 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says find the vector and Cartesian equations of the plane that passes through the point 1 0 minus 2 and the normal to the plane is i k plus j k plus k cap Now we know that a plane is determined if it passes through a point and is perpendicular to a given direction So the equation of a plane through a point whose position vector is a and perpendicular to the vector n is vector r minus vector a Into vector n is equal to zero So this is a vector equation of the plane and the equation of a plane perpendicular to a given line with direction ratios a b c and passing through a given point with coordinates x1 y1 z1 is a into x minus x1 plus b into y minus y1 plus c into z minus z1 is equal to zero So this is the Cartesian form of the equation of a plane So this is a key idea behind that question We will take the help of this key idea to solve the above question So let's start the solution Let the plane passes through the point a with coordinates 1 0 minus 2 with position vector a And perpendicular to the vector we have vector a is equal to i cap minus 2 k cap and the vector n is given to us which is i cap plus j cap minus k cap Now according to our key idea the vector equation of the plane is given by vector r minus vector a Into vector n is equal to zero or vector r minus vector a which is i cap minus 2 k cap Into vector n which is i cap plus j cap minus k cap is equal to zero So this is a vector equation of the plane now the Cartesian equation of a plane perpendicular to a given line with direction ratios a b c and passing through a given point x 1 y 1 z 1 is given by a into x minus x 1 plus b into y minus y 1 Plus c into z minus z 1 is equal to zero Now here we have a is equal to 1 v is equal to 1 and c is equal to minus 1 Again the given plane is passing through the point 1 0 minus 2 so x 1 y 1 z 1 are 1 0 minus 2 respectively so the required Cartesian equation of the plane is 1 into x minus 1 plus 1 into y minus 0 minus 1 into z plus 2 is equal to zero or x minus 1 plus y minus z minus 2 is equal to zero or x plus y minus z is equal to 3 So this is a Cartesian equation of the plane that passes through the point 1 0 minus 2 and the normal to the plane is I cap plus j cap minus k cap So the answer for the above question is the vector equation of the plane is vector R minus I cap minus 2k cap into I cap plus j cap minus k cap is equal to zero and The Cartesian form of the equation of the plane is x plus 5 minus z is equal to 3 So this completes our session. I hope the solution is clear to you while harmonizing