.
$2$. Expression $(B)$ $a \times (b \cdot c)$ is meaningless because $(b \cdot c)$ is a scalar, and the cross product of a vector $a$ and a scalar is not defined.
$3$. Expression $(C)$ $a \cdot (b \cdot c)$ is meaningless because $(b \cdot c)$ is a scalar, and the dot product of a vector $a$ and a scalar is not defined.
$4$. Expression $(D)$ $|a|(b \cdot c)$ represents the product of a scalar $|a|$ and a scalar $(b \cdot c)$, which is meaningful.
$5$. Expression $(E)$ $(a \cdot b) \times (b \cdot c)$ is meaningless because $(a \cdot b)$ and $(b \cdot c)$ are both scalars, and the cross product between two scalars is not defined.
$6$. Thus, $(A)$ and $(D)$ are meaningful, while $(B)$, $(C)$, and $(E)$ are meaningless.

(A) Let $\vec{AB} = \vec{b} - \vec{a} = -\hat{i} - \hat{j} - 4\hat{k}$, so $c = |\vec{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{18} = 3\sqrt{2}$.
Let $\vec{AC} = \vec{c} - \vec{a} = -4\hat{i} + 2\hat{j} - 4\hat{k}$, so $b = |\vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{36} = 6$.
The angle bisector of $\angle A$ divides the opposite side $BC$ in the ratio $c:b = 3\sqrt{2}:6 = 1:\sqrt{2}$.
Using the section formula, the position vector $\vec{p}$ of the point dividing $BC$ in ratio $m:n$ is $\frac{m\vec{c} + n\vec{b}}{m+n}$.
Here $m=1, n=\sqrt{2}$, so $\vec{p} = \frac{1\vec{c} + \sqrt{2}\vec{b}}{1+\sqrt{2}} = \frac{(-4\hat{i} + 9\hat{j} + 6\hat{k}) + \sqrt{2}(-\hat{i} + 6\hat{j} + 6\hat{k})}{1+\sqrt{2}} = \frac{(-4-\sqrt{2})\hat{i} + (9+6\sqrt{2})\hat{j} + (6+6\sqrt{2})\hat{k}}{1+\sqrt{2}}$.
Rationalizing the coefficients: $\frac{-4-\sqrt{2}}{1+\sqrt{2}} = \frac{(-4-\sqrt{2})(\sqrt{2}-1)}{1} = -4\sqrt{2} + 4 - 2 + \sqrt{2} = 2 - 3\sqrt{2}$.
$\frac{9+6\sqrt{2}}{1+\sqrt{2}} = \frac{(9+6\sqrt{2})(\sqrt{2}-1)}{1} = 9\sqrt{2} - 9 + 12 - 6\sqrt{2} = 3 + 3\sqrt{2}$.
$\frac{6+6\sqrt{2}}{1+\sqrt{2}} = 6$. Thus, $\vec{p} = (2 - 3\sqrt{2})\hat{i} + (3 + 3\sqrt{2})\hat{j} + 6\hat{k}$.

(A) Let the vector be $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
The direction cosines are $l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$.
Given $\beta = \pi/4$ and $\gamma = \pi/2$, so $m = \cos(\pi/4) = 1/\sqrt{2}$ and $n = \cos(\pi/2) = 0$.
Since $l^2 + m^2 + n^2 = 1$, we have $l^2 + (1/\sqrt{2})^2 + 0^2 = 1$, which gives $l^2 + 1/2 = 1$, so $l^2 = 1/2$, hence $l = \pm 1/\sqrt{2}$.
The vector is $\vec{r} = |\vec{r}|(l\hat{i} + m\hat{j} + n\hat{k}) = 3\sqrt{2}(\pm \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}) = \pm 3\hat{i} + 3\hat{j}$.

(B) Given $|\vec{p}| = 3$, $|\vec{q}| = 2$ and the angle $\theta = \pi/3$.
$\vec{p} \cdot \vec{q} = |\vec{p}||\vec{q}| \cos(\pi/3) = 3 \times 2 \times (1/2) = 3$.
Calculate $|\vec{a}|^2 = |3\vec{p} - \vec{q}|^2 = 9|\vec{p}|^2 + |\vec{q}|^2 - 6(\vec{p} \cdot \vec{q}) = 9(9) + 4 - 6(3) = 81 + 4 - 18 = 67$.
So, $|\vec{a}| = \sqrt{67}$.
Calculate $|\vec{b}|^2 = |\vec{p} + 3\vec{q}|^2 = |\vec{p}|^2 + 9|\vec{q}|^2 + 6(\vec{p} \cdot \vec{q}) = 9 + 9(4) + 6(3) = 9 + 36 + 18 = 63$.
So, $|\vec{b}| = \sqrt{63}$.
The ratio $|\vec{a}| : |\vec{b}| = \sqrt{67} : \sqrt{63}$.

(C) Let the origin be at the center $O$ of the regular hexagon. The position vectors of the vertices are $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{e}, \vec{f}$.
Since $O$ is the center, $\vec{a} + \vec{d} = \vec{0}$, $\vec{b} + \vec{e} = \vec{0}$, and $\vec{c} + \vec{f} = \vec{0}$.
Also, $\vec{d} = -\vec{a}$.
The given sum is $\vec{S} = (\vec{b}-\vec{a}) + (\vec{c}-\vec{a}) + (\vec{d}-\vec{a}) + (\vec{e}-\vec{a}) + (\vec{f}-\vec{a})$.
$\vec{S} = (\vec{a} + \vec{b} + \vec{c} + \vec{d} + \vec{e} + \vec{f}) - 6\vec{a} - \vec{a} = \vec{0} - 6\vec{a} = -6\vec{a} = 6\vec{d}$.
Given $\vec{S} = p\vec{AD} = p(\vec{d}-\vec{a}) = p(2\vec{d}) = 2p\vec{d}$.
Comparing $6\vec{d} = 2p\vec{d}$, we get $p = 3$.
Given $\vec{S} = q\vec{AO} = q\vec{a} = q(-\vec{d})$.
Since $\vec{S} = 6\vec{d}$, then $q(-\vec{d}) = 6\vec{d}$, so $q = -6$. However, checking the vector direction $\vec{S} = 6\vec{d} = 6\vec{AO}$ is incorrect as $\vec{AO} = -\vec{a} = \vec{d}$.
Thus $\vec{S} = 6\vec{d} = 6\vec{AO}$. So $p=3, q=6$.

(A) The vectors representing the sides are:
$\vec{AB} = \vec{b} - \vec{a} = (4 - (-2))\hat{i} + (0 - (-1))\hat{j} = 6\hat{i} + \hat{j}$
$\vec{BC} = \vec{c} - \vec{b} = (3 - 4)\hat{i} + (3 - 0)\hat{j} = -\hat{i} + 3\hat{j}$
$\vec{CD} = \vec{d} - \vec{c} = (-3 - 3)\hat{i} + (2 - 3)\hat{j} = -6\hat{i} - \hat{j}$
$\vec{DA} = \vec{a} - \vec{d} = (-2 - (-3))\hat{i} + (-1 - 2)\hat{j} = \hat{i} - 3\hat{j}$
Since $\vec{AB} = -\vec{CD}$ and $\vec{BC} = -\vec{DA}$, $ABCD$ is a parallelogram.
Calculate dot product of adjacent sides: $\vec{AB} \cdot \vec{BC} = (6)(-1) + (1)(3) = -6 + 3 = -3 \neq 0$. Thus, it is not a rectangle.
Calculate magnitudes: $|\vec{AB}| = \sqrt{6^2 + 1^2} = \sqrt{37}$ and $|\vec{BC}| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$.
Since $|\vec{AB}| \neq |\vec{BC}|$, it is not a rhombus.
Therefore, $ABCD$ is a parallelogram which is neither a rhombus nor a rectangle.

(D) The magnitude of a vector remains invariant under rotation. Therefore, $|\vec{v}|^2 = |\vec{v}'|^2$.
$|\vec{v}|^2 = 3^2 + 4^2 + (-5)^2 = 9 + 16 + 25 = 50$.
$|\vec{v}'|^2 = (a + 1)^2 + (-3)^2 + 5^2 = (a + 1)^2 + 9 + 25 = (a + 1)^2 + 34$.
Equating the two: $(a + 1)^2 + 34 = 50$.
$(a + 1)^2 = 16$.
$a + 1 = \pm 4$.
Case $1$: $a + 1 = 4 \implies a = 3$.
Case $2$: $a + 1 = -4 \implies a = -5$.
Thus, the possible values of $a$ are $3$ or $-5$.

(D) Step $1$: Find the position vector of point $D$ which divides $BC$ in ratio $2:1$. Using section formula, $\vec{d} = \frac{2\vec{c} + 1\vec{b}}{2+1} = \frac{\vec{b} + 2\vec{c}}{3}$.
Step $2$: Point $E$ divides $AD$ in ratio $1:2$. Using section formula, $\vec{e} = \frac{1(\vec{d}) + 2(\vec{a})}{1+2} = \frac{\vec{d} + 2\vec{a}}{3}$.
Step $3$: Substitute $\vec{d}$ into the expression for $\vec{e}$: $\vec{e} = \frac{\frac{\vec{b} + 2\vec{c}}{3} + 2\vec{a}}{3} = \frac{\vec{b} + 2\vec{c} + 6\vec{a}}{9} = \frac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$.