To find the angles of triangle $\Delta PQR$, where the points are given as:

• $P(0, -1, 3)$
• $Q(2, 2, 1)$
• $R(-2, 2, 4)$

we can proceed by using the dot product of vectors to calculate the angles. Here's the step-by-step process:

Step 1: Find vectors PQ, PR, and QR

We first compute the vectors formed by the points.

• Vector $\mathbf{PQ}$: $\mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (2 - 0, 2 - (-1), 1 - 3) = (2, 3, -2)$

• Vector $\mathbf{PR}$: $\mathbf{PR} = \mathbf{R} - \mathbf{P} = (-2 - 0, 2 - (-1), 4 - 3) = (-2, 3, 1)$

• Vector $\mathbf{QR}$: $\mathbf{QR} = \mathbf{R} - \mathbf{Q} = (-2 - 2, 2 - 2, 4 - 1) = (-4, 0, 3)$

Step 2: Find the magnitudes of vectors

Next, we find the magnitudes of the vectors.

• $|\mathbf{PQ}| = \sqrt{2^2 + 3^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$
• $|\mathbf{PR}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$
• $|\mathbf{QR}| = \sqrt{(-4)^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Step 3: Compute dot products

Now, let's compute the dot products of the pairs of vectors.

• $\mathbf{PQ} \cdot \mathbf{PR} = (2)(-2) + (3)(3) + (-2)(1) = -4 + 9 - 2 = 3$
• $\mathbf{PQ} \cdot \mathbf{QR} = (2)(-4) + (3)(0) + (-2)(3) = -8 + 0 - 6 = -14$
• $\mathbf{PR} \cdot \mathbf{QR} = (-2)(-4) + (3)(0) + (1)(3) = 8 + 0 + 3 = 11$

Step 4: Find the angles using the cosine formula

We can now find the cosine of the angles using the dot product formula: $\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}$

• Angle $\angle PQR$: $\cos(\theta) = \frac{\mathbf{PQ} \cdot \mathbf{QR}}{|\mathbf{PQ}| |\mathbf{QR}|} = \frac{-14}{\sqrt{17} \times 5} = \frac{-14}{5\sqrt{17}} \approx -0.852$ $\theta_{PQR} = \cos^{-1}(-0.852) \approx 148.2^\circ$

• Angle $\angle QPR$: $\cos(\theta) = \frac{\mathbf{PQ} \cdot \mathbf{PR}}{|\mathbf{PQ}| |\mathbf{PR}|} = \frac{3}{\sqrt{17} \times \sqrt{14}} = \frac{3}{\sqrt{238}} \approx 0.194$ $\theta_{QPR} = \cos^{-1}(0.194) \approx 78.8^\circ$

• Angle $\angle PRQ$: $\cos(\theta) = \frac{\mathbf{PR} \cdot \mathbf{QR}}{|\mathbf{PR}| |\mathbf{QR}|} = \frac{11}{\sqrt{14} \times 5} = \frac{11}{5\sqrt{14}} \approx 0.742$ $\theta_{PRQ} = \cos^{-1}(0.742) \approx 42.9^\circ$

Final Answer:

The approximate angles of $\Delta PQR$ are:

• $\theta_{PQR} \approx 148.2^\circ$
• $\theta_{QPR} \approx 78.8^\circ$
• $\theta_{PRQ} \approx 42.9^\circ$

Would you like further details on the steps or have any questions?

Related Questions:

1. How do you calculate the cross product of vectors in 3D?
2. What is the geometric meaning of the dot product?
3. How can vector magnitudes help in determining distances in 3D space?
4. What is the law of cosines in vector form?
5. How would the method change for points in 2D space?

Tip: When using the dot product for angles, always ensure the vectors are correctly formed from the point coordinates to avoid mistakes!