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) : $0$