A vector a = αi + 2j + βk (α, β ∈ R) lies in the plane of the vectors, b = i + j and c = i - j + 4k.

Q: Wait, why do we need unit vectors? Can't we just add $\vec{b}$ and $\vec{c}$ directly?

Great question! If you add $\vec{b}$ and $\vec{c}$ directly, you get the diagonal of the parallelogram formed by them. This diagonal only bisects the angle if the sides of the parallelogram are equal in length (a rhombus). Since $|\vec{b}| \neq |\vec{c}|$, the diagonal is skewed. By converting to unit vectors, we effectively create a rhombus, ensuring the diagonal is the true angle bisector.

Q: How do I know if I should use the plus or minus sign in $\lambda (\hat{b} \pm \hat{c})$?

The plus sign gives the internal bisector, and the minus sign gives the external bisector. In JEE problems, if one case doesn't yield an answer that fits the options or constraints, you must check the other. It is a standard procedure to test both.

Q: Why is there a scalar $\lambda$ in the formula?

The unit vector sum $\hat{b} + \hat{c}$ gives you the *direction* of the bisector. However, the vector $\vec{a}$ might have a different magnitude than the sum of the unit vectors. The scalar $\lambda$ allows us to scale the direction vector to match the specific components given in the problem (like the $y$-component being $2$).

Question

A vector $\vec{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k} (\alpha, \beta \in \mathbb{R})$ 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

Options:

(A) $\vec{a}.\hat{i} + 3 = 0$

(B) $\vec{a}.\hat{k} + 4 = 0$

(C) $\vec{a}.\hat{i} + 1 = 0$

(D) $\vec{a}.\hat{k} + 2 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the multidimensional landscape of JEE vector algebra, we are given two vectors originating from the same point:
$$\vec{b} = \hat{i} + \hat{j}$$
$$\vec{c} = \hat{i} - \hat{j} + 4\hat{k}$$

We seek a third vector, $\vec{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$, that bisects the angle between $\vec{b}$ and $\vec{c}$.

### The Trap of Direct Addition

Many students incorrectly assume the bisector is simply $\vec{b} + \vec{c}$. This only holds true if the parallelogram formed by the vectors is a rhombus, requiring equal magnitudes.

Calculating the magnitudes:
$$|\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}$$
$$|\vec{c}| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{18} = 3\sqrt{2}$$

Since $|\vec{b}| 
eq |\vec{c}|$, the diagonal is skewed. To find the true bisector, we must normalize the vectors into unit vectors $\hat{b}$ and $\hat{c}$ to ensure they have equal length.

### The Bifurcation: Internal vs. External

The general equation for an angle bisector is $\vec{a} = \lambda (\hat{b} \pm \hat{c})$. The plus sign ($+$) represents the internal bisector, while the minus sign ($-$) represents the external bisector.

We are given the constraint that the $y$-component of $\vec{a}$ must be $2$. First, we define the unit vectors:
$$\hat{b} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$$
$$\hat{c} = \frac{\hat{i} - \hat{j} + 4\hat{k}}{3\sqrt{2}}$$

Testing the internal bisector:
$$\vec{a} = \lambda \left( \frac{3(\hat{i} + \hat{j}) + (\hat{i} - \hat{j} + 4\hat{k})}{3\sqrt{2}} ight)$$

The $y$-component is $\frac{2\lambda}{3\sqrt{2}}$. Setting this equal to $2$ yields $\lambda = 3\sqrt{2}$, resulting in $\vec{a} = 4\hat{i} + 2\hat{j} + 4\hat{k}$. Since this does not match our requirements, we proceed to the external case.

### The External Victory

We now apply the external bisector formula: $\vec{a} = \mu (\hat{b} - \hat{c})$.

Substituting the unit vectors:
$$\vec{a} = \mu \left( \frac{3(\hat{i} + \hat{j}) - (\hat{i} - \hat{j} + 4\hat{k})}{3\sqrt{2}} ight) = \mu \left( \frac{2\hat{i} + 4\hat{j} - 4\hat{k}}{3\sqrt{2}} ight)$$

The $y$-component is $\frac{4\mu}{3\sqrt{2}}$. Setting this to $2$ gives $\mu = \frac{3\sqrt{2}}{2}$.

Substituting $\mu$ back into the expression:
$$\vec{a} = \frac{3\sqrt{2}}{2} \cdot \frac{2\hat{i} + 4\hat{j} - 4\hat{k}}{3\sqrt{2}} = \hat{i} + 2\hat{j} - 2\hat{k}$$

Thus, the resulting vector is **$\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$**. This satisfies the geometric constraints and confirms the logic of the external bisector.

Animated Solution for Mathematics - Vector Algebra: A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then

Select Answer:

Visualized Solution

Visualizing Vectors $b$ and $c$

The Angle Bisector Concept

Calculating Magnitudes $∣ b ∣$ and $∣ c ∣$

Finding Unit Vectors $\hat{b}$ and $\overset{c}{^}$

The Angle Bisector Formula

Case 1: Internal Angle Bisector

Simplifying the Internal Bisector

Solving for $λ$ in Internal Case

Testing Options for Internal Case

Case 2: External Angle Bisector

Solving for $μ$ in External Case

Final Verification of Options

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Trap of Direct Addition

The Bifurcation

The External Victory

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