Let be the angle between the lines whose direction cosines satisfy the equations l+m-n=0 a…

Question

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

Accepted Answer

Squaring $l+m-n=0$: $l^2+m^2+2lm=n^2$. Combined with $l^2+m^2-n^2=0$: subtracting gives $2lm=0 \implies l=0$ or $m=0$.

Case $l=0$: $m=n$, and $l^2+m^2+n^2=1 \implies 2m^2=1 \implies m=\pm\tfrac{1}{\sqrt2}$. So $(l,m,n)=\left(0,\tfrac1{\sqrt2},\tfrac1{\sqrt2}\right)$.

Case $m=0$: $l=n$, similarly $(l,m,n)=\left(\tfrac1{\sqrt2},0,\tfrac1{\sqrt2}\right)$.
$$\cos\alpha=\vec a\cdot\vec b=\frac12 \implies \sin^2\alpha=\frac34.$$
$$\cos^4\alpha+\sin^4\alpha=\frac{1}{16}+\frac{9}{16}=\frac{10}{16}=\frac58.$$

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