 Hello and welcome to the session. In this session we will discuss about the scalar or dot product of two vectors. The scalar product of two non-zero vectors A and B is given by vector A dot vector B is equal to magnitude of vector A into magnitude of vector B into cos theta where this theta is the angle between vector A and vector B and theta is greater than equal to 0 and less than equal to pi. If in case vector A is equal to 0 or vector B is equal to 0 then theta is not defined and in this case vector A dot vector B is equal to 0. The scalar product vector A dot vector B is a real number. Now we have A and B are two non-zero vectors then vector A dot vector B is equal to 0 if and only if vector A and vector B are perpendicular to each other. Now in this if we have theta is equal to 0 then vector A dot vector B is equal to magnitude of vector A into magnitude of vector B and if we have theta is equal to pi then vector A dot vector B is equal to minus magnitude of vector A into magnitude of vector B and if i cap j cap k cap are mutually perpendicular unit vectors then we have i cap dot i cap is equal to j cap dot j cap is equal to k cap dot k cap is equal to 1 and i cap dot j cap is equal to j cap dot k cap is equal to k cap dot i cap is equal to 0. And also from the formula of the scalar product we see that cos theta is equal to vector A dot vector B upon magnitude of vector A into magnitude of vector B or we get theta is equal to cos inverse vector A dot vector B upon magnitude of vector A into magnitude of vector B. Also scalar product is commutative that is we have vector A dot vector B is equal to vector B dot vector A. Now we shall discuss properties of scalar product. The first property is distributivity of scalar product over addition. According to this we have if vector A vector B vector C are any three vectors then vector A dot vector B plus vector C is equal to vector A dot vector B plus vector A dot vector C. And now the next property is if A and B are any two vectors and lambda be any scalar then we have lambda into vector A dot vector B is equal to lambda into vector A dot vector B equal to vector A dot lambda into vector B. Consider vector A equal to i cap minus 2 j cap plus 3 k cap and vector B equal to 3 i cap minus 2 j cap plus k cap then if theta is the angle between the vectors A and B then we have cos theta is equal to vector A dot vector B upon magnitude of vector A into magnitude of vector B. This is equal to i cap minus 2 j cap plus 3 k cap dot 3 i cap minus 2 j cap plus k cap upon square root of 1 square plus minus 2 square plus 3 square multiplied by 3 square plus minus 2 square plus 1 square. So this becomes equal to 3 plus 4 plus 3 upon square root 14 into square root 14. So this becomes equal to 5 upon 7 that is we get cos theta is equal to 5 upon 7 or we can say theta is equal to cos inverse 5 upon 7. Next we shall discuss projection of a vector on a line. Suppose a vector AB makes an angle theta with a given directed line L in anti clockwise direction then we have projection of vector AB on line L is given by the vector P where this vector P is called the projection vector and magnitude of vector P is given as magnitude of vector AB into cos theta magnitude of vector P simply called the projection of the vector AB on line L and direction of P is the same or opposite to that of line L depending upon whether cos theta is positive or negative. If P cap is a unit vector along the line L then projection of vector A on line L is given by vector A dot P cap also projection of vector A on vector B is given by vector A dot B cap which is also equal to vector A dot vector B upon magnitude of vector B that is equal to 1 upon magnitude of vector B into vector A dot vector B and if we have theta is equal to 0 then projection vector of vector AB will be vector AB itself and if theta is equal to pi then projection vector of vector AB will be vector BA also if we have theta equal to pi by 2 or 3 pi by 2 then projection vector of vector AB will be 0 vector. Consider vector A equal to I cap minus J cap and vector B equal to I cap plus J cap then projection of vector A on vector B is given by 1 upon magnitude of vector B into vector A dot vector B which is equal to 1 upon square root of 1 square plus 1 square into I cap minus J cap dot I cap plus J cap this is equal to 1 minus 1 upon square root 2 which is equal to 0. So we get projection of vector A on vector B is 0. So this completes the session hope you have understood the scalar product of two vectors.