Question

Let $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$, $|\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is

• A. $\frac{\sqrt{3{2}$
• B. $\frac{3\sqrt{3{2}$
• C. $3\sqrt{3}$
• D. $4\sqrt{3}$

Hint:

Remember that $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 - 2(\vec{u} \cdot \vec{v}) + |\vec{v}|^2$. For cross products, if $\theta$ is the angle between vectors $\vec{P}$ and $\vec{Q}$, then $|\vec{P} \times \vec{Q}| = |\vec{P}||\vec{Q}|\sin\theta$. Ensure correct calculation of vector magnitudes and cross products.

Answer 1

The Correct Option is B

Step 1: Use the given condition $|\vec{c} - \vec{a}| = 2\sqrt{2}$ to find $|\vec{c}|$.
We have $|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8$.
Expand the dot product: \[ (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 8 \] \[ |\vec{c}|^2 - 2(\vec{a} \cdot \vec{c}) + |\vec{a}|^2 = 8 \] We are given: \[ |\vec{a}| = |2\hat{i} + \hat{j} - 2\hat{k}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3 \] Substitute $|\vec{a}| = 3$ and $\vec{a} \cdot \vec{c} = |\vec{c}|$: \[ |\vec{c}|^2 - 2|\vec{c}| + 9 = 8 \] \[ |\vec{c}|^2 - 2|\vec{c}| + 1 = 0 \] \[ (|\vec{c}| - 1)^2 = 0 \] Therefore, $|\vec{c}| = 1$. 
Step 2: Calculate the cross product $\vec{a} \times \vec{b}$.
Given: \[ \vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}, \quad \vec{b} = \hat{i} + \hat{j} \] \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} \] \[ = \hat{i}(1 \cdot 0 - (-2) \cdot 1) - \hat{j}(2 \cdot 0 - (-2) \cdot 1) + \hat{k}(2 \cdot 1 - 1 \cdot 1) \] \[ = 2\hat{i} - 2\hat{j} + \hat{k} \] 
Step 3: Calculate the magnitude $|\vec{a} \times \vec{b}|$.
\[ |\vec{a} \times \vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3 \] 
Step 4: Use the formula for $|(\vec{a} \times \vec{b}) \times \vec{c}|$.
\[ |(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| \cdot |\vec{c}| \cdot \sin\theta \] Given $\theta = \frac{2\pi}{3}$: \[ \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \] \[ |(\vec{a} \times \vec{b}) \times \vec{c}| = 3 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \] 
Step 5: Final Answer
\[ \boxed{\frac{3\sqrt{3}}{2}} \]