A vector $$\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k}$$ $$(\alpha, \beta \in R)$$ lies in the plane of the vectors, $$\vec{b} = \hat{i} + \hat{j}$$ and $$\vec{c} = \hat{i} - \hat{j} + 4\hat{k}$$. If $$\vec{a}$$ bisects the angle between $$\vec{b}$$ and $$\vec{c}$$, then

We have the three vectors

$$\vec{b}=1\,\hat{i}+1\,\hat{j}+0\,\hat{k},\qquad \vec{c}=1\,\hat{i}-1\,\hat{j}+4\,\hat{k},\qquad \vec{a}=\alpha\,\hat{i}+2\,\hat{j}+\beta\,\hat{k}\;,\; (\alpha,\beta\in\mathbb R).$$

The question tells us two facts about $$\vec{a}:$$

(i) It lies in the plane spanned by $$\vec{b}$$ and $$\vec{c}$$ (this will be automatically true once we use the angle-bisector property), and

(ii) It bisects the angle between $$\vec{b}$$ and $$\vec{c}$$.

In vector geometry the direction of the internal bisector of the angle between two non-zero vectors $$\vec{p}$$ and $$\vec{q}$$ is proportional to $$\dfrac{\vec{p}}{|\vec{p}|}+\dfrac{\vec{q}}{|\vec{q}|},$$ while the direction of the external bisector is proportional to $$\dfrac{\vec{p}}{|\vec{p}|}-\dfrac{\vec{q}}{|\vec{q}|}.$$

We therefore write the two possible bisectors as

$$\vec{a}_{\text{(int)}}=\lambda\left( \frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|}\right), \qquad \vec{a}_{\text{(ext)}}=\mu\left( \frac{\vec{b}}{|\vec{b}|}-\frac{\vec{c}}{|\vec{c}|}\right), $$

where $$\lambda$$ and $$\mu$$ are real scalars still to be determined.

Step 1 - Find the magnitudes of $$\vec{b}$$ and $$\vec{c}$$.

$$|\vec{b}|=\sqrt{1^{2}+1^{2}+0^{2}}=\sqrt{2},$$

$$|\vec{c}|=\sqrt{1^{2}+(-1)^{2}+4^{2}} =\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}.$$

Step 2 - Write the unit vectors.

$$\frac{\vec{b}}{|\vec{b}|} =\left(\frac{1}{\sqrt2}\right)\hat{i} +\left(\frac{1}{\sqrt2}\right)\hat{j} +0\,\hat{k},$$

$$\frac{\vec{c}}{|\vec{c}|} =\left(\frac{1}{3\sqrt2}\right)\hat{i} -\left(\frac{1}{3\sqrt2}\right)\hat{j} +\left(\frac{4}{3\sqrt2}\right)\hat{k}.$$

Step 3 - Internal bisector.

Add the unit vectors:

$$\frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|} =\Bigl(\frac{1}{\sqrt2}+\frac{1}{3\sqrt2}\Bigr)\hat{i} +\Bigl(\frac{1}{\sqrt2}-\!\frac{1}{3\sqrt2}\Bigr)\hat{j} +\Bigl(0+\frac{4}{3\sqrt2}\Bigr)\hat{k}$$

$$=\frac{4}{3\sqrt2}\hat{i} +\frac{2}{3\sqrt2}\hat{j} +\frac{4}{3\sqrt2}\hat{k}.$$

Multiply by $$\lambda$$ and equate to $$\vec{a}:$$

$$\alpha=\lambda\frac{4}{3\sqrt2},\qquad 2=\lambda\frac{2}{3\sqrt2},\qquad \beta=\lambda\frac{4}{3\sqrt2}.$$

From the middle equation

$$\lambda=\frac{2\cdot3\sqrt2}{2}=3\sqrt2.$$

This gives $$\alpha=4$$ and $$\beta=4.$$ But none of the given options involve $$\alpha=4$$ or $$\beta=4,$$ so the internal bisector does not match the given form of $$\vec{a}.$$ Hence $$\vec{a}$$ must be the external bisector.

Step 4 - External bisector.

Subtract the unit vectors:

$$\frac{\vec{b}}{|\vec{b}|}-\frac{\vec{c}}{|\vec{c}|} =\Bigl(\frac{1}{\sqrt2}-\frac{1}{3\sqrt2}\Bigr)\hat{i} +\Bigl(\frac{1}{\sqrt2}+\frac{1}{3\sqrt2}\Bigr)\hat{j} +\Bigl(0-\frac{4}{3\sqrt2}\Bigr)\hat{k}$$

$$=\frac{2}{3\sqrt2}\hat{i} +\frac{4}{3\sqrt2}\hat{j} -\frac{4}{3\sqrt2}\hat{k}.$$

Multiply by $$\mu$$ and equate to $$\vec{a}:$$

$$\alpha=\mu\frac{2}{3\sqrt2},\qquad 2 =\mu\frac{4}{3\sqrt2},\qquad \beta=\mu\!\left(-\frac{4}{3\sqrt2}\right).$$

Use the middle equation to find $$\mu$$:

$$2=\mu\frac{4}{3\sqrt2} \;\;\Longrightarrow\;\; \mu=\frac{2\cdot3\sqrt2}{4} =\frac{3\sqrt2}{2}.$$

Substituting this $$\mu$$ back, we get

$$\alpha=\frac{3\sqrt2}{2}\cdot\frac{2}{3\sqrt2}=1,$$ $$\beta=\frac{3\sqrt2}{2}\left(-\frac{4}{3\sqrt2}\right)=-2.$$

Therefore the actual vector is

$$\vec{a}=1\,\hat{i}+2\,\hat{j}-2\,\hat{k}.$$

The coefficient of $$\hat{k}$$ in $$\vec{a}$$ is $$\beta=-2,$$ and this satisfies the simple relation

$$\vec{a}\cdot\hat{k}+\;2 =\beta+2 =-2+2 =0.$$

This relation matches Option C, which is $$\vec{a}\cdot\hat{k}+2=0.$$

Hence, the correct answer is Option C.