Q.Show that the points $A(2\hat{i}-\hat{j}+\hat{k})$, $B(\hat{i}-3\hat{j}-5\hat{k})$, $C(3\hat{i}-4\hat{j}-4\hat{k})$ are the vertices of a right angled triangle.

Key idea: two sides of a triangle meet at a right angle when their dot product is zero. Test the sides at each vertex.

Step 1 — Side vectors.

$\vec{AB} = B - A = -\hat{i} - 2\hat{j} - 6\hat{k}$

$\vec{BC} = C - B = 2\hat{i} - \hat{j} + \hat{k}$

$\vec{CA} = A - C = -\hat{i} + 3\hat{j} + 5\hat{k}$

Step 2 — Look for a zero dot product. At vertex $C$ the two sides meeting there are $\vec{CA}$ and $\vec{CB} = -\vec{BC} = -2\hat{i} + \hat{j} - \hat{k}$:

$\vec{CA}\cdot\vec{CB} = (-1)(-2) + (3)(1) + (5)(-1) = 2 + 3 - 5 = 0.$ …

To show three points make a right-angled triangle, we build the vectors along the sides and test whether any two sides that share a vertex are perpendicular. Two vectors are perpendicular exactly when their dot product is zero.

1. Write the side vectors

With $A(2\hat{i}-\hat{j}+\hat{k})$, $B(\hat{i}-3\hat{j}-5\hat{k})$, $C(3\hat{i}-4\hat{j}-4\hat{k})$:

2. Confirm a genuine triangle

$\vec{AB}$ and $\vec{CA}$ are not scalar multiples of one another (the component ratios $-1/-1$, $-2/3$, $-6/5$ disagree), so the points are non-collinear and a triangle really exists.

3. Test each vertex for a right angle

The dot product must be taken between the two sides that meet at the vertex being tested.

At $A$ (sides $\vec{AB}$ and $\vec{AC} = -\vec{CA} = \hat{i} - 3\hat{j} - 5\hat{k}$):

At $B$ (sides $\vec{BA} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{BC}$):

$\vec{BA}\cdot\vec{BC} = (1)(2) + (2)(-1) + (6)(1) = 2 - 2 + 6 = 6 \neq 0.$ …

Method: Proving three points form a right-angled triangle

Use this to test whether points $A,B,C$ (given by position vectors) make a right triangle, and at which vertex.

Steps

Step 1: Build the side vectors.

Compute $\overrightarrow{AB}=B-A$, $\overrightarrow{BC}=C-B$, $\overrightarrow{CA}=A-C$ (head minus tail). Confirm they are non-collinear so a genuine triangle exists.

Step 2: Test each vertex with a dot product of the two sides that MEET there.

The right angle at a vertex is between the two sides emanating from that vertex. E.g. at $C$ use $\overrightarrow{CA}$ and $\overrightarrow{CB}=-\overrightarrow{BC}$: …

Common Mistakes

Mistake 1: Dotting sides that do not meet at the tested vertex (e.g. $\overrightarrow{AB}\cdot\overrightarrow{BC}$).

Why it's wrong: the angle at a vertex is between the two sides emanating from that vertex; $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are not both based at one vertex, so their dot product is not the vertex angle's test. Correct approach: at $C$ use $\overrightarrow{CA}$ and $\overrightarrow{CB}$ (both starting at $C$); a zero result locates the right angle. …