Q.If the vertices A, B, C of a triangle ABC are $(1, 2, 3), (-1, 0, 0), (0, 1, 2),$ respectively, then find $\angle ABC. [\angle ABC$ is the angle between the vectors $\vec{BA}$ and $\vec{BC}].$

Concept: Dot Product Angle — the angle between two vectors is given by

$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}$.

Step 1: Find the vectors

$\vec{BA} = A - B = (1 - (-1),\; 2 - 0,\; 3 - 0) = (2, 2, 3)$

$\vec{BC} = C - B = (0 - (-1),\; 1 - 0,\; 2 - 0) = (1, 1, 2)$

Step 2: Compute dot product and magnitudes

$\vec{BA} \cdot \vec{BC} = 2(1) + 2(1) + 3(2) = 2 + 2 + 6 = 10$

$|\vec{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$ …

The problem asks for $\angle ABC$, which is the angle at vertex B. The key idea is that this angle is formed by the two vectors that start at B and go to the other vertices: one to A and one to C. So we need the angle between $\vec{BA}$ and $\vec{BC}$.

Why the dot product? Because the dot product of two vectors gives us a direct link to the cosine of the angle between them: $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$. This is the cleanest way to find an angle in 3D space without drawing anything.

Let’s work through it step by step.

1. Identify the vectors from B.

   Vertex B is $(-1, 0, 0)$.

   • $\vec{BA}$ goes from B to A: $A - B = (1 - (-1), 2 - 0, 3 - 0) = (2, 2, 3)$.
   • $\vec{BC}$ goes from B to C: $C - B = (0 - (-1), 1 - 0, 2 - 0) = (1, 1, 2)$.

2. Compute the dot product $\vec{BA} \cdot \vec{BC}$.

   Multiply corresponding components and add:

   $(2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10$.

3. Find the magnitudes (lengths) of each vector.

   • $|\vec{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.
   • $|\vec{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.

4. Apply the dot product formula. …

Method: Finding the Angle at a Vertex Using the Dot Product

Use this for any 'angle between two vectors' or 'angle of a triangle at a vertex' problem, in 2D or 3D.

Steps

Step 1: Build the two vectors emanating FROM the vertex.

For the angle at $B$, use $\vec{BA}=A-B$ and $\vec{BC}=C-B$ — both must start at $B$. Using $\vec{AB}$ instead measures the wrong vertex's angle.

Step 2: Apply the cosine formula.

$\cos\theta = \frac{\vec{BA}\cdot\vec{BC}}{|\vec{BA}|\,|\vec{BC}|}$ …

Common Mistakes

Mistake 1: Using $\vec{AB}$ and $\vec{CB}$ instead of vectors from $B$.

Why it's wrong: the angle at $B$ is between vectors that START at $B$, namely $\vec{BA}$ and $\vec{BC}$; using $\vec{AB}$ points the wrong way and can flip the sign of $\cos\theta$. Correct approach: form $\vec{BA}=A-B$ and $\vec{BC}=C-B$.

Mistake 2: Dividing by only one magnitude. …