Key Concepts and Formulas

• Vector Components: The components of a vector $\vec{AB}$ from point $A(x_1, y_1, z_1)$ to point $B(x_2, y_2, z_2)$ are given by $\vec{AB} = (x_2-x_1, y_2-y_1, z_2-z_1)$.
• Magnitude of a Vector: The magnitude of a vector $\vec{v} = (v_x, v_y, v_z)$ is given by $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.
• Angle Between Two Vectors (Dot Product Formula): The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is determined by their dot product and magnitudes: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$ where $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z$.

Step-by-Step Solution

Step 1: Identify the Vectors Forming the Angle The question asks for the angle $\angle \mathrm{QPR}$. This angle is formed at vertex P. To find this angle using the dot product formula, we must consider the two vectors that originate from P and extend to Q and R, respectively. These are $\vec{PQ}$ and $\vec{PR}$.

• The coordinates of the vertices are:
  • $P = (3, 2, 3)$
  • $Q = (4, 6, 2)$
  • $R = (7, 3, 2)$

Step 2: Calculate the Component Form of the Vectors We find the components of vectors $\vec{PQ}$ and $\vec{PR}$ by subtracting the coordinates of the initial point (P) from the coordinates of the terminal point (Q or R).

• Vector $\vec{PQ}$: $\vec{PQ} = Q - P = (4-3, 6-2, 2-3) = (1, 4, -1)$

• Vector $\vec{PR}$: $\vec{PR} = R - P = (7-3, 3-2, 2-3) = (4, 1, -1)$

Step 3: Calculate the Dot Product of the Vectors Next, we compute the dot product of $\vec{PQ}$ and $\vec{PR}$. For vectors $\vec{a} = (a_x, a_y, a_z)$ and $\vec{b} = (b_x, b_y, b_z)$, their dot product is $a_x b_x + a_y b_y + a_z b_z$.

• Using $\vec{PQ} = (1, 4, -1)$ and $\vec{PR} = (4, 1, -1)$: $\vec{PQ} \cdot \vec{PR} = (1)(4) + (4)(1) + (-1)(-1)$ $\vec{PQ} \cdot \vec{PR} = 4 + 4 + 1 = 9$

Step 4: Calculate the Magnitudes of the Vectors We now find the magnitudes of $\vec{PQ}$ and $\vec{PR}$ using the formula $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.

• Magnitude of $\vec{PQ}$: $|\vec{PQ}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18}$

• Magnitude of $\vec{PR}$: $|\vec{PR}| = \sqrt{4^2 + 1^2 + (-1)^2} = \sqrt{16 + 1 + 1} = \sqrt{18}$

Step 5: Substitute Values into the Angle Formula and Solve for $\theta$ Let $\theta = \angle \mathrm{QPR}$. We use the dot product formula to find $\cos \theta$: $\cos \theta = \frac{\vec{PQ} \cdot \vec{PR}}{|\vec{PQ}| |\vec{PR}|}$ Substitute the calculated values: $\cos \theta = \frac{9}{\sqrt{18} \cdot \sqrt{18}}$ $\cos \theta = \frac{9}{18}$ $\cos \theta = \frac{1}{2}$ To find the angle $\theta$, we take the inverse cosine: $\theta = \cos^{-1}\left(\frac{1}{2}\right)$ We know that the angle whose cosine is $1/2$ is $\frac{\pi}{3}$ radians (or $60^\circ$). $\theta = \frac{\pi}{3}$

Common Mistakes & Tips

• Incorrect Vector Selection: For an angle $\angle ABC$, always use vectors $\vec{BA}$ and $\vec{BC}$ (both originating from the vertex B) or $\vec{AB}$ and $\vec{CB}$ (both terminating at B). Using vectors like $\vec{AB}$ and $\vec{BC}$ will give the angle between $\vec{AB}$ and a vector parallel to $\vec{BC}$ originating from B, which is not $\angle ABC$.
• Arithmetic Errors: Be meticulous with subtraction of coordinates, multiplication in the dot product, and squaring/square-rooting in magnitude calculations, especially with negative numbers.
• Unit Circle Knowledge: Familiarity with common trigonometric values (e.g., $\cos(\pi/3) = 1/2$) is essential for quickly determining the angle from its cosine value.

Summary

To find the angle $\angle QPR$, we first determined the position vectors $\vec{PQ}$ and $\vec{PR}$ by subtracting the coordinates of P from Q and R, respectively. Then, we calculated their dot product and individual magnitudes. Finally, we applied the dot product formula for the angle between two vectors, $\cos \theta = \frac{\vec{PQ} \cdot \vec{PR}}{|\vec{PQ}| |\vec{PR}|}$, which yielded $\cos \theta = \frac{1}{2}$. This resulted in the angle $\theta = \frac{\pi}{3}$.

The final answer is $\boxed{\frac{\pi}{3}}$, which corresponds to option (D).