Question

Suppose $\alpha \hat{i} + \alpha \hat{j} + \gamma \hat{k}$, $\hat{i} + \hat{k}$ and $\gamma \hat{i} + \gamma \hat{j} + \beta \hat{k}$ are coplanar where $\alpha, \beta, \gamma$ are positive constants. Then the product $\alpha\beta$ is

• A. $\gamma$
• B. $\gamma^2$
• C. $2\gamma$
• D. $2\gamma^2$
• E. $3\gamma$

Hint:

For coplanarity, always jump directly to determinant = 0.

Answer 1

The Correct Option is B

Concept: Three vectors are coplanar if their scalar triple product is zero. \[ \vec{A}\cdot(\vec{B}\times \vec{C}) = 0 \]

Step 1: Write vectors.
\[ \vec{A} = (\alpha,\alpha,\gamma) \] \[ \vec{B} = (1,0,1) \] \[ \vec{C} = (\gamma,\gamma,\beta) \]

Step 2: Form determinant.
\[ \begin{vmatrix} \alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta \end{vmatrix} = 0 \]

Step 3: Expand determinant.
\[ \alpha(0\cdot\beta - 1\cdot\gamma) - \alpha(1\cdot\beta - 1\cdot\gamma) + \gamma(1\cdot\gamma - 0) = 0 \] \[ = -\alpha\gamma - \alpha(\beta - \gamma) + \gamma^2 \] \[ = -\alpha\gamma - \alpha\beta + \alpha\gamma + \gamma^2 \] \[ = -\alpha\beta + \gamma^2 = 0 \]

Step 4: Solve.
\[ \alpha\beta = \gamma^2 \] \[ \boxed{\gamma^2} \]