Q.Show that the vectors $2\hat{i}-\hat{j}+\hat{k}, \hat{i}-3\hat{j}-5\hat{k}$ and $3\hat{i}-4\hat{j}-4\hat{k}$ form the vertices of a right angled triangle.

Concept: Collinearity Condition & Dot Product for Right Angles

We need to show the three vectors represent sides of a right triangle. First, treat them as position vectors of points $A$, $B$, $C$. Then find the side vectors and check if any two are perpendicular.

Step 1: Label the vectors

Let

$A = 2\hat{i} - \hat{j} + \hat{k}$,

$B = \hat{i} - 3\hat{j} - 5\hat{k}$,

$C = 3\hat{i} - 4\hat{j} - 4\hat{k}$.

Step 2: Find side vectors

$\overrightarrow{AB} = B - A = (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$

$\overrightarrow{BC} = C - B = (3-1)\hat{i} + (-4+3)\hat{j} + (-4+5)\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$

$\overrightarrow{CA} = A - C = (2-3)\hat{i} + (-1+4)\hat{j} + (1+4)\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$

Step 3: Check dot products …

We are given three position vectors (points) in space. To show they form a right-angled triangle, we need to prove two things: first, that the three points are not collinear (so they actually make a triangle), and second, that one of the angles between the sides is $90^\circ$.

The key idea is the Collinearity Condition: three points $A$, $B$, $C$ are collinear if and only if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel (i.e., one is a scalar multiple of the other). If they are not parallel, the points form a triangle. Then, for a right angle, the dot product of the two side vectors meeting at that vertex must be zero.

Let’s label the given vectors as points:

• $A = 2\hat{i} - \hat{j} + \hat{k}$
• $B = \hat{i} - 3\hat{j} - 5\hat{k}$
• $C = 3\hat{i} - 4\hat{j} - 4\hat{k}$

We will compute the side vectors and check.

1. Find the side vectors of the triangle.

   Take $A$ as a reference vertex. Then:

   • $\overrightarrow{AB} = B - A = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k})$ $= (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k}$ $= -\hat{i} - 2\hat{j} - 6\hat{k}$
   • $\overrightarrow{AC} = C - A = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (2\hat{i} - \hat{j} + \hat{k})$ $= (3-2)\hat{i} + (-4+1)\hat{j} + (-4-1)\hat{k}$ $= \hat{i} - 3\hat{j} - 5\hat{k}$

   So the two sides from $A$ are $\overrightarrow{AB} = -\hat{i} - 2\hat{j} - 6\hat{k}$ and $\overrightarrow{AC} = \hat{i} - 3\hat{j} - 5\hat{k}$.

2. Check if the points are collinear.

   Are $\overrightarrow{AB}$ and $\overrightarrow{AC}$ parallel? For two vectors to be parallel, one must be a scalar multiple of the other. Compare components:

   If $\overrightarrow{AB} = k \cdot \overrightarrow{AC}$, then:

   • For $\hat{i}$: $-1 = k \cdot 1 \implies k = -1$
   • For $\hat{j}$: $-2 = k \cdot (-3) \implies -2 = 3$? No, $-2 \neq 3$. So no single $k$ works. Hence the vectors are not parallel, and $A$, $B$, $C$ are non-collinear — they form a triangle.

3. Check for a right angle at vertex $A$.

   The angle at $A$ is between $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Compute their dot product:

This is not zero, so angle $A$ is not $90^\circ$.

4. Check the other vertices. Let’s try vertex $B$. Compute $\overrightarrow{BA}$ and $\overrightarrow{BC}$:
   • $\overrightarrow{BA} = A - B = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k})$ $= \hat{i} + 2\hat{j} + 6\hat{k}$
   • $\overrightarrow{BC} = C - B = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k})$ …

Method: Detecting a Right Angle in a Triangle from Its Vertices

Use this to show three points form a right-angled triangle and to locate which vertex holds the right angle.

Steps

Step 1: Treat the points as position vectors and form the side vectors.

For vertices $A,B,C$ build $\vec{AB}$, $\vec{BC}$, $\vec{CA}$ by subtracting coordinates.

Step 2: Test each vertex with a dot product.

The angle at a vertex is $90^\circ$ exactly when the two side vectors meeting there have dot product $0$. Check the pair at each vertex in turn: …

Common Mistakes

Mistake 1: Testing only one vertex and stopping.

Why it's wrong: the right angle can sit at any of the three vertices, so a non-zero dot product at one vertex doesn't rule out a right triangle. Correct approach: test the vector pair at each vertex until a dot product is $0$.

Mistake 2: Using the wrong pair of side vectors at a vertex. …