Question

Let \( \vec{a}=2 \hat{i}+\hat{j}-2 \hat{k} \) and \( \vec{b}=\hat{i}+\hat{j} \). Let \( \overrightarrow{ c } \) be a vector such that \( |\overrightarrow{ c }-\overrightarrow{ a }|=3,|(\overrightarrow{ a } \times \overrightarrow{ b }) \times \overrightarrow{ c }|=3 \) and the angle between \( \overrightarrow{ c } \) and \( \overrightarrow{ a } \times \overrightarrow{ b } \) be \( 30^{\circ} \). Then \( \overrightarrow{ a } \cdot \overrightarrow{ c } \) is equal to: [JEE (Main) 2017\( ] \)

Answer 1

\(\vec a \times \vec b = \begin{vmatrix} \hat i &\hat j &\hat k\\2&1&-2\\1&1&0\end{vmatrix} = 2\hat i - 2\hat j + \hat k\)

\(|(\vec a \times \vec b)\times \vec c| = 3\)

⇒ \(|\vec a \times \vec b|. |\vec c|sin30° = 3\)

⇒ \(|\vec c| = \frac6{|\vec a \times \vec b|} = \frac6{\sqrt{4 + 4+ 1}} = \frac 63 = 2\)

\(\because |\vec c - \vec a|^2 = 9\)

⇒ \(|\vec c|^2 + |\vec a|^2 - 2\vec a .\vec c = 9\)

⇒ \(4 + 9-2\vec a.\vec c = 9\)

⇒ \(2\vec a.\vec c = 4\)

⇒ \(\vec a.\vec c = 2\)