Question

If L, M, N are directional cosins of a vector then what is l²+m²+n²​

Answer 1

Answer:

We have l+m+n= 0 … (1) & l^2+m^2-n^2=0… (2) From (1), n= - (l +m). Putting this value of in (2) we get l^2+m^2 - (l+m)^2 = 0 or -2lm=0 or lm=0 which means either l=0 or m=0

If l=0, from (1) & (2) m + n =0 and m^2 - n^2 =0 which means m = -n and m^2 = n^2. Again, since I,m,n are direction cosines of a line, we have l^2 + m^2 + n^2 = 1 or 0 + m^2 + m^2 = 1 (because l =0 and m = - n) or 2 *m^2=1 or m^2=1/2 or m = +/- sqrt (1/2) & n = -/+ sqrt (1/2)

Considering one sign, the direction cosines of one line are 0, +sqrt (1/2) & -sqrt (1/2) i.e. l1=0, m1=sqrt (1/2) & n1 = - sqrt (1/2)

If m=0, similarly we get the direction cosines of the other line as sqrt (1/2), 0 & - sqrt (1/2). Now, if A be the angle between the two lines, then cos A = l1 * l2 + m1 * m2 + n1 * n2 = 1/2 which means A = pi/ 3

Explanation: