Question

Let \( P(3, 2, 3) \), \( Q(4, 6, 2) \), and \( R(7, 3, 2) \) be the vertices of \( \triangle PQR \). Then, the angle \( \angle QPR \) is

• A. \( \frac{\pi}{6} \)
• B. \( \cos^{-1} \left( \frac{7}{18} \right) \)
• C. \( \cos^{-1} \left( \frac{1}{18} \right) \)
• D. \( \frac{\pi}{3} \)

Answer 1

The Correct Option is D

To find the angle \(\angle QPR\) of \(\triangle PQR\) with given vertices \( P(3, 2, 3) \), \( Q(4, 6, 2) \), and \( R(7, 3, 2) \), we will use the cosine rule in three dimensions. The cosine rule is applicable for any triangle, and it's given as:

\(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\) 

where \(\mathbf{a} = \overrightarrow{QP}\) and \(\mathbf{b} = \overrightarrow{RP}\).

First, we calculate vectors \(\mathbf{a}\) and \(\mathbf{b}\):

• \(\overrightarrow{QP} = P - Q = (3 - 4, 2 - 6, 3 - 2) = (-1, -4, 1)\)
• \(\overrightarrow{RP} = P - R = (3 - 7, 2 - 3, 3 - 2) = (-4, -1, 1)\)

Next, calculate the dot product \(\mathbf{a} \cdot \mathbf{b}\):

\(\mathbf{a} \cdot \mathbf{b} = (-1)(-4) + (-4)(-1) + (1)(1) = 4 + 4 + 1 = 9\)

Now, we calculate the magnitudes \(\|\mathbf{a}\|\) and \(\|\mathbf{b}\|\):

• \(\|\mathbf{a}\| = \sqrt{(-1)^2 + (-4)^2 + 1^2} = \sqrt{1 + 16 + 1} = \sqrt{18}\)
• \(\|\mathbf{b}\| = \sqrt{(-4)^2 + (-1)^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18}\)

Substitute these values into the cosine rule:

\(\cos \theta = \frac{9}{\sqrt{18} \times \sqrt{18}} = \frac{9}{18} = \frac{1}{2}\)

Therefore, \(\theta = \cos^{-1}\left(\frac{1}{2}\right)\). We know:

• \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\)

Hence, the angle \(\angle QPR\) is \(\frac{\pi}{3}\).

Therefore, the correct answer is \(\frac{\pi}{3}\).

Answer 2

Solution: To find the angle ∠QPR, we calculate the direction ratios of PR and PQ.

Step 1. Direction Ratio of PR:  
  PR = (7 − 3, 3 − 2, 2 − 3) = (4, 1, −1)

Step 2. Direction Ratio of PQ:
  PQ = (4 − 3, 6 − 2, 2 − 3) = (1, 4, −1)

Step 3. Calculating cosθ:
  \(\cosθ = \frac{4·1 + 1·4 + (−1)·(−1)}{\sqrt{18}·\sqrt{18}} = \frac{4 + 4 + 1}{18} = \frac{9}{18} = \frac{1}{2}\)
Step 4. Therefore:

  \(θ = \cos⁻¹\left(\frac{1}{2}\right) = \frac{π}{3}\)

The Correct Answer is:\( \frac{π}{3} \)