Given conditions for direction cosines

Let the direction cosines of the required lines be $(l,m,n)$.

They satisfy: $l+m-n=0$ $l^2+m^2-n^2=0$

Also, since $(l,m,n)$ are direction cosines, $l^2+m^2+n^2=1$

Use the first equation

From $l+m-n=0$ we get $n=l+m$

Substitute into the second equation

Using $n=l+m$ in $l^2+m^2-n^2=0$ gives $l^2+m^2-(l+m)^2=0$ $l^2+m^2-(l^2+m^2+2lm)=0$ $-2lm=0$ $lm=0$

So either:

• $l=0$, or
• $m=0$

Find the two lines

Case 1: $l=0$

Then $n=m$.

Using $l^2+m^2+n^2=1$: $0+m^2+m^2=1$ $2m^2=1$ $m=\pm \frac{1}{\sqrt{2}}, \quad n=\pm \frac{1}{\sqrt{2}}$

So one line has direction ratios proportional to $(0,1,1)$

Case 2: $m=0$

Then $n=l$.

Using $l^2+m^2+n^2=1$: $l^2+0+l^2=1$ $2l^2=1$ $l=\pm \frac{1}{\sqrt{2}}, \quad n=\pm \frac{1}{\sqrt{2}}$

So the other line has direction ratios proportional to $(1,0,1)$

Find the angle between the two lines

Let the direction vectors be $\vec{a}=(0,1,1), \qquad \vec{b}=(1,0,1)$

Then $\cos\alpha=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$

Compute dot product: $\vec{a}\cdot\vec{b}=0\cdot 1+1\cdot 0+1\cdot 1=1$

Magnitudes: $|\vec{a}|=\sqrt{0^2+1^2+1^2}=\sqrt{2}$ $|\vec{b}|=\sqrt{1^2+0^2+1^2}=\sqrt{2}$

Hence, $\cos\alpha=\frac{1}{2}$

Therefore, $\cos^2\alpha=\frac{1}{4}, \qquad \sin^2\alpha=1-\frac{1}{4}=\frac{3}{4}$

Compute $\sin^4\alpha+\cos^4\alpha$

$\sin^4\alpha+\cos^4\alpha=(\sin^2\alpha+\cos^2\alpha)^2-2\sin^2\alpha\cos^2\alpha$ $=1-2\left(\frac{3}{4}\cdot \frac{1}{4}\right)$ $=1-2\cdot \frac{3}{16}$ $=1-\frac{3}{8}$ $=\frac{5}{8}$

Compare with options

The correct value is $\boxed{\frac{5}{8}}$

So the correct option is D.