If p (3, 2, 3), Q (4, 6, 2), R (7, 3, 2) are the vertices of `DeltaPQR`, then find `angleQPR=`

Options

A. `cos^(-1)(1/18)`
B. `pi/6`
C. `pi/3`
D. `cos^(-1)(7/18)`

Answer

To find the angle \( \angle QPR \) given the vertices of triangle \( P(3, 2, 3) \), \( Q(4, 6, 2) \), and \( R(7, 3, 2) \), we will follow these steps:

### Step 1: Find the direction ratios of vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \)

1. **Calculate vector \( \overrightarrow{PQ} \)**:
\[ \overrightarrow{PQ} = Q - P = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1) \]

2. **Calculate vector \( \overrightarrow{PR} \)**:
\[ \overrightarrow{PR} = R - P = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1) \]

### Step 2: Use the dot product to find the cosine of the angle \( \angle QPR \)

The cosine of the angle \( \theta \) between two vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) can be found using the formula:
\[ \cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|} \]

1. **Calculate the dot product \( \overrightarrow{PQ} \cdot \overrightarrow{PR} \)**:
\[ \overrightarrow{PQ} \cdot \overrightarrow{PR} = (1)(4) + (4)(1) + (-1)(-1) = 4 + 4 + 1 = 9 \]

2. **Calculate the magnitudes \( |\overrightarrow{PQ}| \) and \( |\overrightarrow{PR}| \)**:
\[ |\overrightarrow{PQ}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2} \]
\[ |\overrightarrow{PR}| = \sqrt{4^2 + 1^2 + (-1)^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2} \]

### Step 3: Substitute into the cosine formula

Now substitute the values into the cosine formula:
\[ \cos \theta = \frac{9}{(3\sqrt{2})(3\sqrt{2})} = \frac{9}{18} = \frac{1}{2} \]

### Step 4: Find the angle \( \theta \)

Since \( \cos \theta = \frac{1}{2} \), we find:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]

### Conclusion

Thus, the angle \( \angle QPR \) is:
\[ \angle QPR = \frac{\pi}{3} \]