 Hello and welcome to the session. In this session, first we will discuss angle between two planes. Basically, the angle between two planes is defined as the angle between their normals. In vector form, if the equation of the planes is given by vector r dot vector n1 equal to d1 vector r dot vector n2 equal to d2. And if theta is the angle between two planes, then we have cos theta is equal to modulus vector n1 dot vector n2 upon magnitude of vector n1 into magnitude of vector n2. That is, we get theta is equal to cos inverse modulus vector n1 dot vector n2 upon magnitude of vector n1 into magnitude of vector n2. If the equation of the planes is given in the Cartesian form as a1x plus b1y plus c1z plus d1 equal to 0 and a2x plus b2y plus c2z plus d2 equal to 0. And again, if theta is the angle between these two planes, then we have cos theta is equal to modulus a1a2 plus b1b2 plus c1c2 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 we have that the planes are at right angles, that is, we have theta is equal to 90 degrees. So, we get cos theta is equal to 0, that is, cos theta is equal to a1a2 plus b1b2 plus c1c2 is equal to 0. And if we have that the planes are parallel, a1 upon a2 is equal to b1 upon b2 is equal to c1 upon c2. Let the two planes in vector form be given by vector r dot i cap plus j cap plus 2k cap equal to 5 and vector r dot 2i cap minus j cap plus k cap equal to 8. Let theta be the angle between these two planes, then we are supposed to find theta. From the first equation of the plane we have vector n1 is equal to i cap plus j cap plus 2k cap and from the second equation of the plane we have vector n2 is equal to 2i cap minus j cap plus k cap. Now, vector n1 dot vector n2 is equal to i cap plus j cap plus 2k cap dot 2i cap minus j cap plus k cap and this comes out to be equal to 3. Then magnitude of vector n1 is equal to square root of 1 square plus 1 square plus 2 square and that is equal to square root 6. Then magnitude of vector n2 is equal to square root of 2 square plus minus 1 square and that is equal to root 6. Now, cos theta is equal to modulus of vector n1 dot vector n2 that is 3 upon magnitude of vector n1 that is root 6 into magnitude of vector n2 that is root 6 and this is equal to modulus of 3 upon 6 that is equal to modulus of 1 upon 2 and so we have cos theta is equal to 1 upon 2 which gives us theta equal to pi upon 3. This is the angle between the given two planes. Next we discuss distance of a point from a plane. Consider the equation of the plane in vector form be given by vector r dot n cap equal to d and let p be a point with position vector vector a. So distance of point p from the plane given by this equation is modulus d minus vector a dot n cap and if the equation of the plane is given in the form vector r dot vector n equal to d where this vector n is the normal to the plane then the perpendicular distance is given by modulus vector a dot vector n minus d upon magnitude of vector n and length of the perpendicular from origin o to the plane vector r dot vector n equal to d is given by modulus d upon magnitude of vector n. Since we have vector a is equal to 0. If in the Cartesian form the equation of the plane is given as ax plus by plus z plus d equal to 0 and let the coordinates of the point p which are the position vector a b x1, y1, z1, y2, y3, y4, y5, y5, y6, then we have distance from point p to the plane given by this equation is modulus x1 plus by1 plus z1 plus d upon square root of a square plus b square plus c square. Let's find out the distance of the point p which coordinates 2 1 minus 1 from the plane given by the equation x minus 2 y plus 4 z equal to 9. From the coordinates of this point p we have x1 equal to 2, y1 equal to 1, z1 equal to minus 1 and from the equation of this plane we have a equal to 1, b equal to minus 2, c equal to 4 and d equal to minus 9. Then the required distance that is the distance of the point p from the given equation of the plane is equal to modulus ax1 plus by1 plus cz1 plus d upon square root of a square plus b square plus c square. This is equal to modulus minus 13 upon square root 21 and so this is equal to 13 upon square root 21 is the required distance. Next we have angle between a line and a plane. Basically the angle between a line and a plane is equal to minus 13 upon square root 21 and so this is equal to minus 13 upon square root 21 is the required distance. Next we have angle between a line and a plane. Basically the angle between a line and a plane is the complement of the angle between the line and normal to the plane. Suppose the equation of the line in the vector form be given by vector r equal to vector a plus lambda into vector b and equation of the plane in the vector form be given by vector r dot vector n equal to d. If we have theta is the angle between line and the plane then we have sin theta is equal to modulus vector b dot vector n upon magnitude of vector b into magnitude of vector n. Consider the equation of the line given as vector r equal to i cap plus j cap minus 3 k cap plus lambda into 2 i cap plus 2 j cap plus k cap and equation of the plane be given by vector r dot 6 i cap minus 3 j cap plus 2 k cap equal to 5. From the equation of the line we have vector b is equal to 2 i cap plus 2 j cap plus k cap then from the equation of the plane we have vector n is equal to 6 i cap minus 3 j cap plus 2 k cap. Now vector b dot vector n is equal to 2 i cap plus 2 j cap plus k cap dot 6 i cap minus 3 j cap plus 2 k cap and that comes out to be equal to 8. Then we have magnitude of vector b is equal to square root of 2 square plus 2 square plus 1 square and that is equal to 3. Then magnitude of vector n is equal to square root of 6 square plus minus 3 square plus 2 square and that is equal to 7. Then theta is the angle between the given line and the plane so we have sin theta is equal to modulus vector b dot vector n that is 8 upon magnitude of vector b into magnitude of vector n that is equal to modulus 8 upon 21 so we have sin theta is equal to 8 upon 21 which gives theta is equal to sin inverse 8 upon 21. So this is the angle between the given line and the plane. This completes the session hope you have understood how we find the angle between two planes distance of a point from a plane and angle between a line and a plane.