Question

Let $l_1$ be the line joining $(1, 1, 1)$ and $(3, 1, 3)$ and let $l_2$ be the line joining $(0, 2, -1)$ and $(2, 0, 3)$. What is the angle between $l_1$ and $l_2$?

• A. $30^\circ$
• B. $60^\circ$
• C. $45^\circ$
• D. $90^\circ$

Hint:

To simplify calculations, divide direction vectors by their common factor first.
For example, use $\vec{v_1'} = \hat{i} + \hat{k}$ and $\vec{v_2'} = \hat{i} - \hat{j} + 2\hat{k}$.
Then $\cos \theta = \frac{1 + 0 + 2}{\sqrt{2}\sqrt{6}} = \frac{3}{\sqrt{12}} = \frac{\sqrt{3}}{2}$, which immediately yields $30^\circ$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The angle between two lines is the angle between their direction vectors.
Key Formula or Approach:
Direction vector $\vec{v} = (x_{2}-x_{1})\hat{i} + (y_{2}-y_{1})\hat{j} + (z_{2}-z_{1})\hat{k}$.
$\cos \theta = \frac{|\vec{v}_{1} \cdot \vec{v}_{2}|}{|\vec{v}_{1}| |\vec{v}_{2}|}$.

Step 2: Detailed Explanation:
1. Direction vector $\vec{v_{1}$ for $l_{1}$:}
$\vec{v}_{1} = (3-1)\hat{i} + (1-1)\hat{j} + (3-1)\hat{k} = 2\hat{i} + 0\hat{j} + 2\hat{k}$.
2. Direction vector $\vec{v_{2}$ for $l_{2}$:}
$\vec{v}_{2} = (2-0)\hat{i} + (0-2)\hat{j} + (3-(-1))\hat{k} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.
3. Calculate magnitudes and dot product:
$|\vec{v}_{1}| = \sqrt{2^{2} + 0^{2} + 2^{2}} = \sqrt{8} = 2\sqrt{2}$.
$|\vec{v}_{2}| = \sqrt{2^{2} + (-2)^{2} + 4^{2}} = \sqrt{24} = 2\sqrt{6}$.
$\vec{v}_{1} \cdot \vec{v}_{2} = (2)(2) + (0)(-2) + (2)(4) = 4 + 0 + 8 = 12$.
4. Find angle:
$\cos \theta = \frac{12}{(2\sqrt{2})(2\sqrt{6})} = \frac{12}{4\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
$\theta = \cos^{-1}(\frac{\sqrt{3}}{2}) = 30^{\circ}$.

Step 3: Final Answer:
The angle is $30^{\circ}$.
This matches option (A).