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Step-by-Step Solution
Step 1: Identify the Position Vectors
Let the origin be O. The position vectors of points A and B are given by:
$ \vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k} $
and
$ \vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k} $.
Step 2: Calculate the Magnitudes of the Vectors
The magnitude of $ \vec{OA} $ is:
$$
|\vec{OA}|
= \sqrt{2^2 + 2^2 + 1^2}
= \sqrt{4 + 4 + 1}
= \sqrt{9}
= 3.
$$
The magnitude of $ \vec{OB} $ is:
$$
|\vec{OB}|
= \sqrt{2^2 + 4^2 + 4^2}
= \sqrt{4 + 16 + 16}
= \sqrt{36}
= 6.
$$
Step 3: Find the Direction of the Internal Bisector of $ \angle AOB $
First, write the unit vectors along $ \vec{OA} $ and $ \vec{OB} $:
$$
\frac{\vec{OA}}{|\vec{OA}|}
= \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right),
\quad
\frac{\vec{OB}}{|\vec{OB}|}
= \left(\frac{2}{6}, \frac{4}{6}, \frac{4}{6}\right)
= \left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right).
$$
The direction vector of the internal bisector is the sum of these unit vectors:
$$
\frac{\vec{OA}}{|\vec{OA}|} + \frac{\vec{OB}}{|\vec{OB}|}
= \left(\frac{2}{3} + \frac{1}{3},\, \frac{2}{3} + \frac{2}{3},\, \frac{1}{3} + \frac{2}{3}\right)
= (1, \tfrac{4}{3}, 1).
$$
Multiplying by 3 to clear denominators, we get the direction vector
$ (3,\,4,\,3). $
Step 4: Parametric Form of the Bisector Line from O
Since this bisector line passes through O (the origin) and has direction $(3,4,3)$, its parametric form is:
$$
\vec{r} = \lambda (3,\,4,\,3),
$$
where $ \lambda $ is a real parameter.
Step 5: Parametric Form of the Line AB
The vector $ \vec{AB} $ is:
$$
\vec{AB} = \vec{OB} - \vec{OA} = (2,4,4) - (2,2,1) = (0,\,2,\,3).
$$
Hence, any point on the line AB can be written as:
$$
\vec{r} = \vec{OA} + \mu \, \vec{AB}
= (2,2,1) + \mu \,(0,2,3),
$$
where $ \mu $ is a real parameter. In component form:
$$
\vec{r} = \bigl(2,\; 2 + 2\mu,\; 1 + 3\mu\bigr).
$$
Step 6: Find the Intersection Point C
The point C lies on both lines, so their coordinates must match. Therefore:
$$
(3\lambda,\, 4\lambda,\, 3\lambda)
= \bigl(2,\; 2 + 2\mu,\; 1 + 3\mu\bigr).
$$
Equating components, we get the system:
$$
3\lambda = 2,
\quad
4\lambda = 2 + 2\mu,
\quad
3\lambda = 1 + 3\mu.
$$
• From $ 3\lambda = 2 $, we obtain $ \lambda = \tfrac{2}{3}. $
• Substitute $ \lambda = \tfrac{2}{3} $ into $ 4\lambda = 2 + 2\mu $:
$$
4 \times \frac{2}{3} = 2 + 2\mu
\quad \Longrightarrow \quad
\frac{8}{3} = 2 + 2\mu
\quad \Longrightarrow \quad
2\mu = \frac{8}{3} - 2
= \frac{8 - 6}{3}
= \frac{2}{3},
\quad \text{so}
\quad \mu = \frac{1}{3}.
$$
• Finally, check $ 3\lambda = 1 + 3\mu $:
$$
3\times \frac{2}{3} = 1 + 3\times \frac{1}{3}
\quad \Longrightarrow \quad
2 = 1 + 1
\quad \Longrightarrow \quad
2 = 2.
$$
All conditions are satisfied, so
$ \lambda = \tfrac{2}{3} $ and $ \mu = \tfrac{1}{3}. $
Step 7: Coordinates of C and the Length $OC$
Substituting $ \lambda = \tfrac{2}{3} $ into $ (3\lambda, 4\lambda, 3\lambda) $ gives:
$$
\vec{OC} = \left(3 \times \frac{2}{3},\; 4 \times \frac{2}{3},\; 3 \times \frac{2}{3}\right)
= \bigl(2,\; \tfrac{8}{3},\; 2\bigr).
$$
Hence, the length $ OC $ is:
$$
|\vec{OC}|
= \sqrt{\,2^2 + \left(\tfrac{8}{3}\right)^2 + 2^2\,}
= \sqrt{4 + \frac{64}{9} + 4}
= \sqrt{\frac{36 + 64 + 36}{9}}
= \sqrt{\frac{136}{9}}
= \frac{\sqrt{136}}{3}
= \frac{2\sqrt{34}}{3}.
$$
Final Answer
Therefore, the length of $OC$ is
$ \frac{2}{3}\sqrt{34}. $
