 Hello and welcome to the session. In this session we discussed the following question which says the position vectors of two points A and B are 5i cap plus 3j cap plus 9k cap and i cap minus 2j cap minus 6k cap respectively. Find the vector equation of the plane passing through A and perpendicular to vector AB. We have the equation of a plane passing through a point P which coordinates x1, y1, z1 is A into x minus x1 plus B into y minus y1 plus C into z minus z1 equal to 0 where we have ABC are the direction ratios of the normal to the plane. This is the key idea for this question. Now we proceed with the solution. We are given the position vector of point A is given as 5i cap plus 3j cap plus 9k cap. So this means that the coordinates of the point A would be 539. Then we have the position vector of point B is given as i cap minus 2j cap minus 6k cap and so the coordinates of point B would be 1 minus 2 minus 6. Now next we find out the direction ratios of vector AB. These are given by 5 minus 1, 3 minus of minus 2 so that would be 3 plus 2, 9 minus of minus 6 so 9 plus 6. So that means we get the direction ratios of vector AB are 4, 5, 15. In the equation we have that the plane passes through the point A and is perpendicular to vector AB. So using this key idea we get the equation of a plane passing through point A and perpendicular to vector AB is 4 into x minus the x coordinate of point A which is 5 plus 5 into y minus y coordinate of point A plus 15 into z minus z coordinate of point A is equal to 0. So this means we have 4x minus 20 plus 5y minus 15 plus 15z minus 135 is equal to 0. So this means we have 4x plus 5y plus 15z minus 170 is equal to 0. So this is the equation of the plane in Cartesian form. Now its vector equation is given as vector r dot 4i cap plus 5j cap plus 15k cap minus 170 is equal to 0. So this is the required vector equation of the plane passing through the point A and perpendicular to the vector AB. So this completes the session. Hope you have understood the solution of this question.