Question

prove that vector a dot in bracket vector a cross vector b is equal to 0​

Answer 1

✔The cross product of A and B i.e, (A×B) is perpendicular to A.

The cross product of A and B i.e, (A×B) is perpendicular to A.We know that the scalar product of two vectors P and Q is PQcos(theta)

The cross product of A and B i.e, (A×B) is perpendicular to A.We know that the scalar product of two vectors P and Q is PQcos(theta)Where theta is the angle between P and Q

The cross product of A and B i.e, (A×B) is perpendicular to A.We know that the scalar product of two vectors P and Q is PQcos(theta)Where theta is the angle between P and QHere the scalar product of A and( A×B)

The cross product of A and B i.e, (A×B) is perpendicular to A.We know that the scalar product of two vectors P and Q is PQcos(theta)Where theta is the angle between P and QHere the scalar product of A and( A×B) That is ,A.(A×B) = A .(A×B) cos90

The cross product of A and B i.e, (A×B) is perpendicular to A.We know that the scalar product of two vectors P and Q is PQcos(theta)Where theta is the angle between P and QHere the scalar product of A and( A×B) That is ,A.(A×B) = A .(A×B) cos90So , A.(A×B) = 0. ;(as cos90 = 0.)

The cross product of A and B i.e, (A×B) is perpendicular to A.We know that the scalar product of two vectors P and Q is PQcos(theta)Where theta is the angle between P and QHere the scalar product of A and( A×B) That is ,A.(A×B) = A .(A×B) cos90So , A.(A×B) = 0. ;(as cos90 = 0.)Hence proved

Answer 2

Answer:

Explanation:

Let  A.(AxB) = 0

As , AxB = AB sinФ n

AB sinФ n is a vector which is perpendicular to the plane having A vector and B vector which implies that it is also perpendicular to A vector .  

As we know dot product of two vectors is zero.

Thus , we can say that

A.(AxB) = 0

Hence , proved.

Hope it helps!