Question

The angle between the straight lines, whose direction cosines are given by the equations 2l+2m-n=0 and mn + nl + lm = 0

Answer 1

90°

Answer 2

1. From the linear equation $2l+2m-n=0$, we express $n$ in terms of $l$ and $m$: $n = 2l+2m$.
2. Substitute this expression for $n$ into the second equation $mn+nl+lm=0$: $m(2l+2m) + (2l+2m)l + lm = 0$
3. Expand and simplify the equation: $2lm + 2m^2 + 2l^2 + 2lm + lm = 0$ $2l^2 + 5lm + 2m^2 = 0$
4. Factor the homogeneous quadratic equation: $(2l+m)(l+2m) = 0$
5. This equation yields two possible relationships between $l$ and $m$, corresponding to the direction ratios of two distinct lines:
   • Case 1: $2l+m=0 \implies m = -2l$. Substitute $m=-2l$ into $n=2l+2m$: $n = 2l + 2(-2l) = 2l - 4l = -2l$. The direction ratios $(l, m, n)$ are in the ratio $l : -2l : -2l$. For $l=1$, we get the direction ratios $\vec{d_1} = \langle 1, -2, -2 \rangle$.
   • Case 2: $l+2m=0 \implies l = -2m$. Substitute $l=-2m$ into $n=2l+2m$: $n = 2(-2m) + 2m = -4m + 2m = -2m$. The direction ratios $(l, m, n)$ are in the ratio $-2m : m : -2m$. For $m=1$, we get the direction ratios $\vec{d_2} = \langle -2, 1, -2 \rangle$.
6. Let $\theta$ be the angle between the two lines. The cosine of the angle is given by the dot product of their direction ratios: $\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$
7. Calculate the dot product $\vec{d_1} \cdot \vec{d_2}$: $\vec{d_1} \cdot \vec{d_2} = (1)(-2) + (-2)(1) + (-2)(-2) = -2 - 2 + 4 = 0$.
8. Since the dot product is $0$, $\cos \theta = 0$.
9. Therefore, the angle between the lines is $\theta = \cos^{-1}(0) = 90^\circ$.