If l, m, n are the direction cosines of a normal to the plane 2 x-3 y+6 z+4=0 , then what is the value of 49 (7 l^2+m^2-n^2 ) ?

Step-by-step solution

The plane equation is a x+b y+c z=d Given the plane equation is 2 x-3 y+6 z+4=0 2 x-3 y+6 z=-4 Comparing then we get, aligned a =2, b & =-3, c=6, d=-4 a^2+b^2+c^2 & = (2)^2+(-3)^2+(6)^2 & = 4+9+36 = 49 =7 aligned Direction cosines of normal to the plane aligned l & = a a^2+b^2+c^2 = 2 7 m & = b a^2+b^2+c^2 =- 3 7 n & = c a^2+b^2+c^2 = 6 7 l, m, n & = 2 7 ,- 3 7 , 6 7 49 (7 l^2+m^2-n^2 ) & =49 [7 ( 2 7 )^2+ (- 3 7 )^2- ( 6 7 )^2 ] & =49 [ 28 49 + 9 49 - 36 49 ] & =49 [ 1 49 ]=1 aligned