Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if and only if their scalar triple product is zero: $[\vec{a} \ \vec{b} \ \vec{c}] = 0$. This can be computed as the determinant of the matrix formed by their components.
• Angle Bisector of Two Vectors: A vector $\vec{v}$ that bisects the angle between two non-zero vectors $\vec{b}$ and $\vec{c}$ is parallel to $\frac{\vec{b}}{|\vec{b}|} \pm \frac{\vec{c}}{|\vec{c}|}$. The '+' sign corresponds to the internal angle bisector, and the '-' sign corresponds to the external angle bisector.
• Vector Equality: Two vectors are equal if and only if their corresponding components are equal.

Step-by-Step Solution

Given vectors: $\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$ $\overrightarrow b = \widehat i + \widehat j$ $\overrightarrow c = \widehat j + \widehat k$

Step 1: Apply the Coplanarity Condition The vector $\overrightarrow a$ lies in the plane of $\overrightarrow b$ and $\overrightarrow c$. This means $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar. We use the scalar triple product condition: $[\vec{a} \ \vec{b} \ \vec{c}] = 0$. The components of the vectors are $\vec{a} = (\alpha, 2, \beta)$, $\vec{b} = (1, 1, 0)$, and $\vec{c} = (0, 1, 1)$. The scalar triple product is given by the determinant: $\begin{vmatrix} \alpha & 2 & \beta \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = 0$ Expanding the determinant along the first row: $\alpha(1 \cdot 1 - 0 \cdot 1) - 2(1 \cdot 1 - 0 \cdot 0) + \beta(1 \cdot 1 - 1 \cdot 0) = 0$ $\alpha(1) - 2(1) + \beta(1) = 0$ $\alpha - 2 + \beta = 0$ $\alpha + \beta = 2 \quad \quad (1)$ This equation establishes a relationship between $\alpha$ and $\beta$ required for coplanarity.

Step 2: Apply the Angle Bisector Condition The vector $\overrightarrow a$ bisects the angle between $\overrightarrow b$ and $\overrightarrow c$. First, we find the magnitudes of $\overrightarrow b$ and $\overrightarrow c$: $|\overrightarrow b| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ $|\overrightarrow c| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$ Since $|\overrightarrow b| = |\overrightarrow c|$, the angle bisector vector is parallel to $\overrightarrow b + \overrightarrow c$ (internal bisector) or $\overrightarrow b - \overrightarrow c$ (external bisector).

Case 1: $\overrightarrow a$ is parallel to the internal angle bisector ($\overrightarrow b + \overrightarrow c$). $\overrightarrow b + \overrightarrow c = (\widehat i + \widehat j) + (\widehat j + \widehat k) = \widehat i + 2\widehat j + \widehat k$ If $\overrightarrow a$ is parallel to $\overrightarrow b + \overrightarrow c$, then $\overrightarrow a = k(\overrightarrow b + \overrightarrow c)$ for some scalar $k$. $\alpha \widehat i + 2\widehat j + \beta \widehat k = k(\widehat i + 2\widehat j + \widehat k)$ $\alpha \widehat i + 2\widehat j + \beta \widehat k = k\widehat i + 2k\widehat j + k\widehat k$ Equating the coefficients of $\widehat i$, $\widehat j$, and $\widehat k$: For $\widehat j$: $2 = 2k \implies k = 1$. For $\widehat i$: $\alpha = k \implies \alpha = 1$. For $\widehat k$: $\beta = k \implies \beta = 1$. So, from the internal bisector condition, we get $\alpha = 1$ and $\beta = 1$.

Case 2: $\overrightarrow a$ is parallel to the external angle bisector ($\overrightarrow b - \overrightarrow c$). $\overrightarrow b - \overrightarrow c = (\widehat i + \widehat j) - (\widehat j + \widehat k) = \widehat i - \widehat k$ If $\overrightarrow a$ is parallel to $\overrightarrow b - \overrightarrow c$, then $\overrightarrow a = k'(\overrightarrow b - \overrightarrow c)$ for some scalar $k'$. $\alpha \widehat i + 2\widehat j + \beta \widehat k = k'(\widehat i - \widehat k)$ $\alpha \widehat i + 2\widehat j + \beta \widehat k = k'\widehat i + 0\widehat j - k'\widehat k$ Equating the coefficients of $\widehat i$, $\widehat j$, and $\widehat k$: For $\widehat j$: $2 = 0 \cdot k'$. This implies $2 = 0$, which is a contradiction. Therefore, $\overrightarrow a$ cannot be parallel to the external angle bisector.

So, the only possibility from the angle bisector condition is $\alpha = 1$ and $\beta = 1$.

Step 3: Verify with the Coplanarity Condition We found $\alpha = 1$ and $\beta = 1$ from the angle bisector condition. We must check if these values satisfy the coplanarity condition derived in Step 1, which is $\alpha + \beta = 2$. Substituting $\alpha = 1$ and $\beta = 1$: $1 + 1 = 2$ $2 = 2$ The coplanarity condition is satisfied.

Thus, the values $\alpha = 1$ and $\beta = 1$ satisfy both conditions.

Common Mistakes & Tips

• Forgetting to Normalize: When dealing with angle bisectors, always remember to divide the vectors by their magnitudes unless their magnitudes are equal. If magnitudes are equal, the bisector is parallel to their sum or difference.
• Ignoring External Bisector: Always consider both the internal and external angle bisectors, as the problem statement does not specify which one.
• Confusing Coplanarity Conditions: Ensure you use the correct condition for coplanarity (scalar triple product is zero or one vector is a linear combination of the other two).

Summary We first used the condition that vector $\vec{a}$ lies in the plane of $\vec{b}$ and $\vec{c}$, which translates to their scalar triple product being zero, yielding the relation $\alpha + \beta = 2$. Then, we applied the condition that $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$. Since $|\vec{b}| = |\vec{c}|$, $\vec{a}$ must be parallel to either $\vec{b} + \vec{c}$ or $\vec{b} - \vec{c}$. By comparing components, we found that $\vec{a}$ must be parallel to $\vec{b} + \vec{c}$, leading to $\alpha = 1$ and $\beta = 1$. Finally, we verified that these values satisfy the coplanarity condition.

The final answer is $\boxed{\alpha = 1,\,\,\beta = 1}$.