Consider the lines $l_1:\frac{x-1}{0}=\frac{y-1}{1}=\frac{2-z}{1}$ and $l_2:\frac{x}{2}=\frac{y}{0}=\frac{2z-1}{4}$, then which of the following are correct?

(A) Direction Ratio's of $l_1= <0,1,1 >$
(B) Direction Ratio's of $l_2=<2,0,2>$
(C) Angle between $l_1$, and $l_2 =\frac{\pi}{3}$
(D) Angle between $l_1$, and $l_2 =\frac{2\pi}{3}$

Choose the correct answer from the options given below:

(B) and (D) only

The correct answer is Option (1) → (B) and (D) only

(A) Direction Ratio's of $l_1= <0,1,1 >$ (Incorrect)
(B) Direction Ratio's of $l_2=<2,0,2>$ (Correct)
(C) Angle between $l_1$, and $l_2 =\frac{\pi}{3}$ (Incorrect)
(D) Angle between $l_1$, and $l_2 =\frac{2\pi}{3}$ (Correct)

Given:

Line $l_1$: $\displaystyle \frac{x - 1}{0} = \frac{y - 1}{1} = \frac{2 - z}{1}$

Line $l_2$: $\displaystyle \frac{x}{2} = \frac{y}{0} = \frac{2z - 1}{4}$

(A) Direction ratios of $l_1$:

$\displaystyle \frac{x - 1}{0} = \frac{y - 1}{1} = \frac{2 - z}{1}$ ⟹ Direction ratios: $0, 1, -1$

(A) is incorrect

(B) Direction ratios of $l_2$:

$\displaystyle \frac{x}{2} = \frac{y}{0} = \frac{2z - 1}{4}$

Middle term $\frac{y}{0}$ implies $y$ is constant ⇒ line is parallel to the XZ-plane.

Let $\displaystyle \frac{x}{2} = \frac{2z - 1}{4} = \lambda$

Then: $x = 2\lambda$, $2z - 1 = 4\lambda \Rightarrow z = 2\lambda + \frac{1}{2}$

So direction ratios = $\langle 2, 0, 2 \rangle$

(B) is correct

Now angle between lines $l_1$ and $l_2$:

Direction ratios:

• $\vec{d_1} = \langle 0, 1, -1 \rangle$
• $\vec{d_2} = \langle 2, 0, 2 \rangle$

Angle θ is given by:

$\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$

Dot product:

$\vec{d_1} \cdot \vec{d_2} = 0 \cdot 2 + 1 \cdot 0 + (-1) \cdot 2 = -2$

$|\vec{d_1}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$
$|\vec{d_2}| = \sqrt{2^2 + 0^2 + 2^2} = \sqrt{8}$

$\cos \theta = \frac{-2}{\sqrt{2} \cdot \sqrt{8}} = \frac{-2}{\sqrt{16}} = \frac{-2}{4} = -\frac{1}{2}$

$\Rightarrow \theta = \cos^{-1} \left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

(D) is correct, (C) is incorrect