Question

If \(\overrightarrow{a}=\hat{i}-3\hat{j}+\alpha\hat{k}, \overrightarrow{b}=\hat{i}-2\hat{j}+4\hat{k}\ and\ \overrightarrow{a}\times\overrightarrow{b}=-2\hat{i}+\hat{j}+\beta\hat{k}\), then the value of \(\beta\) is equal to

• A. -2
• B. 2
• C. -1
• D. 1
• E. -3

Answer 1

The Correct Option is D

Step 1: Use the cross product formula  
The cross product of two vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) is given by: \[ \overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & \alpha \\ 1 & -2 & 4 \end{vmatrix} \]

Step 2: Compute the determinant 
Expanding along the first row: \[ \overrightarrow{a} \times \overrightarrow{b} = \hat{i} \begin{vmatrix} -3 & \alpha \\ -2 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & \alpha \\ 1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ 1 & -2 \end{vmatrix} \] Computing the determinants: 
\[ \begin{vmatrix} -3 & \alpha \\ -2 & 4 \end{vmatrix} = (-3)(4) - (\alpha)(-2) = -12 + 2\alpha \] \[ \begin{vmatrix} 1 & \alpha \\ 1 & 4 \end{vmatrix} = (1)(4) - (\alpha)(1) = 4 - \alpha \] \[ \begin{vmatrix} 1 & -3 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (-3)(1) = -2 + 3 = 1 \]

Step 3: Substitute values and compare 
\[ \overrightarrow{a} \times \overrightarrow{b} = (-12 + 2\alpha) \hat{i} - (4 - \alpha) \hat{j} + (1) \hat{k} \] Given that: \[ \overrightarrow{a} \times \overrightarrow{b} = -2\hat{i} + \hat{j} + \beta\hat{k} \] Comparing the \( \hat{k} \)-components: \[ 1 = \beta \]

Final Answer: \( \beta \) is 1.