Question

If \( \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 4\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{c} = \hat{i} + \alpha \hat{j} + \beta \hat{k} \) are coplanar and \( |\vec{c}| = \sqrt{3} \), then

• A. \( \alpha = \sqrt{2}, \beta = 1 \)
• B. \( \alpha = 1, \beta = \pm1 \)
• C. \( \alpha = \pm1, \beta = 1 \)
• D. \( \alpha = \pm1, \beta = -1 \)
• E. \( \alpha = -1, \beta = \pm1 \)

Hint:

Coplanarity ⇒ always use determinant = 0.

Answer 1

The Correct Option is B

Concept:
• Coplanar vectors ⇒ scalar triple product = 0 \[ \vec{a}\cdot(\vec{b}\times\vec{c}) = 0 \]

Step 1: Write determinant.
\[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0 \]

Step 2: Expand determinant.
\[ 1\begin{vmatrix}3 & 4 \alpha & \beta\end{vmatrix} \\ -1\begin{vmatrix}4 & 4 1 & \beta\end{vmatrix} \\ +1\begin{vmatrix}4 & 3 1 & \alpha\end{vmatrix} = 0 \] \[ = (3\beta - 4\alpha) - (4\beta - 4) + (4\alpha - 3) \]

Step 3: Simplify.
\[ 3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0 \] \[ -\beta + 1 = 0 \Rightarrow \beta = 1 \]

Step 4: Use magnitude condition.
\[ |\vec{c}| = \sqrt{1 + \alpha^2 + \beta^2} = \sqrt{3} \] \[ 1 + \alpha^2 + 1 = 3 \Rightarrow \alpha^2 = 1 \] \[ \alpha = \pm1 \]

Step 5: Final Answer.
\[ \boxed{\alpha = \pm1, \beta = 1} \]