Animated Solution for Mathematics - Vector Algebra: 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:

Understanding the Angle Bisector

Imagine you are standing in a three-dimensional space. You have two vectors, $\vec{b}$ and $\vec{c}$, originating from the same point. They define a plane, like a sheet of paper floating in the air.

Now, you are asked to find a third vector, $\vec{a}$, that lies perfectly in this plane and cuts the angle between $\vec{b}$ and $\vec{c}$ exactly in half. This is the essence of an angle bisector.

To translate this geometric intuition into the rigid language of algebra, we must utilize the magic of unit vectors.

Many students instinctively want to just add $\vec{b}+\vec{c}$. But pause for a moment. If $\vec{b}$ has a magnitude of 10 and $\vec{c}$ has a magnitude of 2, their sum will be heavily skewed toward $\vec{b}$.

It won't be a bisector; it will be a vector biased toward the longer side. To ensure fairness, we must normalize them.

We calculate the unit vectors:

By stripping away their magnitudes and keeping only their directions, we create a scenario where both vectors have a length of 1. Now, when we add them, the resultant vector $\vec{a}=\lambda (\hat{b}+\hat{c})$ acts as a perfect, impartial bisector.

Let us perform the arithmetic with precision. For $\vec{b}=\hat{i}+\hat{j}$, the magnitude is:

Thus, $\hat{b}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$. For $\vec{c}=\hat{i}-\hat{j}+4\hat{k}$, the magnitude is:

Thus, $\hat{c}=\frac{\hat{i}-\hat{j}+4\hat{k}}{3\sqrt{2}}$.

Internal vs. External

We have two potential bisectors: the internal one, ${\vec{a}}_1=\lambda (\hat{b}+\hat{c})$, and the external one, ${\vec{a}}_2=\mu (\hat{b}-\hat{c})$. We don't know which one the problem requires, so we test the internal one first.

By finding a common denominator of $3\sqrt{2}$, we combine the terms. The internal bisector simplifies to a vector proportional to $4\hat{i}+2\hat{j}+4\hat{k}$.

Comparing the $\hat{j}$ component to the given $\vec{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}$, we find $\lambda =3\sqrt{2}$. However, testing this against the options reveals no match. We must pivot.

We turn to the external bisector, ${\vec{a}}_2=\mu (\hat{b}-\hat{c})$. This time, the subtraction yields a vector proportional to $2\hat{i}+4\hat{j}-4\hat{k}$.

Equating the $\hat{j}$ component to 2, we find $\mu =\frac{3\sqrt{2}}{2}$. Substituting this back, the vector simplifies beautifully to $\vec{a}=\hat{i}+2\hat{j}-2\hat{k}$.

Now, we check the dot product $\vec{a}\cdot \hat{k}$. The result is $-2$. Looking at the options, $\vec{a}\cdot \hat{k}+2=-2+2=0$.

We have found our match! This problem teaches us that in the face of ambiguity, systematic testing is not just a strategy; it is the path to truth.