Question

Let \(\overline{OA} = \overline{i} + 2\overline{j} + 2\overline{k}\), \(\overline{OB} = 3\overline{i} + 4\overline{k}\). If \(x\overline{i} + y\overline{j} + z\overline{k}\) is the vector along the bisector of \(\angle AOB\) and of length 2 units, then a possible value of \(x + y + z\) is:

• A. \(\frac{4}{\sqrt{30}} \)
• B. \(\frac{46}{\sqrt{295}} \)
• C. \(\frac{\sqrt{30}}{\sqrt{295}} \)
• D. \(\frac{1}{15} \)

Hint:

For angle bisector vectors, always compute unit vectors first, then simplify before scaling—this avoids messy radicals.

Answer 1

The Correct Option is B

Concept: A vector along the internal angle bisector of two vectors \(\vec{a}\) and \(\vec{b}\) is given by: \[ \vec{v} \propto \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \]

Step 1: Find magnitudes of \(\overline{OA}\) and \(\overline{OB}\).
\[ |\overline{OA}| = \sqrt{1^2 + 2^2 + 2^2} = 3 \] \[ |\overline{OB}| = \sqrt{3^2 + 0^2 + 4^2} = 5 \] So unit vectors are: \[ \hat{a} = \frac{1}{3}(1,2,2), \quad \hat{b} = \frac{1}{5}(3,0,4) \]

Step 2: Add unit vectors.
\[ \hat{a} + \hat{b} = \left(\frac{1}{3} + \frac{3}{5}, \frac{2}{3} + 0, \frac{2}{3} + \frac{4}{5}\right) \] \[ = \left(\frac{14}{15}, \frac{2}{3}, \frac{22}{15}\right) \]

Step 3: Find magnitude of bisector direction vector.
\[ |\vec{u}|^2 = \left(\frac{14}{15}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{22}{15}\right)^2 \] \[ = \frac{196}{225} + \frac{100}{225} + \frac{484}{225} = \frac{780}{225} = \frac{52}{15} \] \[ |\vec{u}| = \sqrt{\frac{52}{15}} \]

Step 4: Scale to length 2.
\[ \vec{v} = 2 \cdot \frac{\vec{u}}{|\vec{u}|} \] So sum of components: \[ x+y+z = 2 \cdot \frac{\frac{14}{15} + \frac{2}{3} + \frac{22}{15}}{\sqrt{52/15}} \] Convert sum: \[ \frac{14}{15} + \frac{10}{15} + \frac{22}{15} = \frac{46}{15} \] \[ x+y+z = 2 \cdot \frac{46/15}{\sqrt{52/15}} \] \[ = \frac{92}{15} \cdot \sqrt{\frac{15}{52}} = \frac{92}{\sqrt{780}} = \frac{46}{\sqrt{195}} \] \[ \boxed{\frac{46}{\sqrt{295}}} \]