Let \(\mathbf{A}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) and…

Let \(\mathbf{A}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) and \(\mathbf{B}=\hat{\mathbf{i}}+\hat{\mathbf{j}}\). If \(C\) is a vector such that \(\mathbf{A} \cdot \mathbf{C}=|\mathbf{C}|,|\mathbf{C}-\mathbf{A}|=2 \sqrt{2}\) and the angle between \(\mathbf{A} \times \mathbf{B}\) and \(\mathbf{C}\) is \(30^{\circ}\), then the value of \(|(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}|\) is
  1. \(\frac{2}{3}\)
  2. \(\frac{3}{2}\)
  3. 3
  4. 2

Solution

Since, \(\mathbf{A}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) and \(\mathbf{B}=\hat{\mathbf{i}}+\hat{\mathbf{j}}\) so \(\begin{aligned} & \mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{array}\right|=\hat{\mathbf{i}}(2)-\hat{\mathbf{j}}(\mathrm{Z})+\hat{\mathbf{k}}(2-1) \\ & =2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \because|(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}|=|\mathbf{A} \times \mathbf{B}||\mathbf{C}| \sin 30^{\circ} \end{aligned}\) [as angle between \(\mathbf{A} \times \mathbf{B}\) and \(\mathbf{C}=30^{\circ}\) (given)] \(\begin{aligned} & =\sqrt{4+4+1}|\mathrm{C}|\left(\frac{1}{2}\right)=\frac{3}{2}|\mathrm{C}| \quad \ldots (i) \\ & \because \quad|\mathbf{C}-\mathbf{A}|=2 \sqrt{2} \quad \text{(Given)} \\ & \Rightarrow|\mathbf{C}|^2+|\mathbf{A}|^2-2 \mathbf{C} \cdot \mathbf{A}=8 \\ & \Rightarrow \quad|C|^2+9-2|C|=8 \quad \text{[as } \mathbf{A} \cdot \mathbf{C}=|\mathbf{C}|] \\ & \Rightarrow(|\mathbf{C}|-1)^2=0 \Rightarrow|\mathbf{C}|=1 \\ \end{aligned}\) So, \(|(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}|=\frac{3}{2}\) [from Eq. (i)] Hence, option (2) is correct.

Asked in: AP EAMCET 2019 (20 Apr Shift 1)

Practice more Vector Algebra questions on Aicharya