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 \(\frac{\pi}{6}\), then the value of \(|(\vec{a}\times\vec{b})\times\vec{c}|\) is :

• A. \(\frac{2}{3}\)
• B. 4
• C. 3
• D. \(\frac{3}{2}\)

Hint:

Find \(|\vec{c}|\) from the two given conditions, then use \(|\vec{u}\times\vec{c}| = |\vec{u}||\vec{c}|\sin\frac{\pi}{6}\).

Answer 1

The Correct Option is D

\(|\vec{a}| = \sqrt{4+1+4} = 3\).
\(|\vec{c}-\vec{a}|^2 = |\vec{c}|^2 - 2\vec{a}\cdot\vec{c} + |\vec{a}|^2 = |\vec{c}|^2 - 2|\vec{c}| + 9 = 8\).
So \(|\vec{c}|^2 - 2|\vec{c}| + 1 = 0\), i.e. \((|\vec{c}|-1)^2 = 0\), giving \(|\vec{c}| = 1\).
\(\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-2\\1&1&0\end{vmatrix} = 2\hat{i} - 2\hat{j} + \hat{k}\), so \(|\vec{a}\times\vec{b}| = 3\).
\(|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}||\vec{c}|\sin\frac{\pi}{6} = 3\cdot 1\cdot\frac{1}{2} = \frac{3}{2}\).
Answer: (4) \(\frac{3}{2}\)