 Hello students, let's work out the following problem. It says prove that vector A cross vector B whole square is equal to mod of vector A square into mod of vector B square minus vector A dot vector B whole square. So let's now move on to the solution. Now we know that vector A cross vector B is given by the formula mod of vector A into mod of vector B into sine theta n cap where n is the unit vector perpendicular to both vector A and vector B. Now we have to find vector A cross vector B whole square. So we have, squaring both sides we have mod of vector A square mod of vector B square sine square theta vector n cap square. So this is mod of vector A square mod of vector B square sine square theta can be written as 1 minus cos square theta and n cap square is n dot n. Now it is vector A square mod of vector B square minus mod of vector A square mod of vector B square cos square theta and n cap dot n cap is 1 as we know that the dot product of unit vector with itself is 1. Now again this is equal to mod of vector A square mod of vector B square and this is vector A dot vector B whole square as we know that A dot vector B is given by the formula vector A mod of vector A into mod of vector B into cos theta. So this is the square of vector A dot vector B whole square and this is what we have to prove. So this completes the question and the session. Bye for now. Take care. Have a good day.