Question

Let \(\mathbf{a} = 2\hat{\mathbf{i}} + \hat{\mathbf{j}} - 2\hat{\mathbf{k}}\) and \(\mathbf{b} = \hat{\mathbf{i}} + \hat{\mathbf{j}}\). If \(\mathbf{c}\) 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 \(|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|\) is equal to

• A. \(\frac{2}{3}\)
• B. \(\frac{3}{2}\)
• C. 2
• D. 3

Hint:

\(|\mathbf{p} \times \mathbf{q}| = |\mathbf{p}||\mathbf{q}|\sin\theta\).

Answer 1

The Correct Option is B

To solve the problem, we need to find the magnitude of the vector \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\). 

1. First, compute the cross product \(\mathbf{a} \times \mathbf{b}\):

Let \(\mathbf{a} = 2\hat{\mathbf{i}} + \hat{\mathbf{j}} - 2\hat{\mathbf{k}}\) and \(\mathbf{b} = \hat{\mathbf{i}} + \hat{\mathbf{j}}\).

The cross product is calculated as follows:

\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} \]

This determinant expansion gives us:

 

\mathbf{a} \times \mathbf{b} = (\hat{\mathbf{i}}(1(0) - (-2)(1)) - \hat{\mathbf{j}}(2(0) - (-2)(1)) + \hat{\mathbf{k}}(2(1) - 1(1)) = 2\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + \hat{\mathbf{k}}.

 

1. Now, given that \(\mathbf{a} \cdot \mathbf{c} = |\mathbf{c}|\\), this indicates that \(\mathbf{c}\) is in the same direction as \(\mathbf{a}\), i.e., \(\mathbf{c} = k \mathbf{a}\) where \(k\) is a constant.
2. The condition \(|\mathbf{c} - \mathbf{a}| = 2\sqrt{2}\) leads to:

We know:

|\mathbf{c} - \mathbf{a}|^2 = (k-1)^2 (4 + 1 + 4).\

So:

(4 + 1 + 4)(k-1)^2 = 8.\

Simplifying gives:

9(k-1)^2 = 8 \Rightarrow (k-1)^2 = \frac{8}{9}\

Thus, \(k = 1 \pm \frac{2\sqrt{2}}{3}\\).

1. The angle between \(\mathbf{a} \times \mathbf{b}\) and \(\mathbf{c}\) is \(30^\circ\).

The magnitude of the cross product \(|\mathbf{b}||\mathbf{c}|\sin\theta\) is:

|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}| = |\mathbf{a} \times \mathbf{b}||\mathbf{c}|\sin30^\circ.\

Calculate \(|\mathbf{a} \times \mathbf{b}|\\):

|\mathbf{a} \times \mathbf{b}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3.\

1. The magnitude of \(|(\mathbf{a} \times \mathbf{b}) \times (k \mathbf{a})| = |\mathbf{a} \times \mathbf{b}||\mathbf{a}| |\sin\theta|\\), where

\(\theta = 30^\circ\).

Therefore:

|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}| = 3 \cdot \frac{3}{\sqrt{3}} \cdot \frac{1}{2} = \frac{3}{2}.\

Thus, the magnitude of \(|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|\\) is \(\frac{3}{2}\).