Three Dimensional Geometry Part 2

Given, l+m−n=0. . . (i)
and I2+m2−n2=0‌‌... (ii)
On squaring Eq. (i), we get
(l+m)2=n2
⇒‌‌l2+m2+2lm=n2 . . . (iii)
From Eqs. (ii) and (iii),
I2+m2−n2=0
I2+m2+2lm=n2
‌

−−‌‌−‌‌−

−n2−2lm=−n2

⇒‌‌2lm=0⇒Im=0
⇒‌‌I=0‌ or ‌m=0
‌ Case I When I = 0 ‌
⇒‌‌0+m−n=0
⇒‌‌m=n
‌ and ‌I2+m2+n2=1
⇒‌‌m2+m2=1‌‌[∵n=m and l=0]
⇒‌‌m2=‌

1

2

‌‌‌m=±‌

1

√2

=n
∴‌‌(I,m,n)=(0,‌

1

√2

,‌

1

√2

)‌ or ‌(0,‌

−1

√2

,‌

−1

√2

)
Case II When m=0
then, l+m−n=0
⇒‌‌I=n and l2+m2+n2=1
[∵n=I and m=0]
⇒‌‌I2+0+I2=1
I=±‌

1

√2

‌‌ [∵n=l and m=0]
⇒ ∴‌‌(I,m,n)=(‌

1

√2

,0,‌

1

√2

) or (‌

−1

√2

,0,‌

−1

√2

)
⇒‌‌a=(0,‌

1

√2

,‌

1

√2

) and b=(‌

1

√2

,0,‌

1

√2

)
Then cos‌α=‌

a⋅b

|a||b|

=‌

1

2

∴sin‌α=±‌

√3

2

Now, cos4α+sin4α=‌

1

16

+‌

9

16

=‌

10

16

=‌

5

8