Question

Let $\vec{a} = 2\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}}$ and $\vec{b} = \hat{\text{i}} + \hat{\text{j}}$. 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 $60^\circ$, then $|(\vec{a} \times \vec{b}) \times \vec{c}| =$

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

Hint:

Recognizing the perfect square step $(|\vec{c}| - 1)^2 = 0$ simplifies the problem completely! It immediately reveals that vector $\vec{c}$ is a unit vector ($|\vec{c}| = 1$), reducing the final calculation to a simple substitution step.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the magnitude of a vector cross product, $|(\vec{a} \times \vec{b}) \times \vec{c}|$. We are given explicit component values for vectors $\vec{a}$ and $\vec{b}$, alongside relational scalar conditions involving vector $\vec{c}$.

Step 2: Key Formula or Approach:
The magnitude of a cross product between two vectors $\vec{U}$ and $\vec{V}$ with an angle $\theta$ between them is given by:
$$|\vec{U} \times \vec{V}| = |\vec{U}| \cdot |\vec{V}| \cdot \sin\theta$$ We will set $\vec{U} = \vec{a} \times \vec{b}$ and $\vec{V} = \vec{c}$. We can determine the unknown magnitude $|\vec{c}|$ by squaring the vector difference equation:
$$|\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c})$$

Step 3: Detailed Explanation:
1. First, let's find the cross product vector $\vec{a} \times \vec{b}$ using a matrix determinant:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$$ $$\vec{a} \times \vec{b} = \hat{\text{i}}(0 - (-2)) - \hat{\text{j}}(0 - (-2)) + \hat{\text{k}}(2 - 1) = 2\hat{\text{i}} - 2\hat{\text{j}} + \hat{\text{k}}$$ Calculate the magnitude of this cross product vector:
$$|\vec{a} \times \vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$ 2. Next, calculate the magnitude of vector $\vec{a}$:
$$|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = 3$$ 3. Now use the condition $|\vec{c} - \vec{a}| = 2\sqrt{2}$. Square both sides of this equation:
$$|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8$$ $$|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c}) = 8$$ Substitute $|\vec{a}|^2 = 3^2 = 9$ and the given identity $\vec{a} \cdot \vec{c} = |\vec{c}|$:
$$|\vec{c}|^2 + 9 - 2|\vec{c}| = 8$$ Rearrange this into a standard quadratic equation format:
$$|\vec{c}|^2 - 2|\vec{c}| + 1 = 0 \implies (|\vec{c}| - 1)^2 = 0 \implies |\vec{c}| = 1$$ 4. Finally, substitute all computed values into the cross product magnitude formula, where the angle is given as $60^\circ$:
$$|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| \cdot |\vec{c}| \cdot \sin(60^\circ)$$ $$= 3 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$$ This matches option (A).

Step 4: Final Answer:
The magnitude of the cross product expression is $\frac{3\sqrt{3}}{2}$, which corresponds to option (A).