Question

The unit vector that bisects the angle between two vectors \( 2\hat{i}+\hat{j}+2\hat{k} \) and \( \hat{i}+2\hat{j}-2\hat{k} \) is

• A. \( \dfrac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} \)
• B. \( \dfrac{\hat{i}-\hat{j}}{\sqrt{2}} \)
• C. \( \dfrac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \)
• D. \( \dfrac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}} \)
• E. \( \dfrac{\hat{i}+\hat{j}}{\sqrt{2}} \)

Hint:

To find the angle bisector between two vectors, use \( \dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{b}}{|\vec{b}|} \) for the internal bisector. Then normalize the result to get a unit vector.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A vector that bisects the angle between two vectors \(\vec{a}\) and \(\vec{b}\) is given by the sum of their unit vectors, \(\hat{a} + \hat{b}\). This resultant vector lies along the angle bisector. To find the unit vector in this direction, we must normalize this resultant vector.
Step 2: Key Formula or Approach:
Let \(\vec{a} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\vec{b} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\). 1. Find the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\). 2. Find the unit vectors \(\hat{a} = \frac{\vec{a}}{|\vec{a}|}\) and \(\hat{b} = \frac{\vec{b}}{|\vec{b}|}\). 3. A vector along the angle bisector is \(\vec{v} = \hat{a} + \hat{b}\). 4. The required unit vector is \(\hat{v} = \frac{\vec{v}}{|\vec{v}|}\).
Step 3: Detailed Explanation:
1. Calculate magnitudes:
\[ |\vec{a}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3 \] \[ |\vec{b}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3 \] Since the magnitudes are equal, the vectors have the same length.
2. Find unit vectors (optional if magnitudes are equal):
In this special case where \(|\vec{a}| = |\vec{b}|\), the sum \(\vec{a} + \vec{b}\) itself lies along the angle bisector (forming a rhombus). So we can simplify the process.
A vector along the angle bisector is \(\vec{v} = \vec{a} + \vec{b}\).
\[ \vec{v} = (2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) + (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \] \[ \vec{v} = (2+1)\mathbf{i} + (1+2)\mathbf{j} + (2-2)\mathbf{k} \] \[ \vec{v} = 3\mathbf{i} + 3\mathbf{j} + 0\mathbf{k} = 3(\mathbf{i} + \mathbf{j}) \] 3. Find the unit vector along \(\vec{v}\):
First, find the magnitude of \(\vec{v}\):
\[ |\vec{v}| = |3\mathbf{i} + 3\mathbf{j}| = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2} \] The unit vector is \(\hat{v} = \frac{\vec{v}}{|\vec{v}|}\):
\[ \hat{v} = \frac{3\mathbf{i} + 3\mathbf{j}}{3\sqrt{2}} = \frac{3(\mathbf{i} + \mathbf{j})}{3\sqrt{2}} = \frac{\mathbf{i} + \mathbf{j}}{\sqrt{2}} \] Step 4: Final Answer:
The unit vector that bisects the angle is \(\frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}}\). This corresponds to option (E).