#Task Q94214
l+m+n=0
m=−(l+n)
2lm+2ln−mn=0
2ln+m(2l−n)=0
2ln−(l+n)(2l−n)=0
2ln−(2l2−nl=2ln−n2)=0
−2l2+nl+n2=0
2l2−nl−n2=0
2l2−2nl+nl−n2=0
2l(l−n)+n(l−n)=0
(l−n)(2l+n)=0
l=n
l=−n/2
m=−(n+n)=−2n
n=−m/2
m=−(−n/2+n)
m=−n/2
l=−m/2=n/1
l/1=m/1=−n/2
l:m:n=1:−2:1
l:m:n=1:1:−2
cosθ=12+22+1212+12+221∗1−2∗1−2∗1=−21
θ=1200
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