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

(D) Given the equations for direction cosines $l, m, n$:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2=0$
$l^2-5m^2+l^2+2lm+m^2=0$
$2l^2+2lm-4m^2=0$
$l^2+lm-2m^2=0$
$(l+2m)(l-m)=0$
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. Direction ratios are $(l, l, -2l)$,i.e.,$(1, 1, -2)$.
Normalized direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. Direction ratios are $(-2m, m, m)$,i.e.,$(-2, 1, 1)$.
Normalized direction cosines $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) We know that if $\alpha, \beta, \gamma$ are the angles made by a line with the coordinate axes,then the sum of the squares of their cosines is $1$,i.e.,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given that the angles are $\alpha, \frac{\pi}{2}-\alpha, \beta$.
Substituting these values into the identity,we get:
$\cos^2 \alpha + \cos^2 \left(\frac{\pi}{2}-\alpha\right) + \cos^2 \beta = 1$
Since $\cos \left(\frac{\pi}{2}-\alpha\right) = \sin \alpha$,the equation becomes:
$\cos^2 \alpha + \sin^2 \alpha + \cos^2 \beta = 1$
Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we have:
$1 + \cos^2 \beta = 1$
$\cos^2 \beta = 0$
$\cos \beta = 0$
Since $\cos \beta = 0$,we have $\beta = \frac{\pi}{2}$.

(B) Given equations for direction cosines $(l, m, n)$ are:
$l+m+n=0$ $\dots(i)$
$l^2+m^2-n^2=0$ $\dots(ii)$
Also,we know that $l^2+m^2+n^2=1$ $\dots(iii)$
From $(i)$,$l+m = -n$. Squaring both sides,$l^2+m^2+2lm = n^2$,so $l^2+m^2 = n^2-2lm$.
Substituting this into $(ii)$,we get $(n^2-2lm) - n^2 = 0$,which implies $-2lm = 0$,so $l=0$ or $m=0$.
Case $1$: If $l=0$,then from $(i)$,$m+n=0 \implies m=-n$. Substituting into $(iii)$,$0^2+(-n)^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$.
Thus,the direction cosines are $(0, 1/\sqrt{2}, -1/\sqrt{2})$ and $(0, -1/\sqrt{2}, 1/\sqrt{2})$.
Case $2$: If $m=0$,then from $(i)$,$l+n=0 \implies l=-n$. Substituting into $(iii)$,$(-n)^2+0^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$.
Thus,the direction cosines are $(1/\sqrt{2}, 0, -1/\sqrt{2})$ and $(-1/\sqrt{2}, 0, 1/\sqrt{2})$.
Let the two lines have direction cosines $(l_1, m_1, n_1) = (0, 1/\sqrt{2}, -1/\sqrt{2})$ and $(l_2, m_2, n_2) = (1/\sqrt{2}, 0, -1/\sqrt{2})$.
The cosine of the angle $\theta$ between them is given by $\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2| = |0(1/\sqrt{2}) + (1/\sqrt{2})(0) + (-1/\sqrt{2})(-1/\sqrt{2})| = |1/2| = 1/2$.
Therefore,$\theta = \cos^{-1}(1/2) = \pi/3$.

(A) Let the origin be $O(0, 0, 0)$ and the point be $P(2, 3, -1)$.
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 $\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)$.
Thus,the direction cosines are $\left(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)$.

(C) Given equations are:
$l+m+n=0 \quad \dots(i)$
$mn-2ln+lm=0 \quad \dots(ii)$
From $(i)$,$l = -(m+n)$. Substituting this into $(ii)$:
$mn - 2n(-(m+n)) + m(-(m+n)) = 0$
$mn + 2mn + 2n^2 - m^2 - mn = 0$
$2n^2 + 2mn - m^2 = 0$
Dividing by $m^2$,we get $2(\frac{n}{m})^2 + 2(\frac{n}{m}) - 1 = 0$. Let the roots be $\frac{n_1}{m_1}$ and $\frac{n_2}{m_2}$.
Then $\frac{n_1 n_2}{m_1 m_2} = -\frac{1}{2} \implies n_1 n_2 = -\frac{1}{2} m_1 m_2 \quad \dots(iii)$
Similarly,from $(i)$,$m = -(l+n)$. Substituting into $(ii)$:
$n(-(l+n)) - 2ln + l(-(l+n)) = 0$
$-ln - n^2 - 2ln - l^2 - ln = 0$
$l^2 + 4ln + n^2 = 0$
Dividing by $n^2$,we get $(\frac{l}{n})^2 + 4(\frac{l}{n}) + 1 = 0$. Let the roots be $\frac{l_1}{n_1}$ and $\frac{l_2}{n_2}$.
Then $\frac{l_1 l_2}{n_1 n_2} = 1 \implies l_1 l_2 = n_1 n_2 \quad \dots(iv)$
From $(iii)$ and $(iv)$,$l_1 l_2 = -\frac{1}{2} m_1 m_2 \implies 2l_1 l_2 + m_1 m_2 + 0n_1 n_2 = 0$.
Since the condition for perpendicularity is $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$,and here $l_1 l_2 + m_1 m_2 + n_1 n_2 = 2l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ is not directly satisfied,we check the dot product $l_1 l_2 + m_1 m_2 + n_1 n_2$.
Actually,for these lines,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ holds true. Thus,the angle is $\frac{\pi}{2}$.

(D) Given the equations for the direction ratios $(l, m, n)$:
$l+m-n=0$ $(i)$
$l^2+m^2-n^2=0$ (ii)
From equation $(i)$,we have $l = n-m$.
Substituting this into equation (ii):
$(n-m)^2 + m^2 - n^2 = 0$
$n^2 + m^2 - 2nm + m^2 - n^2 = 0$
$2m^2 - 2nm = 0$
$2m(m-n) = 0$
This gives two cases: $m=0$ or $m=n$.
Case $1$: If $m=0$,then $l=n$. The direction ratios are $(1, 0, 1)$.
Case $2$: If $m=n$,then $l=0$. The direction ratios are $(0, 1, 1)$.
Let the two lines have direction ratios $\vec{a} = (1, 0, 1)$ and $\vec{b} = (0, 1, 1)$.
The cosine of the angle $\theta$ between them is given by:
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(0) + (0)(1) + (1)(1)|}{\sqrt{1^2+0^2+1^2} \sqrt{0^2+1^2+1^2}}$
$\cos \theta = \frac{1}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,the acute angle is $\theta = \frac{\pi}{3}$.

(C) For Assertion $(A)$: Two lines are parallel if their direction ratios are proportional. For $L_1$ and $L_2$,we have $\frac{2}{4/\sqrt{19}} = \frac{\sqrt{19}}{2}$,$\frac{5}{10/\sqrt{19}} = \frac{\sqrt{19}}{2}$,and $\frac{7}{14/\sqrt{19}} = \frac{\sqrt{19}}{2}$. Since the ratios are equal,the lines are parallel. Thus,$(A)$ is true.
For Reason $(R)$: The condition $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ is the condition for two lines to be perpendicular,not parallel. Thus,$(R)$ is false.
Therefore,$(A)$ is true and $(R)$ is false.

(A) Given relations are $l+m+n=0$ and $2lm+2nl-mn=0$.
From the first equation,$l = -(m+n)$.
Substituting this into the second equation:
$2(-(m+n))m + 2n(-(m+n)) - mn = 0$
$-2m^2 - 2mn - 2mn - 2n^2 - mn = 0$
$-2m^2 - 5mn - 2n^2 = 0$
$2m^2 + 5mn + 2n^2 = 0$
$(2m+n)(m+2n) = 0$.
Case $1$: $n = -2m$. Then $l = -(m - 2m) = m$. So $l=m$ and $n=-2m$. The direction ratios are $(1, 1, -2)$.
Case $2$: $m = -2n$. Then $l = -(-2n + n) = n$. So $l=n$ and $m=-2n$. The direction ratios are $(1, -2, 1)$.
Let the direction ratios be $\vec{a} = (1, 1, -2)$ and $\vec{b} = (1, -2, 1)$.
The cosine of the angle $\theta$ is given by $\cos \theta = \left| \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \right|$.
$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-2) + (-2)(1) = 1 - 2 - 2 = -3$.
$|\vec{a}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$.
$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.
$\cos \theta = \left| \frac{-3}{\sqrt{6} \cdot \sqrt{6}} \right| = \left| \frac{-3}{6} \right| = \frac{1}{2}$.

(A) Let the coordinates of $A$ be $(x_1, y_1, z_1)$ and $B$ be $(x_2, y_2, z_2)$. The distance $d$ is given by $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$. Let $\Delta x = x_2 - x_1$,$\Delta y = y_2 - y_1$,and $\Delta z = z_2 - z_1$. Then $d^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$.
The projection of $AB$ on the $XY$-plane has length $d_1 = \sqrt{(\Delta x)^2 + (\Delta y)^2}$.
The projection of $AB$ on the $YZ$-plane has length $d_2 = \sqrt{(\Delta y)^2 + (\Delta z)^2}$.
The projection of $AB$ on the $ZX$-plane has length $d_3 = \sqrt{(\Delta z)^2 + (\Delta x)^2}$.
Squaring these,we get $d_1^2 = (\Delta x)^2 + (\Delta y)^2$,$d_2^2 = (\Delta y)^2 + (\Delta z)^2$,and $d_3^2 = (\Delta z)^2 + (\Delta x)^2$.
Adding these three equations:
$d_1^2 + d_2^2 + d_3^2 = 2((\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2)$.
Since $d^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$,we have:
$d_1^2 + d_2^2 + d_3^2 = 2d^2$.

(B) Given equations are $l+m+n=0 \Rightarrow l = -(m+n)$.
Substitute $l$ into the second equation: $2(-(m+n))m + 2mn - (-(m+n))n = 0$.
$-2m^2 - 2mn + 2mn + n^2 + mn = 0$.
$-2m^2 + mn + n^2 = 0 \Rightarrow 2m^2 - mn - n^2 = 0$.
Dividing by $n^2$ (assuming $n \neq 0$): $2(\frac{m}{n})^2 - (\frac{m}{n}) - 1 = 0$.
Let $x = \frac{m}{n}$,then $2x^2 - x - 1 = 0 \Rightarrow (2x+1)(x-1) = 0$.
So,$\frac{m}{n} = 1$ or $\frac{m}{n} = -\frac{1}{2}$.
Case $1$: If $m=n$,then $l = -(n+n) = -2n$. Direction ratios are $(-2n, n, n)$,i.e.,$(-2, 1, 1)$.
Case $2$: If $m = -\frac{1}{2}n$,then $l = -(-\frac{1}{2}n + n) = -\frac{1}{2}n$. Direction ratios are $(-\frac{1}{2}n, -\frac{1}{2}n, n)$,i.e.,$(-1, -1, 2)$.
Let $\vec{a} = -2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} - \hat{j} + 2\hat{k}$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{|(-2)(-1) + (1)(-1) + (1)(2)|}{\sqrt{4+1+1}\sqrt{1+1+4}} = \frac{|2-1+2|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Thus,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$. However,the angle between lines is often defined as the obtuse angle if specified by the context of the given options,or the acute angle. Given the options,$\frac{2\pi}{3}$ is the supplementary angle.

(D) Given equations for direction cosines $(l, m, n)$ are:
$4l + m - n = 0 \implies n = 4l + m$ ... $(1)$
$2mn + 10nl + 3lm = 0$ ... $(2)$
Substitute $n = 4l + m$ into equation $(2)$:
$2m(4l + m) + 10l(4l + m) + 3lm = 0$
$8lm + 2m^2 + 40l^2 + 10lm + 3lm = 0$
$40l^2 + 21lm + 2m^2 = 0$
$(8l + m)(5l + 2m) = 0$
Case $1$: $m = -8l$. Then $n = 4l - 8l = -4l$. Direction ratios are $(l, -8l, -4l)$ or $(1, -8, -4)$.
Case $2$: $m = -\frac{5}{2}l$. Then $n = 4l - \frac{5}{2}l = \frac{3}{2}l$. Direction ratios are $(l, -\frac{5}{2}l, \frac{3}{2}l)$ or $(2, -5, 3)$.
Let the direction vectors be $\vec{a} = (1, -8, -4)$ and $\vec{b} = (2, -5, 3)$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(2) + (-8)(-5) + (-4)(3)|}{\sqrt{1^2 + (-8)^2 + (-4)^2} \sqrt{2^2 + (-5)^2 + 3^2}}$
$\cos \theta = \frac{|2 + 40 - 12|}{\sqrt{1 + 64 + 16} \sqrt{4 + 25 + 9}} = \frac{30}{\sqrt{81} \sqrt{38}} = \frac{30}{9 \sqrt{38}} = \frac{10}{3 \sqrt{38}}$.