 Hello and welcome to the session. In this session, first we will discuss equation of a plane in normal form. A plane is determined uniquely if we know the normal to the plane and its distance from the origin that is if we know the equation of a plane in normal form. Consider a plane which is at the perpendicular distance d from the origin where d is not equal to 0. Vector on is the normal from the origin to the plane and n cap is the unit normal vector along on. Let p be any point on the plane which coordinates x, y, z. Position vector of p is vector r with respect to the origin. So equation of a plane which is at a distance d from the origin and n cap is the unit vector normal to the plane through the origin. So the equation is given by vector r dot n cap is equal to d. This is the vector equation of the plane. Then next we have equation of a plane which is at a distance d from the origin and direction cosines of the normal to the plane mn. Then the equation is given by lx plus my plus nz equal to d. This is the Cartesian form of the equation of the plane. If the vector equation of the plane is given as vector r dot ai cap plus vj cap plus ck cap equal to d then we have ax plus by plus cz equal to d is the Cartesian equation of the plane where abc are the direction ratios of the normal to the plane. We need to find the vector equation of the plane which is at a distance of 7 units that is we have d equal to 7 from the origin and which is normal to the vector given by vector n equal to i cap plus 2j cap minus 2k cap. Now n cap is equal to vector n upon magnitude of vector n that is equal to i cap plus 2j cap minus 2k cap upon square root of 1 square plus 2 square plus minus 2 square and so that is equal to 1 upon 3i cap plus 2 upon 3j cap minus 2 upon 3k cap. Now the equation of the plane in vector form is given by vector r dot n cap is equal to d that is vector r dot 1 upon 3i cap plus 2 upon 3j cap minus 2 upon 3k cap is equal to 7 or this could be written as vector r dot i cap plus 2j cap minus 2k cap equal to 21. This is the required equation of the plane. Next we discuss equation of a plane perpendicular to a given vector and passing through a given point. Let a plane pass through a point A with position vector vector A and perpendicular to the vector n. Let r be the position vector of any point P which coordinates x, y, z in the plane. Then we have equation of a plane through a point whose position vector is vector A perpendicular to the vector n is given by vector r minus vector A dot vector n equal to 0. If we have the coordinates of the point A as x1, y1, z1 and coordinates of the point P as x, y, z and we have that A, B, C are the direction ratios of vector n then the equation of the plane is A into x minus x1 plus B into y minus y1 plus C into z minus z1 equal to 0. We need to find the equation of the plane passing through the point P which coordinates 1, 0 minus 2 and perpendicular to the vector n given by i cap plus j cap minus k cap. From the coordinates of the point P we have x1 as 1, y1 as 0 and z1 as minus 2 and from this vector n we get the direction ratios as A equal to 1, B equal to 1 and C equal to minus 1. So the required equation of the plane that passes through this point P and is perpendicular to the vector n is given by A into x minus x1 plus B into y minus y1 plus C into z minus z1 equal to 0. That is we have x minus 1 plus y minus z plus 2 equal to 0. That is finally we get x plus y minus z minus 3 equal to 0. This is the required equation of the plane. Next we discuss equation of a plane passing through 3 non-colonial points. Consider 3 non-colonial points R, S, T on the plane position vector of R is vector A with respect to the origin, position vector of the point S is vector B, position vector of the point T is vector C. Let R be the position vector of the point P which is in the plane. So vector equation of a plane that contains 3 non-colonial points having position vectors vector A, vector B and vector C is given by vector R minus vector A dot vector B minus vector A cross vector C minus vector A equal to 0. If we are given the coordinates of these 3 non-colonial points R, S and T like coordinates of the point R given by x1, y1, z1, point S given by x2, y2, z2 and point T given by x3, y3, z3 then the equation of the plane passing through these 3 non-colonial points is given by determinant x minus x1, y minus y1, z minus z1, x2 minus x1, y2 minus y1, z2 minus z1, x3 minus x1, y3 minus y1, z3 minus z1 equal to 0. We need to find the equation of the plane passing through the 3 non-colonial points given by point A with coordinates 110, B with coordinates 121 and C with coordinates minus 22 minus 1. From the point A we have x1 equal to 1, y1 equal to 1 and z1 equal to 0. From the point B we have x2 equal to 1, y2 equal to 2 and z2 equal to 1. Then from the point C we have x3 equal to minus 2, y3 equal to 2 and z3 equal to minus 1. So the required equation of the plane which passes through the points A, B and C is given by determinant x minus x1, y minus y1, z minus z1, x2 minus x1, y2 minus y1, z2 minus z1 then x3 minus x1, y3 minus y1, z3 minus z1 equal to 0. That is we have determinant x minus 1, y minus 1, z0, 1, 1, minus 3, 1, minus 1 equal to 0. On solving this determinant we get 2x plus 3y minus 3z equal to 5. This is the required equation of the plane. This completes the session. Hope you have understood how we find the equation of a plane in normal form, equation of a plane perpendicular to a given vector and passing through a given point and equation of a plane passing through three non-colonial points.