Find the angle between the lines whose direction cosines are given by l+m+n=0 and 2l2+2m2−n2=0

The angle between two lines with direction cosines (l1, m1, n1) and (l2, m2, n2) is given by the formula:

cos θ = l1l2 + m1m2 + n1*n2

Given the direction cosines of the two lines are l+m+n=0 and 2l^2+2m^2−n^2=0, we can express m and n in terms of l for both lines.

For the first line, we can express m as -l-n and n as -l-m. Substituting these into the equation gives us:

cos θ = ll + (-l-n)(-l-n) + (-l-m)*(-l-m)

Simplifying this gives us:

cos θ = l^2 + l^2 + n^2 + l^2 + m^2

cos θ = 3l^2 + m^2 + n^2

For the second line, we can express m as sqrt(1-l^2-n^2) and n as sqrt(1-l^2-m^2). Substituting these into the equation gives us:

cos θ = 2l^2 + 2*(sqrt(1-l^2-n^2))^2 - (sqrt(1-l^2-m^2))^2

Simplifying this gives us:

cos θ = 2l^2 + 2 - 2l^2 - 2n^2 - 1 + l^2 + m^2

cos θ = l^2 + m^2 - n^2 + 1

Setting the two equations for cos θ equal to each other gives us:

3l^2 + m^2 + n^2 = l^2 + m^2 - n^2 + 1

Solving for l, m, and n gives us:

l = sqrt(1/3), m = sqrt(1/3), n = -sqrt(1/3)

Substituting these values back into the equation for cos θ gives us:

cos θ = 3*(sqrt(1/3))^2 + (sqrt(1/3))^2 + (-sqrt(1/3))^2

cos θ = 1

Therefore, the angle between the two lines is θ = cos^-1(1) = 0 degrees.