System of co-ordinates, Direction cosines and direction ratios, Projection Questions in English

(B) The sum of the squares of the direction cosines of a line is always $1$.
Given direction cosines are $\frac{a}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{c}{\sqrt{83}}$.
Therefore,$\left(\frac{a}{\sqrt{83}}\right)^2 + \left(\frac{5}{\sqrt{83}}\right)^2 + \left(\frac{c}{\sqrt{83}}\right)^2 = 1$.
$\Rightarrow \frac{a^2}{83} + \frac{25}{83} + \frac{c^2}{83} = 1$.
$\Rightarrow a^2 + 25 + c^2 = 83$.
$\Rightarrow a^2 + c^2 = 58$ ...$(i)$.
Given $c - a = 4$,squaring both sides gives $(c - a)^2 = 16$.
$c^2 + a^2 - 2ca = 16$.
Substituting $a^2 + c^2 = 58$ from equation $(i)$:
$58 - 2ca = 16$.
$2ca = 58 - 16 = 42$.
$ca = 21$.

(D) We know that the direction ratios $(a, b, c)$ and direction cosines $(\ell, m, n)$ are related by $a = k\ell, b = km, c = kn$ for some non-zero constant $k$.
Substituting these into the expression:
$\frac{a^2}{b^2+c^2} = \frac{(k\ell)^2}{(km)^2+(kn)^2} = \frac{k^2\ell^2}{k^2(m^2+n^2)} = \frac{\ell^2}{m^2+n^2}$.
Since $\ell^2 + m^2 + n^2 = 1$,we have $m^2 + n^2 = 1 - \ell^2$.
Therefore,$\frac{a^2}{b^2+c^2} = \frac{\ell^2}{1-\ell^2}$.

(B) Given equations are $2l-m+3n=0$ and $l^2+mn-6n^2=0$.
From the first equation,$m=2l+3n$.
Substituting this into the second equation: $l^2+(2l+3n)n-6n^2=0$.
$l^2+2ln+3n^2-6n^2=0 \Rightarrow l^2+2ln-3n^2=0$.
Factoring gives $(l+3n)(l-n)=0$,so $l=n$ or $l=-3n$.
Case $1$: If $l=n$,then $m=2(n)+3n=5n$. The direction ratios are $(n, 5n, n)$,or $(1, 5, 1)$. The direction cosines are $(\frac{1}{\sqrt{1^2+5^2+1^2}}, \frac{5}{\sqrt{27}}, \frac{1}{\sqrt{27}}) = (\frac{1}{3\sqrt{3}}, \frac{5}{3\sqrt{3}}, \frac{1}{3\sqrt{3}})$.
Case $2$: If $l=-3n$,then $m=2(-3n)+3n=-3n$. The direction ratios are $(-3n, -3n, n)$,or $(-3, -3, 1)$. The direction cosines are $(\frac{-3}{\sqrt{(-3)^2+(-3)^2+1^2}}, \frac{-3}{\sqrt{19}}, \frac{1}{\sqrt{19}}) = (\frac{-3}{\sqrt{19}}, \frac{-3}{\sqrt{19}}, \frac{1}{\sqrt{19}})$.
Thus,$l_1=\frac{1}{3\sqrt{3}}, m_1=\frac{5}{3\sqrt{3}}$ and $l_2=\frac{-3}{\sqrt{19}}, m_2=\frac{-3}{\sqrt{19}}$.
Calculating $|l_1 l_2|+|m_1 m_2| = |(\frac{1}{3\sqrt{3}})(\frac{-3}{\sqrt{19}})| + |(\frac{5}{3\sqrt{3}})(\frac{-3}{\sqrt{19}})| = |\frac{-1}{\sqrt{3}\sqrt{19}}| + |\frac{-5}{\sqrt{3}\sqrt{19}}| = \frac{1}{\sqrt{57}} + \frac{5}{\sqrt{57}} = \frac{6}{\sqrt{57}} = \frac{6}{\sqrt{3}\sqrt{19}} = \frac{2\sqrt{3}}{\sqrt{19}}$.

(D) Let the direction cosines ($DC$'s) of $AB$ and $AC$ be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively.
First,find the direction ratios ($DR$'s) of $AB$ and $AC$:
$DR's$ of $AB = (3-1, 1-(-2), -3-3) = (2, 3, -6)$.
$DR's$ of $AC = (-3-1, 1-(-2), 3-3) = (-4, 3, 0)$.
Now,calculate the direction cosines by dividing the $DR$'s by their magnitudes:
Magnitude of $AB = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
So,$l_1 = \frac{2}{7}, m_1 = \frac{3}{7}, n_1 = \frac{-6}{7}$.
Magnitude of $AC = \sqrt{(-4)^2 + 3^2 + 0^2} = \sqrt{16 + 9 + 0} = \sqrt{25} = 5$.
So,$l_2 = \frac{-4}{5}, m_2 = \frac{3}{5}, n_2 = 0$.
Since $\cos A = l_1 l_2 + m_1 m_2 + n_1 n_2$:
$\cos A = \left(\frac{2}{7}\right)\left(\frac{-4}{5}\right) + \left(\frac{3}{7}\right)\left(\frac{3}{5}\right) + \left(\frac{-6}{7}\right)(0)$
$\cos A = \frac{-8}{35} + \frac{9}{35} + 0 = \frac{1}{35}$.

(D) Given that $l, m, n$ are the direction cosines of a line,we have $l^2+m^2+n^2=1$ $(i)$.
From the given relations $l-m+n=0$,we get $l=m-n$.
Substituting this into the second relation $lm+mn-4nl=0$:
$(m-n)m + mn - 4n(m-n) = 0$
$m^2 - mn + mn - 4mn + 4n^2 = 0$
$m^2 - 4mn + 4n^2 = 0$
$(m-2n)^2 = 0 \Rightarrow m=2n$.
Substituting $m=2n$ into $l=m-n$,we get $l=2n-n=n$.
Now,substitute $l=n$ and $m=2n$ into the identity $l^2+m^2+n^2=1$:
$n^2 + (2n)^2 + n^2 = 1$
$n^2 + 4n^2 + n^2 = 1$
$6n^2 = 1 \Rightarrow n = \pm \frac{1}{\sqrt{6}}$.
Thus,$l = \pm \frac{1}{\sqrt{6}}$ and $m = \pm \frac{2}{\sqrt{6}}$.
The direction cosines are $\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$ or $\left(-\frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}\right)$.

(C) Let the direction cosines of the line be $(l, m, n) = (\cos \alpha, \cos \beta, \cos \gamma)$.
We know that for any line,the sum of the squares of its direction cosines is $l^2 + m^2 + n^2 = 1$.
Given that the line makes equal angles with the coordinate axes,we have $\alpha = \beta = \gamma$.
Therefore,$\cos \alpha = \cos \beta = \cos \gamma$.
Substituting this into the identity,we get $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$.
$3 \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{3}$.
Thus,$\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Hence,the direction cosines are $(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}})$.

(A) Given that the direction ratios of $\vec{OP}$ are $(a, b, c) = (1, -2, -2)$.
First,we calculate the magnitude of the direction ratios vector: $\sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The direction cosines $(l, m, n)$ are obtained by dividing the direction ratios by their magnitude:
$l = \frac{1}{3}, m = \frac{-2}{3}, n = \frac{-2}{3}$.
The coordinates of point $P$ at a distance $r = 3$ from the origin are given by $(lr, mr, nr)$.
$P = (\frac{1}{3} \times 3, \frac{-2}{3} \times 3, \frac{-2}{3} \times 3) = (1, -2, -2)$.

(B) Let the vector $\overrightarrow{AB} = l\hat{i} + m\hat{j} + n\hat{k}$.
The projection of $\overrightarrow{AB}$ on the $XY$ plane is $\sqrt{l^2 + m^2} = \sqrt{15}$,so $l^2 + m^2 = 15$ (Equation $1$).
The projection of $\overrightarrow{AB}$ on the $YZ$ plane is $\sqrt{m^2 + n^2} = \sqrt{46}$,so $m^2 + n^2 = 46$ (Equation $2$).
The projection of $\overrightarrow{AB}$ on the $ZX$ plane is $\sqrt{n^2 + l^2} = 7$,so $n^2 + l^2 = 49$ (Equation $3$).
Adding equations $(1)$,$(2)$,and $(3)$:
$2(l^2 + m^2 + n^2) = 15 + 46 + 49 = 110$
$l^2 + m^2 + n^2 = 55$.
To find the projection on the $y$-axis,we need the value of $|m|$.
Using $l^2 + m^2 + n^2 = 55$ and $n^2 + l^2 = 49$:
$m^2 = 55 - 49 = 6$.
Wait,re-evaluating the system:
$l^2 + m^2 = 15$
$m^2 + n^2 = 46$
$n^2 + l^2 = 49$
$(l^2 + m^2 + n^2) = (15 + 46 + 49) / 2 = 110 / 2 = 55$.
$n^2 = 55 - 15 = 40$.
$l^2 = 55 - 46 = 9 \Rightarrow l = 3$.
$m^2 = 55 - 49 = 6 \Rightarrow m = \sqrt{6}$.
Re-checking the calculation: $l^2+m^2 = 9+6 = 15$ (Correct). $m^2+n^2 = 6+40 = 46$ (Correct). $n^2+l^2 = 40+9 = 49$ (Correct).
Therefore,the projection on the $y$-axis is $|m| = \sqrt{6}$.
Given the options provided,there might be a typo in the question values. If $l^2+m^2=15, m^2+n^2=46, n^2+l^2=49$ is correct,the answer is $\sqrt{6}$. If the intended answer is $1$,then $m^2=1$.

(C) Given,the direction cosines of two lines are $l_1 = (\frac{2}{3}, \frac{2}{3}, \frac{1}{3})$ and $l_2 = (\frac{5}{13}, \frac{12}{13}, 0)$.
The direction ratios of the line bisecting the angle between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are proportional to $\langle l_1+l_2, m_1+m_2, n_1+n_2 \rangle$.
Substituting the values,the direction ratios are proportional to $\langle \frac{2}{3} + \frac{5}{13}, \frac{2}{3} + \frac{12}{13}, \frac{1}{3} + 0 \rangle$.
Calculating the sum for each component:
$\frac{2}{3} + \frac{5}{13} = \frac{26+15}{39} = \frac{41}{39}$
$\frac{2}{3} + \frac{12}{13} = \frac{26+36}{39} = \frac{62}{39}$
$\frac{1}{3} + 0 = \frac{13}{39}$
Thus,the direction ratios are proportional to $\langle \frac{41}{39}, \frac{62}{39}, \frac{13}{39} \rangle$.
Multiplying by $39$,we get the direction ratios as $\langle 41, 62, 13 \rangle$.

(A) Let the points be $P = (-2, 4, -5)$ and $Q = (1, 2, 3)$.
The direction ratios of the line segment $\overrightarrow{PQ}$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (1 - (-2), 2 - 4, 3 - (-5)) = (3, -2, 8)$.
The magnitude of the vector $\overrightarrow{PQ}$ is $|\overrightarrow{PQ}| = \sqrt{3^2 + (-2)^2 + 8^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$.
The direction cosines $(l, m, n)$ are obtained by dividing the direction ratios by the magnitude:
$l = \frac{3}{\sqrt{77}}$,$m = \frac{-2}{\sqrt{77}}$,$n = \frac{8}{\sqrt{77}}$.
Thus,the direction cosines are $\left(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right)$.

(B) The direction cosines of a line making angles $\alpha, \beta, \gamma$ with the positive $X, Y$ and $Z$ axes are given by $\cos \alpha, \cos \beta, \cos \gamma$.
Here,$\alpha = 90^{\circ}, \beta = 135^{\circ}, \gamma = 45^{\circ}$.
The direction cosines are:
$l = \cos 90^{\circ} = 0$
$m = \cos 135^{\circ} = \cos(180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} = -\frac{1}{\sqrt{2}}$
$n = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
Thus,the direction cosines are $\left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

(B) The direction cosines of a line are denoted by $l, m, n$.
We know that for any line,the sum of the squares of its direction cosines is always equal to $1$,i.e.,$l^2 + m^2 + n^2 = 1$.
Given the direction cosines are $\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)$.
Substituting these values into the identity:
$\left(\frac{1}{c}\right)^2 + \left(\frac{1}{c}\right)^2 + \left(\frac{1}{c}\right)^2 = 1$
$\frac{1}{c^2} + \frac{1}{c^2} + \frac{1}{c^2} = 1$
$\frac{3}{c^2} = 1$
$c^2 = 3$
$c = \pm \sqrt{3}$

(C) Let the direction angles of the line $AB$ be $\alpha = 45^{\circ}$,$\beta = 120^{\circ}$,and $\gamma = \theta$.
The sum of the squares of the direction cosines of a line is always $1$,i.e.,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values,we have $\cos^2 45^{\circ} + \cos^2 120^{\circ} + \cos^2 \theta = 1$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\cos 120^{\circ} = -\frac{1}{2}$,we get:
$\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{2}\right)^2 + \cos^2 \theta = 1$.
$\frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 1$.
$\frac{3}{4} + \cos^2 \theta = 1$.
$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$.
Since $\theta$ is an acute angle,$\cos \theta = \frac{1}{2}$.
Therefore,$\theta = 60^{\circ}$.

(C) Let the direction cosines of the line be $(l, m, n)$. Since the length of the line is $l$,the projections of the line on the coordinate axes are given by $l_1 = l \cdot |l|$,$l_2 = l \cdot |m|$,and $l_3 = l \cdot |n|$.
Squaring these,we get $l_1^2 = l^2 l^2$,$l_2^2 = l^2 m^2$,and $l_3^2 = l^2 n^2$.
Adding these equations,we get $l_1^2 + l_2^2 + l_3^2 = l^2 (l^2 + m^2 + n^2)$.
Since the sum of the squares of the direction cosines is $l^2 + m^2 + n^2 = 1$,we have $l_1^2 + l_2^2 + l_3^2 = l^2 (1) = l^2$.

(D) Let the direction ratios of the required line be $\langle a, b, c \rangle$.
Since the line is perpendicular to the lines with direction ratios $\langle 1, -2, -2 \rangle$ and $\langle 0, 2, 1 \rangle$,we have:
$1(a) - 2(b) - 2(c) = 0$ (Equation $1$)
$0(a) + 2(b) + 1(c) = 0$ (Equation $2$)
From Equation $2$,we get $c = -2b$.
Substituting $c = -2b$ into Equation $1$:
$a - 2b - 2(-2b) = 0$
$a - 2b + 4b = 0$
$a + 2b = 0 \implies a = -2b$.
Let $b = -1$,then $a = 2$ and $c = 2$.
The direction ratios are $\langle 2, -1, 2 \rangle$.
The magnitude is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
The direction cosines are $\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \rangle$.

(A) The direction cosines of a line making angles $\alpha, \beta, \gamma$ with the $X, Y, Z$-axes are given by $\cos \alpha, \cos \beta, \cos \gamma$.
Given $\alpha = 90^{\circ}$,$\beta = 135^{\circ}$,and $\gamma = 45^{\circ}$.
Calculating the values:
$\cos \alpha = \cos 90^{\circ} = 0$
$\cos \beta = \cos 135^{\circ} = \cos(180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} = -\frac{1}{\sqrt{2}}$
$\cos \gamma = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
Thus,the direction cosines are $\left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

(D) For two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,the condition for them to be perpendicular is $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Statement $4$ claims the sum is $1$,which is incorrect.
Therefore,statement $4$ is false.

(C) The direction cosines $(l, m, n)$ of a line are proportional to its direction ratios $(a, b, c)$.
Given the direction cosines are $\langle -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \rangle$.
We know that direction ratios can be any set of numbers proportional to the direction cosines.
If we multiply the direction cosines by the constant $k = 11$,we get the direction ratios as $\langle -9, 6, -2 \rangle$.
Therefore,the direction ratios are $\langle -9, 6, -2 \rangle$.
Hence,option $C$ is correct.

(C) Two lines are parallel if and only if their direction cosines are proportional.
Given the direction cosines of the two lines are $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,the condition for parallelism is:
$\frac{l_1}{l_2} = \frac{m_1}{m_2} = \frac{n_1}{n_2}$.
Therefore,option $(C)$ is correct.

(B) Let the direction cosines of the two lines be $(l_1, m_1, n_1) = (\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4})$ and $(l_2, m_2, n_2) = (-\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4})$.
The cosine of the angle $\theta$ between the two lines is given by the formula $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Substituting the values:
$\cos \theta = |(\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{\sqrt{3}}{4})(\frac{\sqrt{3}}{4})|$
$\cos \theta = |-\frac{3}{4} + \frac{1}{16} + \frac{3}{16}|$
$\cos \theta = |-\frac{12}{16} + \frac{4}{16}| = |-\frac{8}{16}| = |-\frac{1}{2}| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ}$.
Thus,the angle between the lines is $60^{\circ}$.

(B) The direction cosines of two lines are given as $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$.
Since these are direction cosines,we have the property $a_1^2 + b_1^2 + c_1^2 = 1$ and $a_2^2 + b_2^2 + c_2^2 = 1$.
The angle $\theta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Substituting the given values,we get $\cos \theta = |a_1 a_2 + b_1 b_2 + c_1 c_2|$.
Since the denominator is $\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{1} \cdot \sqrt{1} = 1$,the expression simplifies to $|a_1 a_2 + b_1 b_2 + c_1 c_2|$.

(D) Let the angles made by the line with the positive $x$,$y$,and $z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{4}$.
The direction cosines of the line are $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values: $\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{4}) + \cos^2 \gamma = 1$.
$(\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$.
$\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$.
$\frac{3}{4} + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$.
$\cos \gamma = \frac{1}{2}$ (since the angle is with the positive axis,$\cos \gamma > 0$).
Therefore,$\gamma = \frac{\pi}{3}$.

(D) Let the unit vectors along $OA$ and $OB$ be $\vec{a} = l_1 \hat{i} + m_1 \hat{j} + n_1 \hat{k}$ and $\vec{b} = l_2 \hat{i} + m_2 \hat{j} + n_2 \hat{k}$.
Since $\vec{a}$ and $\vec{b}$ are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.
The internal angular bisector of $\angle AOB$ is along the vector $\vec{v} = \vec{a} + \vec{b}$.
$\vec{v} = (l_1+l_2) \hat{i} + (m_1+m_2) \hat{j} + (n_1+n_2) \hat{k}$.
The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{(l_1+l_2)^2 + (m_1+m_2)^2 + (n_1+n_2)^2}$.
$|\vec{v}|^2 = (l_1^2+m_1^2+n_1^2) + (l_2^2+m_2^2+n_2^2) + 2(l_1l_2 + m_1m_2 + n_1n_2)$.
Since $l_i^2+m_i^2+n_i^2 = 1$ and $l_1l_2 + m_1m_2 + n_1n_2 = \cos \theta$,we have:
$|\vec{v}|^2 = 1 + 1 + 2 \cos \theta = 2(1+\cos \theta) = 4 \cos^2 \frac{\theta}{2}$.
Thus,$|\vec{v}| = 2 \cos \frac{\theta}{2}$.
The direction cosines of the internal bisector are the components of the unit vector $\frac{\vec{v}}{|\vec{v}|}$,which are:
$\frac{l_1+l_2}{2 \cos \frac{\theta}{2}}, \frac{m_1+m_2}{2 \cos \frac{\theta}{2}}, \frac{n_1+n_2}{2 \cos \frac{\theta}{2}}$.

(C) Given equations are:
$l+m+n=0$ ... $(i)$
$2lm+2ln-mn=0$ ... (ii)
From $(i)$,$m+n = -l$. Substituting this into (ii):
$2l(m+n) - mn = 0$
$2l(-l) - mn = 0 \Rightarrow mn = -2l^2$ ... (iii)
We know $l^2+m^2+n^2 = 1$. Also,$(m+n)^2 = m^2+n^2+2mn = (-l)^2 = l^2$.
So,$m^2+n^2 = l^2 - 2mn = l^2 - 2(-2l^2) = 5l^2$.
Substituting into $l^2+m^2+n^2 = 1$:
$l^2 + 5l^2 = 1 \Rightarrow 6l^2 = 1 \Rightarrow l^2 = \frac{1}{6}$.
Let the direction cosines of the two lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.
From $mn = -2l^2$ and $m+n = -l$,$m$ and $n$ are roots of the quadratic $t^2 + lt - 2l^2 = 0$.
$(t+2l)(t-l) = 0 \Rightarrow t = -2l, l$.
Thus,the direction ratios are proportional to $(l, -2l, l)$ and $(l, l, -2l)$.
Normalizing these,the direction cosines are proportional to $(1, -2, 1)$ and $(1, 1, -2)$.
Let $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(1) + (-2)(1) + (1)(-2)|}{\sqrt{1^2+(-2)^2+1^2} \sqrt{1^2+1^2+(-2)^2}} = \frac{|1-2-2|}{\sqrt{6}\sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(B) Let the line make angles $\alpha, \beta, \gamma$ with the $X, Y, Z$-axes respectively. The direction cosines are $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$.
Given $\beta = \frac{\pi}{4}$ and $\gamma = \frac{\pi}{3}$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values: $\cos^2 \alpha + \cos^2 \frac{\pi}{4} + \cos^2 \frac{\pi}{3} = 1$.
$\cos^2 \alpha + (\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 = 1$.
$\cos^2 \alpha + \frac{1}{2} + \frac{1}{4} = 1$.
$\cos^2 \alpha + \frac{3}{4} = 1 \Rightarrow \cos^2 \alpha = \frac{1}{4}$.
$\cos \alpha = \pm \frac{1}{2}$.
Since the angle $\alpha$ is obtuse,$\cos \alpha$ must be negative,so $\cos \alpha = -\frac{1}{2}$.
Therefore,$\alpha = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.

(D) We know that if a line makes angles $\alpha, \beta, \gamma$ with the positive direction of $X$-axis,$Y$-axis,and $Z$-axis respectively,then $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given $\alpha = \tan^{-1} \sqrt{7}$ and $\beta = \tan^{-1} \sqrt{\frac{5}{3}}$.
For $\alpha = \tan^{-1} \sqrt{7}$,we have $\cos \alpha = \frac{1}{\sqrt{1 + (\sqrt{7})^2}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$.
For $\beta = \tan^{-1} \sqrt{\frac{5}{3}}$,we have $\cos \beta = \frac{1}{\sqrt{1 + (\sqrt{5/3})^2}} = \frac{1}{\sqrt{1 + 5/3}} = \frac{1}{\sqrt{8/3}} = \sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{2\sqrt{2}}$.
Substituting these values into the identity $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$:
$\left(\frac{1}{2\sqrt{2}}\right)^2 + \left(\frac{\sqrt{3}}{2\sqrt{2}}\right)^2 + \cos^2 \gamma = 1$
$\frac{1}{8} + \frac{3}{8} + \cos^2 \gamma = 1$
$\frac{4}{8} + \cos^2 \gamma = 1$
$\frac{1}{2} + \cos^2 \gamma = 1$
$\cos^2 \gamma = \frac{1}{2}$
$\cos \gamma = \pm \frac{1}{\sqrt{2}}$
Therefore,$\gamma = \frac{\pi}{4}$ or $\gamma = \frac{3\pi}{4}$.

(D) Given the relations $l+m+n=0$ and $lm=0$.
From $lm=0$,we have either $l=0$ or $m=0$.
Case $1$: If $l=0$,then $m+n=0$,so $n=-m$. The direction cosines are $(0, m, -m)$. Since $l^2+m^2+n^2=1$,we have $0^2+m^2+(-m)^2=1$,which gives $2m^2=1$,so $m = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(0, 1, -1)$.
Case $2$: If $m=0$,then $l+n=0$,so $n=-l$. The direction cosines are $(l, 0, -l)$. Similarly,$l^2+0^2+(-l)^2=1$,which gives $2l^2=1$,so $l = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(1, 0, -1)$.
Let the two lines have direction vectors $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$.
The cosine of the angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$.
$|\vec{a}| = \sqrt{0^2+1^2+(-1)^2} = \sqrt{2}$ and $|\vec{b}| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(D) The direction cosines of a line are denoted by $(l, m, n)$. Given $(l, m, n) = (pq, q, q)$.
We know that for any line,$l^2 + m^2 + n^2 = 1$.
Substituting the given values: $(pq)^2 + q^2 + q^2 = 1$,which simplifies to $p^2q^2 + 2q^2 = 1$,or $q^2(p^2 + 2) = 1$.
Also,the direction cosine $l$ is given by $\cos(\alpha)$,where $\alpha$ is the angle with the $X$-axis.
Given $\alpha = \frac{\pi}{3}$,so $l = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
Since $l = pq$,we have $pq = \frac{1}{2}$,which implies $p^2q^2 = \frac{1}{4}$.
Substitute $p^2q^2 = \frac{1}{4}$ into the equation $p^2q^2 + 2q^2 = 1$:
$\frac{1}{4} + 2q^2 = 1$
$2q^2 = 1 - \frac{1}{4} = \frac{3}{4}$
$q^2 = \frac{3}{8}$.
Now,find $p^2$ using $p^2q^2 = \frac{1}{4}$:
$p^2(\frac{3}{8}) = \frac{1}{4}$
$p^2 = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3}$.
Finally,the ratio $p^2 : q^2 = \frac{2}{3} : \frac{3}{8} = \frac{2}{3} \times \frac{8}{3} = \frac{16}{9}$.

(B) Let the direction cosines of the line be $(l, m, n)$.
Since the line is equally inclined to the coordinate axes,the angles $\alpha, \beta, \gamma$ made by the line with the $x, y, z$ axes respectively are equal,i.e.,$\alpha = \beta = \gamma$.
Therefore,the direction cosines are equal: $l = m = n$.
We know that the sum of the squares of the direction cosines is $l^2 + m^2 + n^2 = 1$.
Substituting $l = m = n$,we get $l^2 + l^2 + l^2 = 1$,which implies $3l^2 = 1$.
Thus,$l^2 = \frac{1}{3}$,so $l = \pm \frac{1}{\sqrt{3}}$.
The angle $\beta$ made by the line with the $y$-axis is given by $\cos \beta = m = \pm \frac{1}{\sqrt{3}}$.
Taking the principal value,$\beta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

(A) Let the points be $O(0, 0, 0)$ and $P(2, 3, -1)$.
The direction ratios of the line $OP$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (2 - 0, 3 - 0, -1 - 0) = (2, 3, -1)$.
The distance $OP$ is given by $\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $\left(\frac{a}{r}, \frac{b}{r}, \frac{c}{r}\right) = \left(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)$.

(D) We know that the direction cosines of a line satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,where $\alpha, \beta, \gamma$ are the angles made by the line with the $X, Y,$ and $Z$-axes respectively.
Given $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{4}$.
Substituting these values into the relation:
$\cos^2 \left(\frac{\pi}{3}\right) + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \gamma = 1$
$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \gamma = 1$
$\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$
$\frac{3}{4} + \cos^2 \gamma = 1$
$\cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$
$\cos \gamma = \pm \frac{1}{2}$
Since $\gamma$ is the angle between $0$ and $\pi$,$\cos \gamma = \frac{1}{2}$ implies $\gamma = \frac{\pi}{3}$ (or $\cos \gamma = -\frac{1}{2}$ implies $\gamma = \frac{2\pi}{3}$).
Comparing with the options,the correct angle is $\frac{\pi}{3}$.

(B) Since the line makes an equal angle $\theta$ with the three coordinate axes,we have $\alpha = \beta = \gamma = \theta$.
Therefore,the direction cosines are $l = m = n = \cos \theta$.
We know that for any line,$l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $3 \cos^2 \theta = 1$,which implies $\cos^2 \theta = \frac{1}{3}$.
Thus,$\cos \theta = \frac{1}{\sqrt{3}}$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get $\sin^2 \theta = 1 - \frac{1}{3} = \frac{2}{3}$,so $\sin \theta = \frac{\sqrt{2}}{\sqrt{3}}$.
Finally,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{2}/\sqrt{3}}{1/\sqrt{3}} = \sqrt{2}$.

(B) The midpoint $M$ of the line segment joining $A(p, -4, 2)$ and $B(3, 2, -4)$ is given by $M = \left( \frac{p+3}{2}, \frac{-4+2}{2}, \frac{2-4}{2} \right) = \left( \frac{p+3}{2}, -1, -1 \right)$.
Since the ray passes through $C(5, q, 1)$ and $M\left( \frac{p+3}{2}, -1, -1 \right)$,its direction ratios are proportional to the differences of the coordinates: $\left( \frac{p+3}{2} - 5, -1 - q, -1 - 1 \right)$.
Given the direction ratios are $(2, 3, c)$,we equate them:
$1) \frac{p+3}{2} - 5 = 2 \implies \frac{p+3}{2} = 7 \implies p+3 = 14 \implies p = 11$.
$2) -1 - q = 3 \implies q = -4$.
$3) -1 - 1 = c \implies c = -2$.
Finally,we calculate $c \cdot (p + 7q) = -2 \cdot (11 + 7(-4)) = -2 \cdot (11 - 28) = -2 \cdot (-17) = 34$.

(A) Let $\alpha, \beta, \gamma$ be the angles made by the ray with the $X, Y$,and $Z$-axes respectively.
The direction cosines of the ray are $\cos \alpha, \cos \beta, \cos \gamma$.
We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given $\beta = \frac{\pi}{3}$ and $\gamma = \frac{\pi}{4}$.
Substituting these values:
$\cos^2 \alpha + \cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{4}) = 1$
$\cos^2 \alpha + (\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 = 1$
$\cos^2 \alpha + \frac{1}{4} + \frac{1}{2} = 1$
$\cos^2 \alpha + \frac{3}{4} = 1$
$\cos^2 \alpha = 1 - \frac{3}{4} = \frac{1}{4}$
Since $\sin^2 \alpha = 1 - \cos^2 \alpha$,we have:
$\sin^2 \alpha = 1 - \frac{1}{4} = \frac{3}{4}$
Therefore,$\sin \alpha = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that the sum of the squares of the direction cosines is always $1$,i.e.,$\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma = 1$.
Using the trigonometric identity $\sin ^2 \theta = 1 - \cos ^2 \theta$,we can rewrite the expression as:
$\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma = (1 - \cos ^2 \alpha) + (1 - \cos ^2 \beta) + (1 - \cos ^2 \gamma)$
$= 3 - (\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma)$
$= 3 - 1 = 2$.
Thus,the value is $2$.

(B) Let the direction ratios of the two lines be $\vec{a} = (2, -2, 1)$ and $\vec{b} = (1, -2, 2)$.
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:
$\cos \theta = \frac{|(2)(1) + (-2)(-2) + (1)(2)|}{\sqrt{2^2 + (-2)^2 + 1^2} \sqrt{1^2 + (-2)^2 + 2^2}}$
Calculating the numerator:
$|2 + 4 + 2| = 8$
Calculating the denominators:
$\sqrt{4 + 4 + 1} = \sqrt{9} = 3$
$\sqrt{1 + 4 + 4} = \sqrt{9} = 3$
Thus,$\cos \theta = \frac{8}{3 \times 3} = \frac{8}{9}$.
Therefore,$\theta = \cos^{-1}\left(\frac{8}{9}\right)$.

(D) Given equations are $2l + m + 2n = 0$ $(1)$ and $3l^2 + 5m^2 - 11n^2 = 0$ $(2)$.
From $(1)$,$m = -2l - 2n$.
Substitute $m$ into $(2)$: $3l^2 + 5(-2l - 2n)^2 - 11n^2 = 0$.
$3l^2 + 5(4l^2 + 8ln + 4n^2) - 11n^2 = 0$.
$3l^2 + 20l^2 + 40ln + 20n^2 - 11n^2 = 0$.
$23l^2 + 40ln + 9n^2 = 0$.
Divide by $n^2$: $23(\frac{l}{n})^2 + 40(\frac{l}{n}) + 9 = 0$.
Let $x = \frac{l}{n}$. Then $23x^2 + 40x + 9 = 0$.
Let the roots be $x_1 = \frac{l_1}{n_1}$ and $x_2 = \frac{l_2}{n_2}$.
Then $x_1 x_2 = \frac{l_1 l_2}{n_1 n_2} = \frac{9}{23}$.
Similarly,substituting $l = -\frac{m+2n}{2}$ into $(2)$ gives $3(\frac{m+2n}{2})^2 + 5m^2 - 11n^2 = 0$,which leads to $3(m^2 + 4mn + 4n^2) + 20m^2 - 44n^2 = 0$,so $23m^2 + 12mn - 32n^2 = 0$.
Dividing by $n^2$,$23(\frac{m}{n})^2 + 12(\frac{m}{n}) - 32 = 0$.
Let $y_1 = \frac{m_1}{n_1}$ and $y_2 = \frac{m_2}{n_2}$. Then $y_1 y_2 = \frac{m_1 m_2}{n_1 n_2} = -\frac{32}{23}$.
For two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Since $l_i^2 + m_i^2 + n_i^2 = 1$,we use the relation $l_1 l_2 + m_1 m_2 + n_1 n_2 = n_1 n_2 (x_1 x_2 + y_1 y_2 + 1) = n_1 n_2 (\frac{9}{23} - \frac{32}{23} + 1) = n_1 n_2 (\frac{-23}{23} + 1) = 0$.
Thus,$\cos \theta = 0$,which means $\theta = \frac{\pi}{2}$.