© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Recall that direction cosines satisfy the condition
For any line in three-dimensional space with direction cosines
$l, m, n$, we have
$$
l^2 + m^2 + n^2 = 1.
$$
Step 2: Use the given equations
We are given two conditions:
$$
l + m - n = 0 \quad \text{(1)}
$$
$$
l^2 + m^2 - n^2 = 0 \quad \text{(2)}
$$
From equation (2),
$$
l^2 + m^2 = n^2.
$$
Substituting into
$l^2 + m^2 + n^2 = 1$
gives
$$
n^2 + n^2 = 1 \quad \Longrightarrow \quad 2n^2 = 1 \quad \Longrightarrow \quad n^2 = \frac{1}{2}.
$$
Thus,
$$
n = \pm \frac{1}{\sqrt{2}}.
$$
Step 3: Express l and m in terms of n
From equation (1),
$l + m = n.$
Considering
$n = \frac{1}{\sqrt{2}},$
we get
$$
l + m = \frac{1}{\sqrt{2}}.
$$
Also from
$l^2 + m^2 = \frac{1}{2}.$
Step 4: Resolve l and m satisfying these conditions
We have
$$
l + m = \frac{1}{\sqrt{2}}, \quad l^2 + m^2 = \frac{1}{2}.
$$
Recall that
$(l + m)^2 = l^2 + m^2 + 2lm.$
Substituting the given values:
$$
\left(\frac{1}{\sqrt{2}}\right)^2
= \frac{1}{2} + 2lm \quad \Longrightarrow \quad \frac{1}{2} = \frac{1}{2} + 2lm,
$$
hence
$$
2lm = 0 \quad \Longrightarrow \quad lm = 0.
$$
Therefore, either
$l = 0 \text{ and } m = \frac{1}{\sqrt{2}},$
or
$m = 0 \text{ and } l = \frac{1}{\sqrt{2}}.
$
Step 5: Form the direction vectors and find cos α
Two possible direction vectors (satisfying $n = \frac{1}{\sqrt{2}}$) are:
$$
\langle 0, \tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}} \rangle
\quad \text{and} \quad
\langle \tfrac{1}{\sqrt{2}}, 0, \tfrac{1}{\sqrt{2}} \rangle.
$$
To find the angle α between these two lines, use the dot product:
$$
\vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos \alpha.
$$
Since each is a direction cosine vector, its magnitude is 1. Then
$$
\vec{a} \cdot \vec{b}
= 0 \cdot \tfrac{1}{\sqrt{2}}
+ \tfrac{1}{\sqrt{2}} \cdot 0
+ \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}}
= \frac{1}{2}.
$$
Hence,
$$
\cos \alpha = \frac{1}{2}.
$$
Therefore,
$$
\alpha = 60^\circ,
$$
and
$$
\sin \alpha = \frac{\sqrt{3}}{2}.
$$
Step 6: Compute $ \sin^4 \alpha + \cos^4 \alpha $
Use the identity:
$$
\sin^4 \alpha + \cos^4 \alpha
= (\sin^2 \alpha + \cos^2 \alpha)^2 - 2 \sin^2 \alpha \cos^2 \alpha.
$$
Since $\sin^2 \alpha + \cos^2 \alpha = 1$, we only need
$2 \sin^2 \alpha \cos^2 \alpha.$
We have:
$$
\sin^2 \alpha = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4},
\quad
\cos^2 \alpha = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.
$$
Hence,
$$
2 \sin^2 \alpha \cos^2 \alpha
= 2 \times \frac{3}{4} \times \frac{1}{4}
= 2 \times \frac{3}{16}
= \frac{3}{8}.
$$
Therefore,
$$
\sin^4 \alpha + \cos^4 \alpha
= 1 - \frac{3}{8}
= \frac{5}{8}.
$$
Final Answer:
$ \displaystyle \frac{5}{8}. $
