Question

If the vector \[ \alpha\hat i+\beta\hat j+\hat k \] is along the bisector of the angle between the vectors \[ 2\hat i-\hat j+2\hat k \] and \[ \hat i+2\hat j+2\hat k, \] then \(2\alpha+6\beta=\)

• A. \(0\)
• B. \(2\)
• C. \(3\)
• D. \(1\)

Hint:

The internal bisector direction between vectors \(\vec a\) and \(\vec b\) is \[ \frac{\vec a}{|\vec a|}+\frac{\vec b}{|\vec b|}. \] Always normalize before adding.

Answer 1

The Correct Option is C

Concept: The internal angle bisector of two vectors \(\vec a\) and \(\vec b\) is directed along \[ \frac{\vec a}{|\vec a|} + \frac{\vec b}{|\vec b|}. \] Thus we first normalize both vectors and then compare the resulting direction with the given vector.

Step 1: Find magnitudes. Let \[ \vec a=2\hat i-\hat j+2\hat k, \] \[ \vec b=\hat i+2\hat j+2\hat k. \] Then \[ |\vec a| = \sqrt{4+1+4} = 3, \] \[ |\vec b| = \sqrt{1+4+4} = 3. \]

Step 2: Find the bisector direction. \[ \frac{\vec a}{3} + \frac{\vec b}{3} = \frac13(\vec a+\vec b). \] Now, \[ \vec a+\vec b = (2+1)\hat i+(-1+2)\hat j+(2+2)\hat k. \] \[ =3\hat i+\hat j+4\hat k. \] Hence the bisector direction is \[ 3\hat i+\hat j+4\hat k. \]

Step 3: Compare with the given vector. Given vector: \[ \alpha\hat i+\beta\hat j+\hat k. \] Since both are parallel, \[ \alpha:\beta:1 = 3:1:4. \] Therefore, \[ \alpha=\frac34, \qquad \beta=\frac14. \]

Step 4: Compute \(2\alpha+6\beta\). \[ 2\alpha+6\beta = 2\left(\frac34\right) + 6\left(\frac14\right). \] \[ = \frac32+\frac62. \] \[ = 3. \] Conclusion: \[ \boxed{3} \]