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,$\vec{PM} = \vec{OM} - \vec{OP} = (\frac{1}{5}\vec{a} + \frac{1}{5}\vec{b}) - \vec{a}$.
$\vec{PM} = \frac{1}{5}\vec{b} + (\frac{1}{5} - 1)\vec{a} = \frac{1}{5}\vec{b} - \frac{4}{5}\vec{a} = \frac{1}{5}(\vec{b} - 4\vec{a})$.