Question

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

• A. 3
• B. -3
• C. 7
• D. -7
• E. 0

Answer 1

Answer: (E)

We are given the vectors \[ \mathbf{a} = \hat{i} - 3\hat{j} + 3\hat{k}, \quad \mathbf{b} = 2\hat{i} + \hat{j} - 3\hat{k}. \] We need to find the value of \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} \). First, compute the cross product \( \mathbf{a} \times \mathbf{b} \). Using the determinant form for the cross product: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 3 \\ 2 & 1 & -3 \end{vmatrix}. \] Expanding the determinant: \[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} -3 & 3 \\ 1 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 2 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix}. \] Calculating the 2x2 determinants: \[ \mathbf{a} \times \mathbf{b} = \hat{i} \left( (-3)(-3) - (3)(1) \right) - \hat{j} \left( (1)(-3) - (3)(2) \right) + \hat{k} \left( (1)(1) - (-3)(2) \right). \] \[ \mathbf{a} \times \mathbf{b} = \hat{i} \left( 9 - 3 \right) - \hat{j} \left( -3 - 6 \right) + \hat{k} \left( 1 + 6 \right). \] \[ \mathbf{a} \times \mathbf{b} = \hat{i}(6) - \hat{j}(-9) + \hat{k}(7). \] \[ \mathbf{a} \times \mathbf{b} = 6\hat{i} + 9\hat{j} + 7\hat{k}. \] Now, compute the dot product \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} \): \[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = (6\hat{i} + 9\hat{j} + 7\hat{k}) \cdot (2\hat{i} + \hat{j} - 3\hat{k}). \] Using the distributive property of the dot product: \[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = (6)(2) + (9)(1) + (7)(-3). \] \[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 12 + 9 - 21 = 0. \] Thus, the value of \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} \) is \( 0 \).

The correct option is (E) : \(0\)

Answer 2

We are given \(\bar{a} = \hat{i} - 3\hat{j} + 3\hat{k}\) and \(\bar{b} = 2\hat{i} + \hat{j} - 3\hat{k}\). We want to find \((\bar{a} \times \bar{b}) \cdot \bar{b}\).

The cross product of two vectors is always perpendicular to both vectors. That is, \(\bar{a} \times \bar{b}\) is perpendicular to both \(\bar{a}\) and \(\bar{b}\).

Therefore, the dot product of \(\bar{a} \times \bar{b}\) with either \(\bar{a}\) or \(\bar{b}\) must be zero.

So, \((\bar{a} \times \bar{b}) \cdot \bar{b} = 0\).