Find the angle between two planes `2x+y-5z=0 and 4x -3y+7z=0`

Answer

To find the angle between the two planes given by the equations \(2x + y - 5z = 0\) and \(4x - 3y + 7z = 0\), we can use the formula for the angle \(\theta\) between two planes. The formula is:

\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

### Step 1: Identify the coefficients of the planes
For the first plane \(2x + y - 5z = 0\):
- \(a_1 = 2\)
- \(b_1 = 1\)
- \(c_1 = -5\)

For the second plane \(4x - 3y + 7z = 0\):
- \(a_2 = 4\)
- \(b_2 = -3\)
- \(c_2 = 7\)

### Step 2: Substitute the coefficients into the formula
Now, substitute these coefficients into the formula for \(\cos \theta\):

\[ \cos \theta = \frac{(2)(4) + (1)(-3) + (-5)(7)}{\sqrt{(2^2 + 1^2 + (-5)^2)} \cdot \sqrt{(4^2 + (-3)^2 + 7^2)}} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ = 8 - 3 - 35 = 8 - 3 - 35 = -30 \]

### Step 4: Calculate the denominator
Now calculate the denominator:

1. For the first plane:
\[ \sqrt{(2^2 + 1^2 + (-5)^2)} = \sqrt{4 + 1 + 25} = \sqrt{30} \]

2. For the second plane:
\[ \sqrt{(4^2 + (-3)^2 + 7^2)} = \sqrt{16 + 9 + 49} = \sqrt{74} \]

Now combine these results:

\[ \text{Denominator} = \sqrt{30} \cdot \sqrt{74} = \sqrt{30 \cdot 74} = \sqrt{2220} \]

### Step 5: Combine results to find \(\cos \theta\)
Now we can find \(\cos \theta\):

\[ \cos \theta = \frac{-30}{\sqrt{2220}} \]

### Step 6: Find \(\theta\)
To find \(\theta\), take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{-30}{\sqrt{2220}}\right) \]

### Final Answer
Thus, the angle between the two planes is given by:

\[ \theta = \cos^{-1}\left(\frac{-30}{\sqrt{2220}}\right) \]