 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says Now we know that cos theta is equal to mod of l1 l2 plus m1 n2 plus n1 n2 Where l1 m1 n1 and l2 m2 n2 are the direction cosines of two lines And theta is the acute angle between the two lines Now the equation of a plane containing a line is lx plus my plus nz is equal to 0 Where l mn are the direction cosines of the line So this is the key idea behind our question We will take the help of this key idea to solve the above question. So let's start the solution Now we are given the direction cosines for two lines Which are l1 m1 n1 and l2 m2 n2 So according to our key idea the planes which contain the two lines are of the form l1 x plus m1 y plus n1 z is equal to 0 Let us give this equation as number 1 and l2 x plus m2 y Plus n2 z is equal to 0. Let us give this as number 2 where l1 m1 n1 and l2 m2 n2 are direction cosines of the normal to the plane Now we have to show that the direction cosines of the line perpendicular to both of these lines are m1 n2 minus m2 n1 n1 l2 minus n2 l1 l1 m2 minus l2 m1 Let The line that is a third line With direction cosines l3 m3 m3 is perpendicular to Both the lines it is perpendicular to both the planes containing the lines the equation of the plane containing this third line is plus m3 y Plus n3 z is equal to 0 Let us give this as number 3 Now we have to find out the direction cosines of this line That is we have to find out l3 m3 n3 So the direction cosines of the third line can be find out by the cross product of normal vector to the given two planes thus the direction cosines of the line perpendicular to both of the given lines is given by by the cross product of equation 1 and 2 therefore, we have determinant of x y z l1 m1 n1 and l2 m2 n2 is equal to 0 or x into m1 n2 minus n1 m2 plus y into n1 l2 minus l1 n2 plus z into l1 n2 minus m1 l2 is equal to 0 Now we will compare these two equations so on comparing above equation with equation 3 we get l3 is equal to m1 n2 minus m2 n1 and m3 is equal to n1 l2 minus n2 l1 n3 is equal to l1 m2 minus l2 m1 and we have proved that the direction cosines of the line perpendicular to both of these lines are m1 n2 minus m2 n1 n1 l2 minus n2 l1 l1 m2 minus l2 m1 So this completes our session. I hope the solution is clear to you. Bye and have a nice day