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Prove that vector a dot in bracket vector a cross vector b is equal to 0 |
| Answer» OSS 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 QThe 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) cos90The 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 | |