1. Key Concepts and Formulas

• Vector Angle Bisector: A vector $\overrightarrow v$ that bisects the angle between two non-zero vectors $\overrightarrow u$ and $\overrightarrow w$ is parallel to the sum of their unit vectors: $\overrightarrow v \propto (\widehat u + \widehat w)$, where $\widehat u = \frac{\overrightarrow u}{|\overrightarrow u|}$ and $\widehat w = \frac{\overrightarrow w}{|\overrightarrow w|}$. This typically refers to the internal angle bisector.
• Coplanarity of Vectors: Three vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar if their scalar triple product is zero: $[\overrightarrow a \ \overrightarrow b \ \overrightarrow c] = 0$. Equivalently, $\overrightarrow a$ can be expressed as a linear combination of $\overrightarrow b$ and $\overrightarrow c$: $\overrightarrow a = x\overrightarrow b + y\overrightarrow c$ for some scalars $x, y$.
• Dot Product: The dot product of two vectors $\overrightarrow p = p_1 \widehat i + p_2 \widehat j + p_3 \widehat k$ and $\overrightarrow q = q_1 \widehat i + q_2 \widehat j + q_3 \widehat k$ is $\overrightarrow p \cdot \overrightarrow q = p_1 q_1 + p_2 q_2 + p_3 q_3$.

2. Step-by-Step Solution

We are given $\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$, $\overrightarrow b = \widehat i + \widehat j$, and $\overrightarrow c = \widehat i - \widehat j + 4\widehat k$.

Step 1: Calculate the Unit Vectors $\widehat b$ and $\widehat c$ To find the direction of the angle bisector, we first need the unit vectors along $\overrightarrow b$ and $\overrightarrow c$.

• For $\overrightarrow b$: $|\overrightarrow b| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ $\widehat b = \frac{\overrightarrow b}{|\overrightarrow b|} = \frac{1}{\sqrt{2}}(\widehat i + \widehat j)$
• For $\overrightarrow c$: $|\overrightarrow c| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$ $\widehat c = \frac{\overrightarrow c}{|\overrightarrow c|} = \frac{1}{3\sqrt{2}}(\widehat i - \widehat j + 4\widehat k)$ Explanation: Calculating unit vectors is essential for the angle bisector formula, as it ensures that the direction of the bisector is determined by the relative orientations of $\overrightarrow b$ and $\overrightarrow c$, not their magnitudes.

Step 2: Determine the Direction Vector of the Angle Bisector Since $\overrightarrow a$ bisects the angle between $\overrightarrow b$ and $\overrightarrow c$, it must be parallel to the sum of their unit vectors (representing the internal angle bisector). Let $\overrightarrow d$ be the direction vector of the bisector. $\overrightarrow d = \widehat b + \widehat c$ $\overrightarrow d = \frac{1}{\sqrt{2}}(\widehat i + \widehat j) + \frac{1}{3\sqrt{2}}(\widehat i - \widehat j + 4\widehat k)$ To add these, we find a common denominator: $\overrightarrow d = \frac{3(\widehat i + \widehat j)}{3\sqrt{2}} + \frac{\widehat i - \widehat j + 4\widehat k}{3\sqrt{2}}$ $\overrightarrow d = \frac{3\widehat i + 3\widehat j + \widehat i - \widehat j + 4\widehat k}{3\sqrt{2}}$ $\overrightarrow d = \frac{4\widehat i + 2\widehat j + 4\widehat k}{3\sqrt{2}}$ Explanation: The sum of unit vectors gives a vector that lies exactly along the angle bisector. Any vector along this bisector will be a scalar multiple of this sum.

Step 3: Express $\overrightarrow a$ in terms of the Bisector Direction Vector Since $\overrightarrow a$ lies in the plane of $\overrightarrow b$ and $\overrightarrow c$ and bisects the angle between them, $\overrightarrow a$ must be parallel to $\overrightarrow d$. Therefore, $\overrightarrow a = k \overrightarrow d$ for some scalar $k$. $\overrightarrow a = k \left( \frac{4\widehat i + 2\widehat j + 4\widehat k}{3\sqrt{2}} \right)$ $\overrightarrow a = \frac{4k}{3\sqrt{2}}\widehat i + \frac{2k}{3\sqrt{2}}\widehat j + \frac{4k}{3\sqrt{2}}\widehat k$ We are given $\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$. By comparing the coefficients of $\widehat j$: $2 = \frac{2k}{3\sqrt{2}}$ Solving for $k$: $1 = \frac{k}{3\sqrt{2}} \implies k = 3\sqrt{2}$ Explanation: The given $\widehat j$ component of $\overrightarrow a$ (which is 2) provides a direct way to find the scalar multiplier $k$, which scales the direction vector to match $\overrightarrow a$.

Step 4: Find the Components $\alpha$ and $\beta$ of $\overrightarrow a$ Now that we have $k = 3\sqrt{2}$, we can substitute it back into the expressions for the coefficients of $\widehat i$ and $\widehat k$:

• For $\alpha$ (coefficient of $\widehat i$): $\alpha = \frac{4k}{3\sqrt{2}} = \frac{4(3\sqrt{2})}{3\sqrt{2}} = 4$
• For $\beta$ (coefficient of $\widehat k$): $\beta = \frac{4k}{3\sqrt{2}} = \frac{4(3\sqrt{2})}{3\sqrt{2}} = 4$ So, the vector $\overrightarrow a$ is: $\overrightarrow a = 4\widehat i + 2\widehat j + 4\widehat k$ Explanation: With the scalar $k$ determined, we can now fully define the vector $\overrightarrow a$ by finding its unknown components.

Step 5: Verify Coplanarity (Optional but Recommended) The problem states that $\overrightarrow a$ lies in the plane of $\overrightarrow b$ and $\overrightarrow c$. This implies their scalar triple product is zero. $[\overrightarrow a \ \overrightarrow b \ \overrightarrow c] = \begin{vmatrix} 4 & 2 & 4 \\ 1 & 1 & 0 \\ 1 & -1 & 4 \end{vmatrix} = 4(4-0) - 2(4-0) + 4(-1-1) = 16 - 8 - 8 = 0$ The condition is satisfied, confirming our vector $\overrightarrow a$ is valid. Explanation: This step confirms that our derived $\overrightarrow a$ is consistent with the coplanarity requirement, acting as a self-check.

Step 6: Check the Given Options We now test the calculated vector $\overrightarrow a = 4\widehat i + 2\widehat j + 4\widehat k$ against each option.

• (A) $\overrightarrow a .\widehat i + 3 = 0$ $(4\widehat i + 2\widehat j + 4\widehat k) \cdot \widehat i + 3 = (4 \cdot 1 + 2 \cdot 0 + 4 \cdot 0) + 3 = 4 + 3 = 7 \neq 0$.
• (B) $\overrightarrow a .\widehat k - 4 = 0$ $(4\widehat i + 2\widehat j + 4\widehat k) \cdot \widehat k - 4 = (4 \cdot 0 + 2 \cdot 0 + 4 \cdot 1) - 4 = 4 - 4 = 0$.
• (C) $\overrightarrow a .\widehat i + 1 = 0$ $(4\widehat i + 2\widehat j + 4\widehat k) \cdot \widehat i + 1 = (4 \cdot 1 + 2 \cdot 0 + 4 \cdot 0) + 1 = 4 + 1 = 5 \neq 0$.
• (D) $\overrightarrow a .\widehat k + 2 = 0$ $(4\widehat i + 2\widehat j + 4\widehat k) \cdot \widehat k + 2 = (4 \cdot 0 + 2 \cdot 0 + 4 \cdot 1) + 2 = 4 + 2 = 6 \neq 0$.

Option (B) is the only one that holds true for our derived vector $\overrightarrow a$.

3. Common Mistakes & Tips

• Forgetting to Normalize: Using $\overrightarrow b$ and $\overrightarrow c$ directly instead of their unit vectors in the angle bisector formula will lead to an incorrect direction.
• Confusing Internal/External Bisector: The problem usually implies the internal bisector unless stated otherwise. If multiple options seem plausible, re-check the problem wording or assume the standard interpretation.
• Algebraic Errors: Vector algebra, especially with fractions and square roots, can be prone to arithmetic mistakes. Double-check calculations, especially when solving for scalar multipliers.

4. Summary

The problem requires us to find a vector $\overrightarrow a$ that lies in the plane of two given vectors $\overrightarrow b$ and $\overrightarrow c$, and also bisects the angle between them. We first found the unit vectors of $\overrightarrow b$ and $\overrightarrow c$, then formed the direction vector of the internal angle bisector by summing these unit vectors. By equating $\overrightarrow a$ to a scalar multiple of this bisector direction vector and using the known $\widehat j$ component of $\overrightarrow a$, we determined the scalar multiplier. This allowed us to find the components $\alpha$ and $\beta$ of $\overrightarrow a$. Finally, we substituted the derived vector $\overrightarrow a$ into the given options to find the correct one.

The final answer is $\boxed{\overrightarrow a .\widehat k - 4 = 0}$.