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: Write the given vectors in component form.
Let \[ \vec{a}=2\hat{i}+\hat{j}+2\hat{k} \] and \[ \vec{b}=\hat{i}+2\hat{j}-2\hat{k} \] We need the unit vector along the internal angle bisector between these two vectors.

Step 2: Recall the formula for the angle bisector.
The direction of the internal angle bisector between two vectors \( \vec{a} \) and \( \vec{b} \) is given by \[ \frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|} \] So first we find the magnitudes of both vectors.

Step 3: Find the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \).
For \( \vec{a}=2\hat{i}+\hat{j}+2\hat{k} \), \[ |\vec{a}|=\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3 \] For \( \vec{b}=\hat{i}+2\hat{j}-2\hat{k} \), \[ |\vec{b}|=\sqrt{1^2+2^2+(-2)^2}=\sqrt{1+4+4}=\sqrt{9}=3 \] Thus, \[ |\vec{a}|=|\vec{b}|=3 \]

Step 4: Form the sum of the corresponding unit vectors.
Since both magnitudes are equal, \[ \frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|} = \frac{1}{3}\vec{a}+\frac{1}{3}\vec{b} = \frac{1}{3}(\vec{a}+\vec{b}) \] Now, \[ \vec{a}+\vec{b} = (2\hat{i}+\hat{j}+2\hat{k})+(\hat{i}+2\hat{j}-2\hat{k}) \] \[ =3\hat{i}+3\hat{j}+0\hat{k} \] \[ =3(\hat{i}+\hat{j}) \] So the angle bisector has the same direction as \[ \hat{i}+\hat{j} \]

Step 5: Convert this direction into a unit vector.
The magnitude of \[ \hat{i}+\hat{j} \] is \[ \sqrt{1^2+1^2}=\sqrt{2} \] Hence the required unit vector is \[ \frac{\hat{i}+\hat{j}}{\sqrt{2}} \]

Step 6: Verify that it is indeed a unit vector.
Let \[ \vec{u}=\frac{\hat{i}+\hat{j}}{\sqrt{2}} \] Then \[ |\vec{u}|=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2} \] \[ =\sqrt{\frac{1}{2}+\frac{1}{2}}=\sqrt{1}=1 \] So this is a unit vector.

Step 7: Final conclusion.
Therefore, the unit vector bisecting the angle between the given vectors is \[ \boxed{\frac{\hat{i}+\hat{j}}{\sqrt{2}}} \] Hence, the correct option is \[ \boxed{(5)\ \dfrac{\hat{i}+\hat{j}}{\sqrt{2}}} \]