A force of magnitude 6 units acting parallel to \(\overset { \wedge }{ 2i } -\overset { \wedge }{ 2j } +\overset { \wedge }{ k } \) displaces the point of application from (1, 2, 3) to (5, 3, 7). Find the work done.

Forces \(2\overset { \wedge }{ i } +7\overset { \wedge }{ j } \), \(2\overset { \wedge }{ i } -5\overset { \wedge }{ j } +6\overset { \wedge }{ k } \), \(-\overset { \wedge }{ i } +2\overset { \wedge }{ j } -\overset { \wedge }{ k } \) act at a point P whose position vector is \(4\overset { \wedge }{ i } -3\overset { \wedge }{ j } -2\overset { \wedge }{ k } \). Find  the vector moment of the resultant of these forces acting at P about this point Q whose position vector is \(6\overset { \wedge }{ i } +\overset { \wedge }{ j } -3\overset { \wedge }{ k } \)

Find the parametric form of vector equation of a line passing through a point (2, -1, 3) and parallel to line \({ \overset { \rightarrow }{ r } }=\left( \overset { \wedge }{ i } +\overset { \wedge }{ j } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \)

If the planes \({ \overset { \rightarrow }{ r } }.\left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) =7\) and \({ \overset { \rightarrow }{ r } }.\left( \lambda \overset { \wedge }{ i } +2\overset { \wedge }{ j } -7\overset { \wedge }{ k } \right) =26\) are perpendicular. Find the value of λ.

Let \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } \) be unit vectors such \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ c } =0\) and the angle between \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) is \(\frac { \pi }{ 6 } \). Prove that \(\overset { \rightarrow }{ a } =\pm 2\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)