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 \( 30^\circ \), then \( |(\vec{a} \times \vec{b}) \times \vec{c}| \) is equal to

• A. $2/3$
• B. $3/2$
• C. 2
• D. 3

Hint:

Let $\veca=2\hati+\hatj-2\hatk$ and $\vecb=\hati+\hatj$, if $\vecc$ is a vector such that $\veca·\vecc=|\vecc|$, $|\vecc-\veca|=2\sqrt2$ and the angle between $\vecax\vecb$ and $\vecc$ is $30$, then $|(\vecax\vecb)x\vecc|$ is equal to

Answer 1

The Correct Option is B

Step 1: Concept
Use the properties of vector magnitude and the cross product formula.
Step 2: Analysis
$|\vec{c}-\vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a}\cdot\vec{c}) = 8$. Since $|\vec{a}|^2 = 9$ and $\vec{a}\cdot\vec{c} = |\vec{c}|$, we have $|\vec{c}|^2 + 9 - 2|\vec{c}| = 8$.
Step 3: Evaluation
Solving $(|\vec{c}| - 1)^2 = 0$ gives $|\vec{c}| = 1$. $\vec{a}\times\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$, so $|\vec{a}\times\vec{b}| = 3$.
Step 4: Conclusion
$|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}| |\vec{c}| \sin 30^{\circ} = 3 \times 1 \times 1/2 = 3/2$.
Final Answer: (b)