Question

Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l+m-n=0$ and $l+m^2-n^2=0$. Then the value of $\sin^4\alpha + \cos^4\alpha$ is

Answer 1

1/2

Answer 2

The direction cosines $(l, m, n)$ satisfy $l+m-n=0$ and $l+m^2-n^2=0$. From the first equation, $n = l+m$. Substituting into the second equation: $l + m^2 - (l+m)^2 = 0$ $l + m^2 - (l^2 + 2lm + m^2) = 0$ $l - l^2 - 2lm = 0$ $l(1 - l - 2m) = 0$

This implies either $l=0$ or $1-l-2m=0$.

Case 1: $l=0$. If $l=0$, then $n = 0+m = m$. Using the property $l^2+m^2+n^2=1$: $0^2 + m^2 + m^2 = 1$ $2m^2 = 1 \implies m = \pm \frac{1}{\sqrt{2}}$. If $m = \frac{1}{\sqrt{2}}$, then $n = \frac{1}{\sqrt{2}}$. Direction cosines are $(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. If $m = -\frac{1}{\sqrt{2}}$, then $n = -\frac{1}{\sqrt{2}}$. Direction cosines are $(0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. These two sets of direction cosines define one line. Let's pick a direction vector for Line 1: $\mathbf{v_1} = (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.

Case 2: $1-l-2m=0$. This implies $l = 1-2m$. Then $n = l+m = (1-2m)+m = 1-m$. Using the property $l^2+m^2+n^2=1$: $(1-2m)^2 + m^2 + (1-m)^2 = 1$ $(1 - 4m + 4m^2) + m^2 + (1 - 2m + m^2) = 1$ $6m^2 - 6m + 2 = 1$ $6m^2 - 6m + 1 = 0$. The roots for $m$ are $m = \frac{6 \pm \sqrt{36 - 4(6)(1)}}{12} = \frac{6 \pm \sqrt{12}}{12} = \frac{6 \pm 2\sqrt{3}}{12} = \frac{3 \pm \sqrt{3}}{6}$.

Let $m_1 = \frac{3+\sqrt{3}}{6}$. $l_1 = 1 - 2m_1 = 1 - 2(\frac{3+\sqrt{3}}{6}) = 1 - \frac{3+\sqrt{3}}{3} = \frac{3 - (3+\sqrt{3})}{3} = -\frac{\sqrt{3}}{3}$. $n_1 = 1 - m_1 = 1 - \frac{3+\sqrt{3}}{6} = \frac{6 - (3+\sqrt{3})}{6} = \frac{3-\sqrt{3}}{6}$. Direction cosines: $(-\frac{\sqrt{3}}{3}, \frac{3+\sqrt{3}}{6}, \frac{3-\sqrt{3}}{6})$.

Let $m_2 = \frac{3-\sqrt{3}}{6}$. $l_2 = 1 - 2m_2 = 1 - 2(\frac{3-\sqrt{3}}{6}) = 1 - \frac{3-\sqrt{3}}{3} = \frac{3 - (3-\sqrt{3})}{3} = \frac{\sqrt{3}}{3}$. $n_2 = 1 - m_2 = 1 - \frac{3-\sqrt{3}}{6} = \frac{6 - (3-\sqrt{3})}{6} = \frac{3+\sqrt{3}}{6}$. Direction cosines: $(\frac{\sqrt{3}}{3}, \frac{3-\sqrt{3}}{6}, \frac{3+\sqrt{3}}{6})$.

The problem states "the lines", implying two lines. The direction cosines from $l=0$ define one line. The direction cosines from $1-l-2m=0$ define another line. We can pick one representative vector for the second line. Let's pick $\mathbf{v_2} = (-\frac{\sqrt{3}}{3}, \frac{3+\sqrt{3}}{6}, \frac{3-\sqrt{3}}{6})$.

The angle $\alpha$ between the lines is given by $\cos\alpha = \frac{\mathbf{v_1} \cdot \mathbf{v_2}}{|\mathbf{v_1}| |\mathbf{v_2}|}$. Since these are direction cosines, their magnitudes are 1. $\cos\alpha = (0)(-\frac{\sqrt{3}}{3}) + (\frac{1}{\sqrt{2}})(\frac{3+\sqrt{3}}{6}) + (\frac{1}{\sqrt{2}})(\frac{3-\sqrt{3}}{6})$ $\cos\alpha = \frac{1}{\sqrt{2}} \left( \frac{3+\sqrt{3}}{6} + \frac{3-\sqrt{3}}{6} \right)$ $\cos\alpha = \frac{1}{\sqrt{2}} \left( \frac{6}{6} \right) = \frac{1}{\sqrt{2}}$.

Now we need to find $\sin^4\alpha + \cos^4\alpha$. $\cos^2\alpha = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$. $\sin^2\alpha = 1 - \cos^2\alpha = 1 - \frac{1}{2} = \frac{1}{2}$. $\sin^4\alpha = (\sin^2\alpha)^2 = (\frac{1}{2})^2 = \frac{1}{4}$. $\cos^4\alpha = (\cos^2\alpha)^2 = (\frac{1}{2})^2 = \frac{1}{4}$. $\sin^4\alpha + \cos^4\alpha = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.