 Hello and welcome to the session. In this session first we will discuss intercept form of the equation of a plane. Let the plane make A intercept on x axis, B intercept on y axis and C intercept on the z axis. Hence we have that the plane meets the x axis at the point P with coordinates A00, it meets y axis at the point Q with coordinates 0B0 and at z axis the plane meets at the point R with coordinates 00C. So we have equation of a plane that cuts the coordinate axis PQNR is given by x upon A plus y upon B plus where upon C equal to 1. Let's find the equation of the plane which makes the intercept 1 on x axis, 2 on y axis and 3 on z axis that is we have A is equal to 1, B is equal to 2 and C is equal to 3. So the equation of the plane is given by x upon A plus y upon B plus z upon B equal to 1 or this could be written as 6x plus 3y plus 2z equal to 6. This is the required equation of the plane. Next we have plane passing through the intersection of two given planes. Suppose the equation of the two planes in vector form be given by vector R dot vector N1 equal to D1 and vector R dot vector N2 equal to D2. So we have vector equation of a plane that passes through the intersection of two planes given by these two equations is vector R dot vector N1 plus lambda into vector N2 is equal to D1 plus lambda into D2 where this lambda is some nonzero constant. If the equation of two planes is given in Cartesian form as A1x plus B1y plus C1z minus D1 equal to 0 and A2x plus B2y plus C2z minus D2 is equal to 0 then the Cartesian equation of a plane that passes through the intersection of two planes given by these two equations is A1x plus B1y plus C1z minus D1 plus lambda into A2x plus B2y plus C2z minus D2 equal to 0. Suppose the two equation of the planes in vector form be given by vector R dot i cap plus 3j cap minus k cap equal to 5 and vector R dot 2i cap minus j cap plus k cap equal to 3. We need to find the equation of the plane that passes through the intersection of these two given planes and a point A which coordinates 2 1 minus 2. From the first equation of the plane we have vector n1 is equal to i cap plus 3j cap minus k cap and D1 is equal to 5. Now from the second equation of the plane we have vector n2 is equal to 2i cap minus j cap plus k cap and D2 is equal to 3. Now equation of the plane through the intersection of these two planes is vector R dot vector n1 plus lambda into vector n2 equal to D1 plus lambda into D2. That is we get vector R dot 1 plus 2 lambda i cap plus 3 minus lambda j cap minus 1 minus lambda k cap equal to 5 plus 3 lambda. This is the equation of the plane that passes through the intersection of the two given planes. It is also given that this plane passes through the point A which coordinates 2 1 minus 2. So we have vector R equal to 2i cap plus j cap minus 2k cap. Hence we have from this equation of the plane 2i cap plus j cap minus 2k cap dot 1 plus 2 lambda i cap plus 3 minus lambda j cap minus 1 minus lambda k cap equal to 5 plus 3 lambda. So from here we have 2 into 1 plus 2 lambda plus 3 minus lambda minus 2 into lambda minus 1 minus 5 plus 3 lambda. From here we get value of lambda as 1. When we substitute this lambda equal to 1 in this equation of the plane we get vector R dot 3i cap plus 2j cap equal to 8. So this is the required equation of the plane that passes through the intersection of the two given planes and the point A which coordinates 2 1 minus 2. Next we discuss cooperality of two lines. If the two lines be given in the vector form as vector R equal to vector A1 plus lambda into vector B1 and vector R equal to vector A2 plus mu into vector B2. This line passes through say a point A with position vector vector A1 and is parallel to vector B1 and this line passes through say a point B with position vector vector A2 and is parallel to vector B2. Then these two lines would be co-planar if we have vector A2 minus vector A1 dot vector B1 cross vector B2 is equal to 0. If the equation of the lines is given in Cartesian form as x minus x1 upon A1 equal to y minus y1 upon B1 equal to z minus z1 upon C1 then x minus x2 upon A2 equal to y minus y2 upon B2 equal to z minus z2 upon C2 where x1, y1, z1 are the coordinates of the point A through which the line 1 passes and A1, B1, C1 are the direction ratios of the vector B1. Then x2, y2, z2 are the coordinates of the point B through which the line 2 passes and A2, B2, C2 are the direction ratios of the vector B2. Then these two lines would be co-planar if we have determinant x2 minus x1, y2 minus y1, z2 minus z1, A1, B1, C1, A2, B2, C2 is equal to 0. Let the equation of the lines in Cartesian form be given as x upon 1 equal to y minus 2 upon 2 equal to z plus 3 upon 3 and x minus 2 upon 2 equal to y minus x upon 3 equal to z minus 3 upon 4. Let's see if these two lines are co-planar or not. From the first equation of the line we have x1 equal to 0, y1 equal to 2 and z1 equal to minus 3, then A1 equal to 1, B1 equal to 2 and C1 equal to 3. From the second equation of the line we have x2 equal to 2, y2 equal to 6 and z2 equal to 3, A2 equal to 2, B2 equal to 3 and C2 equal to 4. Let's consider the determinant x2 minus x1, y2 minus y1, z2 minus z1, A1, B1, C1, A2, B2, C2. That is this is equal to determinant 2, 4, 6, 1, 2, 3 and 2, 3, 4. On solving this determinant we get this is equal to 0 and so we have the given two lines are co-planar. Thus this completes the session. Hope you have understood the intersect form of the equation of a plane, plane passing through the intersection of the two given planes and co-planarity of two lines.