Find the angle of intersection of the plane 2x-2y-3z=1 with the plane -1x+4y-3z=4.

To find the angle of intersection between two planes, we use the normals of the planes. The angle \( \theta \) between the two planes is the same as the angle between their normal vectors, which can be calculated using the dot product formula.

### Step 1: Identify the normal vectors of the planes
For a plane of the form \( ax + by + cz = d \), the normal vector is \( \vec{n} = (a, b, c) \).

1. **First plane**: \( 2x - 2y - 3z = 1 \)
   - Normal vector \( \vec{n_1} = (2, -2, -3) \)

2. **Second plane**: \( -x + 4y - 3z = 4 \)
   - Normal vector \( \vec{n_2} = (-1, 4, -3) \)

### Step 2: Use the dot product to find the cosine of the angle
The dot product of two vectors \( \vec{n_1} \) and \( \vec{n_2} \) is given by:

\[
\vec{n_1} \cdot \vec{n_2} = (2)(-1) + (-2)(4) + (-3)(-3) = -2 - 8 + 9 = -1
\]

The magnitudes of the vectors are:

\[
|\vec{n_1}| = \sqrt{(2)^2 + (-2)^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17}
\]

\[
|\vec{n_2}| = \sqrt{(-1)^2 + (4)^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}
\]

Now, the cosine of the angle \( \theta \) between the two planes is:

\[
\cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} = \frac{-1}{\sqrt{17} \cdot \sqrt{26}} = \frac{-1}{\sqrt{442}}
\]

### Step 3: Calculate the angle \( \theta \)

\[
\theta = \cos^{-1}\left( \frac{-1}{\sqrt{442}} \right)
\]

Let me compute the exact value of this expression.

The angle of intersection between the two planes is approximately \( 92.73^\circ \).

Would you like any additional details or clarifications?

Here are 5 related questions you might find interesting:
1. How do we calculate the normal vector for a plane given its equation?
2. What is the geometric interpretation of the dot product between two vectors?
3. How can we find the distance between two parallel planes?
4. What does it mean if the dot product of two normal vectors is zero?
5. Can two planes intersect in more than one line?

**Tip:** Always double-check that you're working with the correct normal vectors when solving problems involving angles between planes.

Find the angle of intersection of the plane 2x - 2y - 3z = 1 with the plane -x + 4y - 3z = 4.

Discover how to find the angle of intersection between two planes using the dot product formula. This example features the planes 2x - 2y - 3z = 1 and -x + 4y - 3z = 4, and covers key geometric concepts such as normal vectors and vector magnitudes. Suitable for students in Grades 11-12.

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