Question

Unit vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar. A unit vector \(\vec{d}\) is perpendicular to them. If \((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\frac{1}{6} \hat{i}-\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k}\) and the angle between \(\vec{a}\) and \(\vec{b}\) is \(30^{\circ}\), then \(\vec{c}\) is/are :

• A. \(\pm\left(\frac{-\hat{i}-2 \hat{j}+2 \hat{k}}{3}\right)\)
• B. \(\left(\frac{2 \hat{i}+\hat{j}-\hat{k}}{3}\right)\)
• C. \(\pm\left(\frac{-\hat{i}+2 \hat{j}-2 \hat{k}}{3}\right)\)
• D. \(\left(\frac{-2 \hat{i}-2 \hat{j}+\hat{k}}{3}\right)\)

Hint:

Use \((\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})=\vec{c}\,[(\vec{a}\times\vec{b})\cdot\vec{d}]-\vec{d}\,[(\vec{a}\times\vec{b})\cdot\vec{c}]\); the second term vanishes because \(\vec{a},\vec{b},\vec{c}\) are coplanar.

Answer 1

The Correct Option is C

Let \(\vec{n}=\vec{a}\times\vec{b}\). Since \(\vec{a},\vec{b}\) are unit vectors at \(30^{\circ}\), \(|\vec{n}|=\sin 30^{\circ}=\frac{1}{2}\), and \(\vec{n}\) is perpendicular to the plane containing \(\vec{a},\vec{b},\vec{c}\).
Since \(\vec{d}\) is a unit vector perpendicular to that plane, \(\vec{n}\parallel\vec{d}\), so \(\vec{n}\cdot\vec{d}=\pm\frac{1}{2}\). Also \(\vec{n}\cdot\vec{c}=0\) (coplanar).
Using \(\vec{n}\times(\vec{c}\times\vec{d})=\vec{c}(\vec{n}\cdot\vec{d})-\vec{d}(\vec{n}\cdot\vec{c})\) we get \((\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})=\pm\frac{1}{2}\vec{c}\).
Hence \(\pm\frac{1}{2}\vec{c}=\frac{1}{6}\hat{i}-\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}\), so \(\vec{c}=\pm\left(\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}\right)=\pm\left(\frac{-\hat{i}+2\hat{j}-2\hat{k}}{3}\right)\) (its magnitude is 1, as required).
Answer: (3) \(\pm\left(\frac{-\hat{i}+2 \hat{j}-2 \hat{k}}{3}\right)\)