Dot product of a vector with vector \(\overset { \wedge }{ 3i } -5\overset { \wedge }{ k } \), \(2\overset { \wedge }{ i } +7\overset { \wedge }{ j } \) and \(\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) are respectively -1, 6 and 5. Find the vector.

Find the equation of the plane through the intersection of the planes 2x-3y+ z-4 -0 and x - y + z + 1 = 0 and perpendicular to the plane x + 2y - 3z + 6 = 0

If \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } ,\overset { \rightarrow }{ b } =\overset { \wedge }{ j } -\overset { \wedge }{ k } ,\overset { \rightarrow }{ c } =\overset { \wedge }{ k } -\overset { \wedge }{ i } \) then find \(\left[ \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ b } -\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right] \)

Prove by vector method, that in a right angled triangle the square of the hypotenuse is equal to the sum of the square of the other two sides.

Show that the four points whose position vectors are \(6\overset { \wedge }{ i } -7\overset { \wedge }{ j } ,16\overset { \wedge }{ i } -29\overset { \wedge }{ j } -4\overset { \wedge }{ k } ,3\overset { \wedge }{ i } -6\overset { \wedge }{ j } \) are co-planar