Question

If the position vectors of the points A and B are \(\hat{i}+\hat{j}+\hat{k}\) and \(2\hat{i}+5\hat{j}\) respectively, then find the unit vector along the straight line AB.

Hint:

Remember that the vector from point A to point B is always "Position Vector of B - Position Vector of A". A common mistake is to subtract in the wrong order. Always think "final minus initial".

Answer 1

Step 1: Understanding the Concept:
To find the unit vector along the line AB, we first need to find the vector \(\vec{AB}\) by subtracting the position vector of the initial point A from the position vector of the terminal point B. Then, we find the unit vector by dividing \(\vec{AB}\) by its magnitude.
Step 2: Key Formula or Approach:
1. Let the position vectors of A and B be \(\vec{OA}\) and \(\vec{OB}\) respectively.
2. The vector along the line is \(\vec{AB} = \vec{OB} - \vec{OA}\).
3. The unit vector is \(\widehat{AB} = \frac{\vec{AB}}{|\vec{AB}|}\).
Step 3: Detailed Explanation:
We are given the position vectors: \[ \vec{OA} = \hat{i} + \hat{j} + \hat{k} \] \[ \vec{OB} = 2\hat{i} + 5\hat{j} + 0\hat{k} \] First, calculate the vector \(\vec{AB}\): \[ \vec{AB} = (2\hat{i} + 5\hat{j}) - (\hat{i} + \hat{j} + \hat{k}) \] \[ \vec{AB} = (2-1)\hat{i} + (5-1)\hat{j} + (0-1)\hat{k} \] \[ \vec{AB} = 1\hat{i} + 4\hat{j} - 1\hat{k} \] Next, calculate the magnitude of \(\vec{AB}\): \[ |\vec{AB}| = \sqrt{1^2 + 4^2 + (-1)^2} \] \[ |\vec{AB}| = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2} \] Finally, find the unit vector by dividing \(\vec{AB}\) by its magnitude: \[ \widehat{AB} = \frac{\hat{i} + 4\hat{j} - \hat{k}}{\sqrt{18}} \] \[ \widehat{AB} = \frac{1}{\sqrt{18}}\hat{i} + \frac{4}{\sqrt{18}}\hat{j} - \frac{1}{\sqrt{18}}\hat{k} \] Step 4: Final Answer:
The unit vector along the straight line AB is \(\frac{1}{\sqrt{18}}(\hat{i} + 4\hat{j} - \hat{k})\).