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:

• A. 1/8
• B. 3/8
• C. 5/8
• D. 7/8

Answer 1

Answer: (C)

5/8

Answer 2

The direction cosines $(l, m, n)$ satisfy $l+m-n=0$ and $l^2+m^2-n^2=0$. Substituting $n=l+m$ into the second equation gives $-2lm=0$, so $l=0$ or $m=0$. If $l=0$, then $n=m$. With $l^2+m^2+n^2=1$, we get $(0, \pm 1/\sqrt{2}, \pm 1/\sqrt{2})$. If $m=0$, then $n=l$. With $l^2+m^2+n^2=1$, we get $(\pm 1/\sqrt{2}, 0, \pm 1/\sqrt{2})$. These define two lines. Let their direction cosines be $(0, 1/\sqrt{2}, 1/\sqrt{2})$ and $(1/\sqrt{2}, 0, 1/\sqrt{2})$. The cosine of the angle $\alpha$ between them is $\cos \alpha = (0)(1/\sqrt{2}) + (1/\sqrt{2})(0) + (1/\sqrt{2})(1/\sqrt{2}) = 1/2$. Then $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - (1/2)^2 = 3/4$. $\sin^4 \alpha + \cos^4 \alpha = (\sin^2 \alpha)^2 + (\cos^2 \alpha)^2 = (3/4)^2 + (1/2)^2 = 9/16 + 1/4 = 9/16 + 4/16 = 13/16$.

Let's recheck the calculation. $\cos \alpha = 1/2$. $\sin \alpha = \sqrt{1 - (1/2)^2} = \sqrt{3}/2$. $\sin^4 \alpha = (\sqrt{3}/2)^4 = (3/4)^2 = 9/16$. $\cos^4 \alpha = (1/2)^4 = 1/16$. $\sin^4 \alpha + \cos^4 \alpha = 9/16 + 1/16 = 10/16 = 5/8$.