Question

If a= (1, 1, 1), b= (x, y, z), c= (0, 1, -1) are three vectors such that \( a \cdot b = 3, a \times b = c \), then b is

• A. \( (2/3, 1/3, 2/3) \)
• B. \( (1/3, 2/3, 1) \)
• C. \( (2/3, 2/3, 5/3) \)
• D. \( (5/3, 2/3, 2/3) \)

Hint:

The dot product \( a \cdot b \) gives a scalar equation relating the components. The cross product \( a \times b = c \) gives vector equation, which yields up to three scalar equations by comparing components. Solving these equations simultaneously determines the unknown vector components.

Answer 1

The Correct Option is D

Step 1: Write down the given vectors and conditions. \[ a = (1, 1, 1) = \hat{i} + \hat{j} + \hat{k} \] \[ b = (x, y, z) = x\hat{i} + y\hat{j} + z\hat{k} \] \[ c = (0, 1, -1) = 0\hat{i} + \hat{j} - \hat{k} \] Conditions: 1. \( a \cdot b = 3 \) 2. \( a \times b = c \) Step 2: Use the dot product condition \( a \cdot b = 3 \). \[ % Option (1)(x) + (1)(y) + (1)(z) = 3 \] \[ x + y + z = 3 \quad \cdots (1) \] Step 3: Use the cross product condition \( a \times b = c \). Calculate \( a \times b \): \[ a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & 1
x & y & z \end{vmatrix} \] \[ a \times b = \hat{i}(1 \cdot z - 1 \cdot y) - \hat{j}(1 \cdot z - 1 \cdot x) + \hat{k}(1 \cdot y - 1 \cdot x) \] \[ a \times b = (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} \] \[ a \times b = (z - y)\hat{i} + (x - z)\hat{j} + (y - x)\hat{k} \] We are given \( a \times b = c = 0\hat{i} + 1\hat{j} - 1\hat{k} \). Equating the components: \[ z - y = 0 \quad \Rightarrow \quad z = y \quad \cdots (2) \] \[ x - z = 1 \quad \Rightarrow \quad x = z + 1 \quad \cdots (3) \] \[ y - x = -1 \quad \Rightarrow \quad y = x - 1 \quad \cdots (4) \] (Note that (2) and (3) imply (4): \( y = z = x - 1 \)). Step 4: Solve the system of linear equations (1), (2), and (3). From (2), \( z = y \). Substitute this into (3): \[ x = y + 1 \] Now substitute \( z = y \) and \( x = y + 1 \) into equation (1): \[ (y + 1) + y + y = 3 \] \[ 3y + 1 = 3 \] \[ 3y = 2 \] \[ y = \frac{2}{3} \] Step 5: Find x and z using the value of y. \[ z = y = \frac{2}{3} \] \[ x = y + 1 = \frac{2}{3} + 1 = \frac{2}{3} + \frac{3}{3} = \frac{5}{3} \] Step 6: State the vector b and compare with options. \[ b = (x, y, z) = \left(\frac{5}{3}, \frac{2}{3}, \frac{2}{3}\right) \] This result matches option (D).