Finding Angle C in Triangle ABC Using Position Vectors

The problem asks us to find the measure of angle C in a triangle ABC, given the position vectors of its vertices A, B, and C.

The position vectors of the vertices are given as:

• Position vector of A, \(\vec{a} = \hat{\text{j}} + \hat{\text{k}} = 0\hat{\text{i}} + 1\hat{\text{j}} + 1\hat{\text{k}}\)
• Position vector of B, \(\vec{b} = 3\hat{\text{i}} + \hat{\text{j}} + 5\hat{\text{k}}\)
• Position vector of C, \(\vec{c} = 3\hat{\text{j}} + 3\hat{\text{k}} = 0\hat{\text{i}} + 3\hat{\text{j}} + 3\hat{\text{k}}\)

To find angle C of the triangle, we need to consider the vectors forming the sides of the angle, which are \(\vec{CA}\) and \(\vec{CB}\). The angle C is the angle between these two vectors.

We can find the vector representing a side by subtracting the position vector of the initial point from the position vector of the terminal point. So, vector \(\vec{CA}\) is found by \(\vec{a} - \vec{c}\), and vector \(\vec{CB}\) is found by \(\vec{b} - \vec{c}\).

Calculating Vectors CA and CB

Let's calculate the vector \(\vec{CA}\):

\(\vec{CA} = \vec{a} - \vec{c} = (0\hat{\text{i}} + 1\hat{\text{j}} + 1\hat{\text{k}}) - (0\hat{\text{i}} + 3\hat{\text{j}} + 3\hat{\text{k}})\)

\(\vec{CA} = (0-0)\hat{\text{i}} + (1-3)\hat{\text{j}} + (1-3)\hat{\text{k}} = 0\hat{\text{i}} - 2\hat{\text{j}} - 2\hat{\text{k}}\)

\(\vec{CA} = -2\hat{\text{j}} - 2\hat{\text{k}}\)

Next, let's calculate the vector \(\vec{CB}\):

\(\vec{CB} = \vec{b} - \vec{c} = (3\hat{\text{i}} + 1\hat{\text{j}} + 5\hat{\text{k}}) - (0\hat{\text{i}} + 3\hat{\text{j}} + 3\hat{\text{k}})\)

\(\vec{CB} = (3-0)\hat{\text{i}} + (1-3)\hat{\text{j}} + (5-3)\hat{\text{k}} = 3\hat{\text{i}} - 2\hat{\text{j}} + 2\hat{\text{k}}\)

\(\vec{CB} = 3\hat{\text{i}} - 2\hat{\text{j}} + 2\hat{\text{k}}\)

Using the Dot Product to Find Angle C

The angle between two vectors \(\vec{u}\) and \(\vec{v}\) can be found using the dot product formula:

\(\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(\theta)\)

Where \(\theta\) is the angle between the vectors. In our case, the angle is C, and the vectors are \(\vec{CA}\) and \(\vec{CB}\).

\(\cos(C) = \frac{\vec{CA} \cdot \vec{CB}}{|\vec{CA}| |\vec{CB}|}\)

Let's calculate the dot product \(\vec{CA} \cdot \vec{CB}\):

\(\vec{CA} \cdot \vec{CB} = (-2\hat{\text{j}} - 2\hat{\text{k}}) \cdot (3\hat{\text{i}} - 2\hat{\text{j}} + 2\hat{\text{k}})\)

Remember that \(\hat{\text{i}} \cdot \hat{\text{i}} = \hat{\text{j}} \cdot \hat{\text{j}} = \hat{\text{k}} \cdot \hat{\text{k}} = 1\) and \(\hat{\text{i}} \cdot \hat{\text{j}} = \hat{\text{i}} \cdot \hat{\text{k}} = \hat{\text{j}} \cdot \hat{\text{k}} = 0\).

\(\vec{CA} \cdot \vec{CB} = (0)(3) + (-2)(-2) + (-2)(2)\)

\(\vec{CA} \cdot \vec{CB} = 0 + 4 - 4 = 0\)

The dot product \(\vec{CA} \cdot \vec{CB}\) is 0.

If the dot product of two non-zero vectors is zero, the vectors are orthogonal (perpendicular), and the angle between them is \(\frac{\pi}{2}\) (or 90 degrees).

Let's confirm that the vectors \(\vec{CA}\) and \(\vec{CB}\) are non-zero by calculating their magnitudes:

\(|\vec{CA}| = \sqrt{(0)^2 + (-2)^2 + (-2)^2} = \sqrt{0 + 4 + 4} = \sqrt{8}\)

\(|\vec{CB}| = \sqrt{(3)^2 + (-2)^2 + (2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}\)

Since both magnitudes are non-zero, the vectors are not zero vectors.

Using the dot product formula for \(\cos(C)\):

\(\cos(C) = \frac{0}{\sqrt{8} \sqrt{17}} = 0\)

The angle whose cosine is 0 is \(\frac{\pi}{2}\) radians (or 90 degrees).

\(C = \cos^{-1}(0) = \frac{\pi}{2}\)

Therefore, angle C is equal to \(\frac{\pi}{2}\).

The angle C in triangle ABC is \(\frac{\pi}{2}\).

Revision Table: Vector Operations for Angle Calculation

Additional Information: Vectors in Geometry Problems

Vectors are very useful tools for solving geometry problems, especially those involving points, lines, and angles in 2D or 3D space. Position vectors help represent points easily.

To find angles in a triangle using vectors, you can represent the sides as vectors. For example, in triangle ABC:

• Vector \(\vec{AB} = \vec{b} - \vec{a}\)
• Vector \(\vec{BC} = \vec{c} - \vec{b}\)
• Vector \(\vec{CA} = \vec{a} - \vec{c}\)

Note that \(\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}\), forming a closed loop.

To find angle A, you would find the angle between vectors \(\vec{AB}\) and \(\vec{AC}\) (or \(\vec{AB}\) and \(-\vec{CA}\)). Specifically, the angle at vertex A is the angle between \(\vec{AB}\) and \(\vec{AC}\). The dot product formula for angle A would be:

\(\cos(A) = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|}\)

Similarly, for angle B, you would find the angle between \(\vec{BA}\) and \(\vec{BC}\):

\(\cos(B) = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|}\)

And as calculated above, for angle C, we used vectors \(\vec{CA}\) and \(\vec{CB}\):

\(\cos(C) = \frac{\vec{CA} \cdot \vec{CB}}{|\vec{CA}| |\vec{CB}|}\)

The choice of vectors pointing away from the vertex (\(\vec{CA}\) and \(\vec{CB}\) for angle C) is standard for calculating the internal angle of the triangle using the dot product. If you used vectors pointing towards the vertex (e.g., \(\vec{AC}\) and \(\vec{BC}\)), the dot product result would be the same magnitude but potentially opposite sign, and you would need to be careful about the directionality.

In this specific problem, the dot product of the vectors forming angle C turned out to be 0, which directly indicates that angle C is a right angle (\(\frac{\pi}{2}\) radians).