Let the points on the plane P be equidistant from the points `(-4,2,1)and (2,-2,3)` . Then the acute angle between the plane P and the plane `2x+y+3z` = 1 is

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `(5pi)/12`

Answer

To solve the problem, we need to find the acute angle between the plane \( P \) (which consists of points equidistant from the points \((-4, 2, 1)\) and \((2, -2, 3)\)) and the plane given by the equation \( 2x + y + 3z = 1 \).

### Step 1: Find the equation of the plane \( P \)

The points on the plane \( P \) are equidistant from the points \((-4, 2, 1)\) and \((2, -2, 3)\). This means that for any point \( (x, y, z) \) on the plane \( P \), the distances to both points must be equal.

Using the distance formula, we set up the equation:

\[ \sqrt{(x + 4)^2 + (y - 2)^2 + (z - 1)^2} = \sqrt{(x - 2)^2 + (y + 2)^2 + (z - 3)^2} \]

### Step 2: Square both sides to eliminate the square roots

Squaring both sides gives:

\[ (x + 4)^2 + (y - 2)^2 + (z - 1)^2 = (x - 2)^2 + (y + 2)^2 + (z - 3)^2 \]

### Step 3: Expand both sides

Expanding both sides:

Left side:
\[ (x^2 + 8x + 16) + (y^2 - 4y + 4) + (z^2 - 2z + 1) = x^2 + y^2 + z^2 + 8x - 4y - 2z + 21 \]

Right side:
\[ (x^2 - 4x + 4) + (y^2 + 4y + 4) + (z^2 - 6z + 9) = x^2 + y^2 + z^2 - 4x + 4y - 6z + 17 \]

### Step 4: Set the expanded equations equal and simplify

Setting the two sides equal gives:

\[ x^2 + y^2 + z^2 + 8x - 4y - 2z + 21 = x^2 + y^2 + z^2 - 4x + 4y - 6z + 17 \]

Canceling \(x^2\), \(y^2\), and \(z^2\) from both sides:

\[ 8x - 4y - 2z + 21 = -4x + 4y - 6z + 17 \]

### Step 5: Rearranging terms

Rearranging gives:

\[ 12x - 8y + 4z + 4 = 0 \]

Dividing through by 4 simplifies to:

\[ 3x - 2y + z + 1 = 0 \]

### Step 6: Identify the normal vector of plane \( P \)

The normal vector \( \mathbf{n_1} \) of plane \( P \) is \( (3, -2, 1) \).

### Step 7: Identify the normal vector of the given plane

The normal vector \( \mathbf{n_2} \) of the plane \( 2x + y + 3z = 1 \) is \( (2, 1, 3) \).

### Step 8: Calculate the cosine of the angle between the two planes

The cosine of the angle \( \theta \) between the two planes can be found using the formula:

\[ \cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} \]

Calculating the dot product:

\[ \mathbf{n_1} \cdot \mathbf{n_2} = 3 \cdot 2 + (-2) \cdot 1 + 1 \cdot 3 = 6 - 2 + 3 = 7 \]

Calculating the magnitudes:

\[ |\mathbf{n_1}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]
\[ |\mathbf{n_2}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]

Thus, we have:

\[ \cos \theta = \frac{7}{\sqrt{14} \cdot \sqrt{14}} = \frac{7}{14} = \frac{1}{2} \]

### Step 9: Find the angle \( \theta \)

Since \( \cos \theta = \frac{1}{2} \), we find:

\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Step 10: Conclusion

The acute angle between the plane \( P \) and the plane \( 2x + y + 3z = 1 \) is \( 60^\circ \).