Angle Between Two 3D Lines from Their Direction Vectors

We are given two lines in vector form:

$$\begin{aligned} \text{Line 1: } & \vec r = (3, 2, -4) + \lambda (1, 2, 2) \\ \text{Line 2: } & \vec r = (5, -2, 0) + \mu (3, 2, 6) \end{aligned}$$

Each line is written as:

$$\vec r = \vec a + t\,\vec b$$

where:

• $\vec a$ is a fixed point on the line,
• $\vec b$ is the direction vector of the line,
• $t$ is a scalar parameter.

So for our lines:

$$\vec b_1 = (1, 2, 2), \quad \vec b_2 = (3, 2, 6)$$

The angle between the two lines is the same as the angle between their direction vectors $\vec b_1$ and $\vec b_2$.

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Step 1: Recall the dot product formula

For any two vectors $\vec u$ and $\vec v$:

$$\vec u \cdot \vec v = |\vec u|\,|\vec v|\cos\theta$$

where $\theta$ is the angle between them.

Rearranging:

$$\cos\theta = \dfrac{\vec u \cdot \vec v}{|\vec u|\,|\vec v|}$$

We will apply this to $\vec b_1$ and $\vec b_2$.

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Step 2: Compute the dot product $\vec b_1 \cdot \vec b_2$

$$\vec b_1 = (1,2,2), \quad \vec b_2 = (3,2,6)$$

Dot product:

$$\begin{aligned} \vec b_1 \cdot \vec b_2 &= 1\cdot 3 + 2\cdot 2 + 2\cdot 6 \\ &= 3 + 4 + 12 \\ &= 19 \end{aligned}$$

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Step 3: Compute the magnitudes $|\vec b_1|$ and $|\vec b_2|$

Magnitude of a vector $(x,y,z)$ is:

$$|(x,y,z)| = \sqrt{x^2 + y^2 + z^2}$$

For $\vec b_1 = (1,2,2)$:

$$|\vec b_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

For $\vec b_2 = (3,2,6)$:

$$|\vec b_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$

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Step 4: Use the cosine formula

$$\cos\theta = \left|\dfrac{\vec b_1 \cdot \vec b_2}{|\vec b_1|\,|\vec b_2|}\right| = \left|\dfrac{19}{3\times 7}\right| = \dfrac{19}{21}$$

(We take the absolute value because the angle between two lines is usually taken as the acute angle.)

So:

$$\theta = \cos^{-1}\left(\dfrac{19}{21}\right)$$

That is the angle between the given pair of lines.

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Visual idea (what the canvas will show)

To make this easy to see:

1. We place the tail of both direction vectors at the center of the canvas (representing the origin).
2. We draw:
   • A cyan arrow for $\vec b_1 = (1,2,2)$
   • A pink arrow for $\vec b_2 = (3,2,6)$
3. We draw the plane spanned by these two vectors as a soft neon grid.
4. We highlight the angle between them with a yellow arc and display the numeric value of $\theta$.
5. A slider lets you rotate the camera around the origin, so you can see the vectors and the angle from different viewpoints.

This helps you connect the algebraic formula (dot product) with the geometric picture (angle between arrows in 3D).