Question

If \( x, y \) and \( z \) are non-zero real numbers and \( a = x\hat{i} + 2\hat{j}, b = y\hat{j} + 3\hat{k} \) and \( c = x\hat{i} + y\hat{j} + z\hat{k} \) are such that \( a \times b = z\hat{i} - 3\hat{j} + \hat{k} \), then \( [abc] \) is equal to

• A. 3
• B. 10
• C. 9
• D. 6

Hint:

For vector problems involving cross and dot products, remember to break down the vectors into their components and use the determinant formula for cross products. This will help you to solve efficiently.

Answer 1

The Correct Option is C

We are given three vectors: \( a = x\hat{i} + 2\hat{j}, b = y\hat{j} + 3\hat{k}, c = x\hat{i} + y\hat{j} + z\hat{k} \), and the condition \( a \times b = z\hat{i} - 3\hat{j} + \hat{k} \). First, we calculate \( a \times b \): \[ a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
x & 2 & 0
0 & y & 3 \end{vmatrix} = \hat{i} \begin{vmatrix} 2 & 0
y & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} x & 0
0 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} x & 2
0 & y \end{vmatrix} \] \[ = \hat{i} (2 \times 3 - 0 \times y) - \hat{j} (x \times 3 - 0 \times 0) + \hat{k} (x \times y - 2 \times 0) \] \[ = \hat{i} (6) - \hat{j} (3x) + \hat{k} (xy) = 6\hat{i} - 3x\hat{j} + xy\hat{k} \] Now compare this with \( a \times b = z\hat{i} - 3\hat{j} + \hat{k} \). Equating the components: \[ 6 = z, \quad -3x = -3, \quad xy = 1 \] From \( -3x = -3 \), we get \( x = 1 \). Substituting \( x = 1 \) in \( xy = 1 \), we get \( y = 1 \). Now, from \( 6 = z \), we get \( z = 6 \). Thus, the value of \( [abc] \) is: \[ [abc] = |a \cdot (b \times c)| = 9 \]