Question

Show that (3\(\hat{i}\) - 4\(\hat{j}\) - 4\(\hat{k}\)), (2\(\hat{i}\) - \(\hat{j}\) + \(\hat{k}\)) and (\(\hat{i}\) - 3\(\hat{j}\) - 5\(\hat{k}\)) are the position vectors of vertices of a right angle triangle.

Hint:

To verify if vectors form a right-angled triangle, calculating dot products of the side vectors is often faster than calculating magnitudes and checking the Pythagorean theorem. If any dot product is zero, the sides are perpendicular.

Answer 1

Step 1: Understanding the Concept:
To show that three given position vectors form the vertices of a right-angled triangle, we can find the vectors representing the sides of the triangle.
If any two of these side vectors are perpendicular to each other, the triangle is a right-angled triangle.
Two vectors are perpendicular if their dot product (scalar product) is zero.
Step 2: Key Formula or Approach:
Let the position vectors of the vertices A, B, and C be \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) respectively.
The vectors representing the sides are \(\vec{AB} = \vec{b} - \vec{a}\), \(\vec{BC} = \vec{c} - \vec{b}\), and \(\vec{AC} = \vec{c} - \vec{a}\).
The condition for a right-angled triangle is that the dot product of any two side vectors is zero, for example, \(\vec{AB} \cdot \vec{AC} = 0\).
Step 3: Detailed Explanation or Calculation:
Let the given position vectors be:
\(\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}\)
\(\vec{b} = 2\hat{i} - \hat{j} + \hat{k}\)
\(\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}\)
Now, let's find the vectors for the sides of the triangle:
\(\vec{AB} = \vec{b} - \vec{a} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k})\)
\(\vec{AB} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}\)
\(\vec{BC} = \vec{c} - \vec{b} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k})\)
\(\vec{BC} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5 - 1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}\)
\(\vec{AC} = \vec{c} - \vec{a} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k})\)
\(\vec{AC} = (1-3)\hat{i} + (-3 - (-4))\hat{j} + (-5 - (-4))\hat{k} = -2\hat{i} + \hat{j} - \hat{k}\)
Next, we calculate the dot products of these side vectors:
\(\vec{AB} \cdot \vec{BC} = (-\hat{i} + 3\hat{j} + 5\hat{k}) \cdot (-\hat{i} - 2\hat{j} - 6\hat{k}) = (-1)(-1) + (3)(-2) + (5)(-6) = 1 - 6 - 30 = -35\)
\(\vec{BC} \cdot \vec{AC} = (-\hat{i} - 2\hat{j} - 6\hat{k}) \cdot (-2\hat{i} + \hat{j} - \hat{k}) = (-1)(-2) + (-2)(1) + (-6)(-1) = 2 - 2 + 6 = 6\)
\(\vec{AB} \cdot \vec{AC} = (-\hat{i} + 3\hat{j} + 5\hat{k}) \cdot (-2\hat{i} + \hat{j} - \hat{k}) = (-1)(-2) + (3)(1) + (5)(-1) = 2 + 3 - 5 = 0\)
Step 4: Final Answer:
Since the dot product \(\vec{AB} \cdot \vec{AC} = 0\), the sides represented by vectors \(\vec{AB}\) and \(\vec{AC}\) are perpendicular to each other.
Therefore, the triangle formed by the given position vectors is a right-angled triangle, with the right angle at vertex A.