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 Question:
We are given two lines in 3D space, $l_1$ and $l_2$, each defined by a pair of points.
We need to find the angle between these two lines, which is determined by the angle between their respective direction vectors.

Step 2: Key Formula or Approach: The direction vector $\vec{v}$ of a line joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:
\[ \vec{v} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k} \]
The cosine of the angle $\theta$ between two direction vectors $\vec{v_1}$ and $\vec{v_2}$ is given by:
\[ \cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|} \]

Step 3: Detailed Explanation:

• Let us determine the direction vector $\vec{v_1}$ for the line $l_1$ joining $(1, 1, 1)$ and $(3, 1, 3)$:
\[ \vec{v_1} = (3 - 1)\hat{i} + (1 - 1)\hat{j} + (3 - 1)\hat{k} \] \[ \vec{v_1} = 2\hat{i} + 0\hat{j} + 2\hat{k} \]
• Let us determine the direction vector $\vec{v_2}$ for the line $l_2$ joining $(0, 2, -1)$ and $(2, 0, 3)$:
\[ \vec{v_2} = (2 - 0)\hat{i} + (0 - 2)\hat{j} + (3 - (-1))\hat{k} \] \[ \vec{v_2} = 2\hat{i} - 2\hat{j} + 4\hat{k} \]
• Now, we calculate the dot product $\vec{v_1} \cdot \vec{v_2}$:
\[ \vec{v_1} \cdot \vec{v_2} = (2)(2) + (0)(-2) + (2)(4) \] \[ \vec{v_1} \cdot \vec{v_2} = 4 + 0 + 8 = 12 \]
• Next, we find the magnitude of each vector:
\[ |\vec{v_1}| = \sqrt{2^2 + 0^2 + 2^2} = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2} \] \[ |\vec{v_2}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \]
• We substitute these values into the angle formula:
\[ \cos \theta = \frac{12}{(2\sqrt{2})(2\sqrt{6})} = \frac{12}{4\sqrt{12}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]
• Since $\cos \theta = \frac{\sqrt{3}}{2}$, the angle $\theta$ is:
\[ \theta = 30^\circ \]

Step 4: Final Answer:
The angle between $l_1$ and $l_2$ is $30^\circ$.