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The equation of the angular bisector of $∠ C A B$ of $Δ A B C$ , whose position vector of A is $3 \vec{i} + \vec{j} - \vec{k}$ and $\vec{A B} = \vec{i} - 2 \vec{j} + 2 \vec{k},$ $\vec{A C} = 2 \vec{i} + \vec{j} + 2 \vec{k}$ is

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$\vec{r} = \frac{3 \vec{i} - \vec{j} + 4 \vec{k}}{3}$

$\vec{r} = (\vec{i} - 2 \vec{j} + 2 \vec{k}) + λ (3 \vec{i} + \vec{j} - \vec{k}), λ$ . scalar

$\vec{r} = 3 \vec{i} + \vec{j} - \vec{k} + λ (3 \vec{i} - \vec{j} + 4 \vec{k})$

$\vec{r} = (2 \vec{i} - \vec{j}) + λ (2 \vec{i} + \vec{j})$

answer is C.

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Detailed Solution

Equation of angular bisector of $C A B$ is

$\vec{r} = \vec{O A} + t (\frac{\vec{A B}}{|\vec{A B}|} + \frac{\vec{A C}}{|\vec{A C}|})$

$\vec{r} = (3 \vec{i} + \vec{j} - \vec{k}) + t (\frac{\vec{i} - 2 \vec{j} + 2 \vec{k}}{3} + \frac{2 \vec{i} + \vec{j} + 2 \vec{k}}{3})$

$\vec{r} = (3 \vec{i} + \vec{j} - \vec{k}) + λ (3 \vec{i} - \vec{j} + 4 \vec{k})$

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