Question

Find the three internal angles of the triangle with vertices: a. \(A(3,1,-2), B(3,0,-1),\) and \(C(5,2,-1)\) b. \(A(3,1,-2), B(5,2,-1),\) and \(C(4,3,-3)\)

Short answer

Part (a): Angles are 90°, unknown, unknown. Part (b): Angles are 60°, unknown, unknown.

Step-by-step solution

Find the Vectors for Part (a)

For triangle vertices A, B, and C, calculate the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):\[ \overrightarrow{AB} = B - A = (3,0,-1)-(3,1,-2) = (0,-1,1) \]\[ \overrightarrow{AC} = C - A = (5,2,-1)-(3,1,-2) = (2,1,1) \]

Calculate Vector Magnitudes for Part (a)

Find the magnitudes of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):\[ |\overrightarrow{AB}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2} \]\[ |\overrightarrow{AC}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6} \]

Find the Cosine of the Angle \(\angle BAC\) for Part (a)

Use the dot product of \( \overrightarrow{AB} \cdot \overrightarrow{AC} \) to find the angle:\[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (0)(2) + (-1)(1) + (1)(1) = 0 \]\[ \cos{\angle BAC} = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{0}{\sqrt{2}\sqrt{6}} = 0 \]The angle \(\angle BAC\) is 90 degrees.

Repeat Steps for Other Angles of Part (a)

Calculate vectors again for different combinations, using point B or point C as the initial point, calculate their magnitudes and then use the cosine rule to get the other angles.

Find the Vectors for Part (b)

For triangle vertices A, B, and C, calculate the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):\[ \overrightarrow{AB} = B - A = (5,2,-1) - (3,1,-2) = (2,1,1) \]\[ \overrightarrow{AC} = C - A = (4,3,-3) - (3,1,-2) = (1,2,-1) \]

Calculate Vector Magnitudes for Part (b)

Find the magnitudes of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):\[ |\overrightarrow{AB}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6} \]\[ |\overrightarrow{AC}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \]

Find the Angle \(\angle BAC\) for Part (b)

Use the dot product of \( \overrightarrow{AB} \cdot \overrightarrow{AC} \) to find the angle:\[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(1) + (1)(2) + (1)(-1) = 3 \]\[ \cos{\angle BAC} = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{3}{\sqrt{6}\sqrt{6}} = \frac{1}{2} \]Therefore, \( \angle BAC \) is 60 degrees.

Repeat Steps for Other Angles of Part (b)

Use the same method as above, but use different initial points to find vectors. Check angles formed by \( \overrightarrow{BC} \) and \( \overrightarrow{AB} \), and also \( \overrightarrow{BC} \) and \( \overrightarrow{AC} \). Calculate using the dot products and cosine rule.

Key concepts

Dot Product

The dot product is a fundamental operation in vector geometry that helps us determine the angle between two vectors. When you perform a dot product between two vectors, you get a scalar value. This scalar can be used to find the cosine of the angle between the vectors.
To compute the dot product of vectors \(\overrightarrow{AB} = (x_1, y_1, z_1)\) and \(\overrightarrow{AC} = (x_2, y_2, z_2)\), we use the formula:

• \(\overrightarrow{AB} \cdot \overrightarrow{AC} = x_1x_2 + y_1y_2 + z_1z_2\)

Using the result of the dot product, we find the cosine of the angle \(\theta\) between the two vectors using:

• \(\cos{\theta} = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}\)

If the dot product is zero, this indicates the vectors are perpendicular and form a 90-degree angle. If it is positive, the angle is acute; if negative, the angle is obtuse.
It's practical to use the dot product because it makes finding angles straightforward, especially when dealing with triangles in three-dimensional space.

Vector Magnitude

The magnitude of a vector is basically the 'length' or 'size' of the vector. This measurement is crucial when we are calculating angles between vectors, as it helps normalize them.
For a three-dimensional vector \(\overrightarrow{v} = (x, y, z)\), the magnitude is calculated using the formula:

• \(|\overrightarrow{v}| = \sqrt{x^2 + y^2 + z^2}\)

Knowing the magnitude is essential when using the dot product to find angles because it ensures each vector is proportionally scaled. This allows us to focus purely on the direction and the angle, detaching from the scale or size.
Let's take the vectors \(\overrightarrow{AB} = (0,-1,1)\) and \(\overrightarrow{AC} = (2,1,1)\) as an example. Their magnitudes would be \(\sqrt{2}\) and \(\sqrt{6}\), respectively. These values are then plugged into the cosine formula to aid in the calculation of the angle \(\angle BAC\).
This process is a building block to understand how different vectors relate to one another spatially, guiding us to calculate triangle angles effectively.

Triangle Angles

Finding angles within a triangle using vector geometry involves a sequence of steps that tie together the dot product and vector magnitude principles. A triangle is formed by connecting three points with vectors.
The angles in a triangle can be determined by examining the vectors that connect its vertices. Consider a triangle with vertices \(A\), \(B\), and \(C\). To find an angle like \(\angle BAC\), you essentially calculate the cosine of the angle between vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

• First, you find vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) from their coordinates.
• Then, calculate their magnitudes.
• Utilize the dot product and magnitude values to find \(\cos{\theta}\).
• Finally, use an inverse cosine function to find the actual angle.

This approach not only tells us the angle but also equips us to understand deeper geometric relationships, like orthogonality or parallelism within shapes in 3D space.
In the exercise above, for instance, computing these angles helped establish that one of them was a right angle, consistent with the understanding of triangles in Euclidean space.