 Hello and welcome to the session. In this session we will discuss angle between two lines. We have two lines L1, L2 passing through the origin, let P be a point on line L1 and Q be a point on line L2. We have directed lines OP and OQ, let theta be the acute angle between OP and OQ. Now suppose that A1, B1, C1 are direction ratios of line L1, A2, B2, C2 are direction ratios of line L2. Now the directed line segment OP is a vector with components A1, B1, C1 and directed line OQ is a vector with components A2, B2, C2. As theta is the angle between the directed lines OP and OQ, so this theta is given by cos theta equal to modulus A1, A2 plus B1, B2 plus C1, C2 upon square root of A1 square plus B1 square plus C1 square multiplied by square root of A2 square plus B2 square plus C2 square. If instead of direction ratios for the lines L1 and L2 we are given the direction cosines for the lines that is we have L1, M1, N1 are direction cosines for the line L1 and L2, M2, N2 are direction cosines for the line L2 and theta is the acute angle between the lines L1 and L2 then we have cos theta is equal to modulus L1, L2 plus M1, M2 plus N1, M2. If the equation of the two lines is given by vector r equal to vector A1 plus lambda into vector B1 and vector r equal to vector A2 plus mu into vector B2 where these lambda and mu are some real numbers and if theta is the angle between these two lines then we have cos theta is equal to modulus vector B1 dot vector B2 upon magnitude of vector B1 into magnitude of vector B2 If in Cartesian form the equation of the lines is given by x minus x1 upon A1 equal to y minus y1 upon B1 equal to z minus z1 upon C1 and x minus x2 upon A2 equal to y minus y2 upon B2 equal to z minus z2 upon C2 where A1, B1, C1 are direction ratios of one line and A2, B2, C2 are direction ratios of the second line So if theta is the angle between these two lines then we have cos theta is given by modulus A1, A2 plus B1, B2 plus C1, C2 upon square root of A1 square plus B1 square plus C1 square multiplied by square root of A2 square plus B2 square plus C2 square and if the equation of the lines is given in the form x minus x1 upon L1 equal to y minus y1 upon M1 equal to z minus z1 upon N1 and x minus x2 upon L2 equal to y minus y2 upon M2 equal to z minus z2 upon N2 where L1, M1, N1 are direction cosines of one line and L2, M2, N2 are direction cosines of the other line and if theta is the angle between these two lines then we have cos theta is equal to modulus L1, L2 plus M1, M2 plus N1, N2 One very important thing if we have two lines with direction ratios A1, B1, C1 and A2, B2, C2 then the two lines are perpendicular that is we have theta equal to 90 degrees if A1, A2 plus B1, B2 plus C1, C2 is equal to 0 and the two lines would be parallel that is theta is equal to 0 degrees if we have A1 upon A2 is equal to B1 upon B2 is equal to C1 upon C2 Suppose we are given the equation of two lines as x plus 4 upon 3 equal to y minus 1 upon 5 equal to z plus 3 upon 4 and x plus 1 upon 1 equal to y minus 4 upon 1 equal to z minus 5 upon 2 we have to find the angle between these two lines From the first equation of line we have A1 equal to 3, B1 equal to 5, C1 equal to 4 that is these are the direction ratios of this line Then from the second equation of line we have A2 equal to 1, B2 equal to 1 and C2 equal to 2 that is the direction ratios of the second line Now the angle between the two lines is given by theta so cos theta is equal to modulus A1, A2 plus B1, B2 plus C1, C2 upon square root of A1, square plus B1, square plus C1, square multiplied by square root of A2 square plus B2 square plus C2 square this comes out to be equal to 8 root 3 upon 15 that is cos theta is equal to 8 root 3 upon 15 so from here we get the angle between the two lines theta is equal to cos inverse 8 root 3 upon 15 This completes the session hope you have understood how we find the angle between the two lines