#### Find the equations of the 2 lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi}{3}$ each.

Given the equation of the line, we need to find the equations of two lines through the origin which intersect the given line.

According to the theorem, equation of a line with direction ratios d1 = (b1 , b2 , b3 ) that passes through the point (x1 , y1 , z1 ) is  expressed as:

$\frac{x-x_{1}}{b_{1}}=\frac{y-y_{1}}{b_{2}}=\frac{z-z_{1}}{b_{3}}$

We also know, the angle between two lines with direction ratios d1 and d2 respectively is given by:

$\theta = \cos^{-1}\left ( \frac{d_{1}d_{2}}{\left |d_{1} \right |\left |d_{2} \right |} \right )$

We use these theorems to find the equations of the two lines.

Let the equation of a line be:

$\theta = \cos^{-1}\left ( \frac{d_{1}d_{2}}{\left |d_{1} \right |\left |d_{2} \right |} \right )$

Given that it passes through the origin, (0, 0, 0)

Therefore, equation of both lines passing through the origin will be :

$\frac{x}{b_{1}}=\frac{y}{b_{2}}=\frac{z}{b_{3}}=\lambda \, \, .....(i)$

Let,

$\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\mu \, \, .....(ii)$

Direction ratio of the line = (2, 1, 1)

$\Rightarrow d_{1} = (2, 1, 1).... (iii)$

If we represent the direction ratio in terms of a position vector,

$d_{1}=2\hat{i}+\hat{j}+\hat{k} .....(iv)$

Any point on the line is given by (x, y, z). From (ii),

$\frac{x-3}{2}=\mu, \frac{y-3}{1}=\mu ,\frac{z}{1}=\mu$

$\\\text{take} \ \frac{x-3}{2}=\mu\\ \Rightarrow x-3=2\mu\\ \Rightarrow x=2\mu+3\\ \\ take \frac{y-3}{1}=\mu \\ \Rightarrow y-3=\mu\\ \Rightarrow y=\mu+3\\ \\ take \frac{z}{1}=\mu\\ \Rightarrow z=\mu$

Hence, any point on line (ii) is $P(2\mu + 3, \mu + 3, \mu)$

Since line (i) passes through the origin, we can say

$\left ( b_{1},b_{2},b_{3} \right )\equiv (2\mu + 3, \mu + 3, \mu)$

$\Rightarrow \: direction\: \: ratio\: of\; line(i)= (2\mu + 3, \mu + 3, \mu)\\ \Rightarrow d_{2}= (2\mu + 3, \mu + 3, \mu)....(v)$

We can represent the direction ratio in terms of position vector like:

$d_{2}= \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \: \: ....(vi)$

From the theorem, we know

$\cos \theta=\frac{d_{1}.d_{2}}{\left |d_{1} \right |\left |d_{2} \right |}$

If we substitute the values of d? and d? from (iv) and (vi) in the above equation, and putting $\theta=\frac{\pi}{3}$ from the question:

$\Rightarrow \cos \frac{\pi}{3}=\frac{\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )}{\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |}$

Solving the numerator,

$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 2\left (2\mu + 3 \right )+1\left ( \mu + 3 \right )+1. \mu$

$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 4\mu + 6+\mu + 3+ \mu$

$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 4\mu +\mu + \mu+6+3$

$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 6\mu+9$

Solving the denominator,

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{2^{2}+1^{2}+1^{2}}\sqrt{ \left (2\mu + 3 \right )^{2}+\left ( \mu + 3 \right )^{2}+ \mu^{2}}$

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{4+1+1}\sqrt{ \left (2\mu \right )^{2}+ 3^{2}+2\left ( 2\mu \right )\left ( 3 \right )+\left ( \mu \right )^{2}+3^{2}+ 2(\mu)(3)+\mu^{2}}$

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{4\mu^{2}+9+12\mu+u^{2}+9+6\mu+\mu^{2}}$

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{4\mu^{2}+u^{2}+\mu^{2}+12\mu+6\mu+9+9}$

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6\mu^{2}+18\mu+18}$

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6\left (\mu^{2}+3\mu+3 \right )}$

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6}\sqrt{\mu^{2}+3\mu+3}$

$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=6\sqrt{\mu^{2}+3\mu+3 }$

And cos π/3 = 1/2

Substituting the values, we get

$\Rightarrow \frac{1}{2}=\frac{6\mu+9}{6\sqrt{\mu^{2}+3\mu+3}}$

Performing cross multiplication,

$\Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=2\left (6\mu+9 \right )\\ \Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=2 \times 3\left (2\mu+3 \right )\\ \Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=6\left (2\mu+3 \right )\\ \Rightarrow \sqrt{\mu^{2}+3\mu+3}=2\mu+3$

Squaring both sides,

$\Rightarrow \left (\sqrt{\mu^{2}+3\mu+3} \right )^{2}=\left (2\mu+3 \right )^{2}\\ \Rightarrow \mu^{2}+3\mu+3=(2\mu)^{2}+3^{2}+2(2\mu)(3)\left [ \because (a+b)^{2}=a^{2}+b^{2}+2ab \right ]\\ \Rightarrow \mu^{2}+3\mu+3=4\mu^{2}+9+12\mu\\ \Rightarrow 4\mu^{2}-\mu^{2}+12\mu-3\mu+9-3=0\\ \Rightarrow 3\mu^{2}+9\mu+6=0$

$\Rightarrow 3\left ( \mu^{2}+3\mu+2 \right ) =0\\ \Rightarrow \mu^{2}+3\mu+2=0\\ \Rightarrow\mu^{2}+2\mu+\mu+2=0\\ \Rightarrow \mu\left ( \mu+2 \right )+\left ( \mu+2 \right )=0\\ \Rightarrow \left ( \mu+1 \right )+\left ( \mu+2 \right )=0\\$

$\Rightarrow \left ( \mu+1 \right )=0 \: \: or\: \left ( \mu+2 \right )=0\\ \Rightarrow \mu=-1 \: \: \: or\: \: \: \mu=-2$

Therefore, from equation (v)

Direction ratio =$(2\mu+ 3, \mu + 3, \mu)$

Putting μ = -1:

Direction Ratio = (2(-1) + 3, (-1) + 3, -1)

⇒ Direction Ratio = (-2 + 3, -1 + 3, -1)

⇒ Direction Ratio = (1, 2, -1) …(vi)

Now putting μ = -2:

Direction Ratio = (2(-2) + 3, (-2) + 3, -2)

⇒ Direction Ratio = (-4 + 3, -2 + 3, -2)

⇒ Direction Ratio = (-1, 1, -2) …(vii)

Using the direction ratios in (vi) and (vii) in equation (i);

$\frac{x}{b_{1}}=\frac{y}{b_{2}}=\frac{z}{b_{3}}=\lambda\\ \\ \\ \frac{x}{1}=\frac{y}{2}=\frac{z}{-1}=\lambda$

And,

$\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}=\lambda$

Therefore, the two required lines are $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}=\lambda$ and $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}=\lambda$